Calculus 2 - Integral

Contents: Integral Indefinite Integral Area Problem Definite Integral & Its Applications Improper Integral Infinite Series First Order Differential Equation Reference: Paul's Online Notes: Integral

Indefinite Integral

Antiderivatives
If F is an antiderivative of f on an interval I, then every antiderivative of f on I has the following form G(x) = F(x) + C, where C is an arbitrary constant.

Rules of Integration

$\begin{align*} \text{1. }&\intop\nolimits c.f(x) \; dx = c \intop\nolimits f(x)\;dx\\ \text{2. } &\intop\nolimits [f(x) \pm g(x)] \; dx = \intop f(x)\;dx \pm \intop g(x)\;dx\\ \end{align*}$

Power Rule and Exponential Function

$\begin{align*} \text{3. } &\intop\nolimits ax^n\;dx = \frac{a}{n+1}x^{n+1}+C, n \neq 1\\ \text{4. } &\intop\nolimits a^x\;dx = \frac{a^x}{\ln{a}}+C,\;a>0\text{ and }a\neq 1\\ \end{align*}$

Integration of Trigonometric Function

$\begin{align*} \text{5. } &\intop\nolimits \cos{x}\;dx = \sin{x}+C\\ \text{6. } &\intop\nolimits \sin{x}\;dx = -\cos{x}+C\\ \text{7. } &\intop\nolimits \sec^2{x}\;dx = \tan{x}+C\\ \text{8. } &\intop\nolimits \sec{x}\tan{x}\;dx = \sec{x}+C\\ \text{9. } &\intop\nolimits \csc{x}\cot{x}\;dx = -\csc{x}+C\\ \text{10. } &\intop\nolimits \csc^2{x}\;dx = -\cot{x}+C\\ \text{11. } &\intop\nolimits \tan{x}\;dx = \ln{|\sec{x}|}+C\\ \text{12. } &\intop\nolimits \cot{x}\;dx = \ln{|\sin{x}|}+C\\ \text{13. } &\intop\nolimits \sec{x}\;dx = \ln{|\sec{x}+\tan{x}|}+C\\ \text{14. } &\intop\nolimits \csc{x}\;dx = \ln{|\csc{x}-\cot{x}|}+C\\ \end{align*}$

Integration of Inverse Trigonometric Function

$\begin{align*} \text{15. } &\intop\nolimits \frac{1}{\sqrt{1-x^2}}\;dx = \sin^{-1}x+C\\ \text{16. } &\intop\nolimits \frac{1}{1+x^2}\;dx = \tan^{-1}x+C\\ \text{17. } &\intop\nolimits \frac{1}{|x|\sqrt{x^2-1}}\;dx = \sec^{-1}x+C\\ \end{align*}$

  Example 1.1 - Integrating Functions

Find:
$\begin{align*} &\text{A. }\int (x^2-1)(x+1)\; dx\\ &\text{B. }\int \frac{4-x^2}{x^2}\; dx\\ &\text{C. }\int (1+\cot^2{x})\; dx\\ \end{align*}$

Solution:

$\begin{align*} \text{A. }&\int (x^2-1)(x+1) \; dx\\ &=\int (x^3+x^2-x-1) \;dx \\ &=\int x^3 \;dx + \int x^2 \;dx - \int x \;dx - \int 1 \;dx \\ &= \frac{1}{3+1}x^{3+1} + \frac{1}{2+1}x^{2+1} - \frac{1}{1+1}x^{1+1} - \frac{0}{0+1}x^{0+1} + C \\ &= \frac{1}{4}x^4 + \frac{1}{3}x^3 - \frac{1}{2}x^2 - x + C\\ \\ \text{B. }&\int \frac{4-x^2}{x^2} \;dx\\ &=\int \left ( \frac{4}{x^2}-1 \right ) \; dx \\ &=\int \left ( \frac{4}{x^2} \right ) \; dx - \int 1 \; dx\\ &= 4 \int \left ( \frac{1}{x^2} \right ) \; dx - \int 1 \; dx\\ &= -\frac{4}{x} - x + C\\ \\ \text{C. }&\int (1+\cot^2{x}) \; dx\\ &=\int (\csc^2{x}) \; dx\\ &=-\cot{x}+C \end{align*}$

Techniques of Integration
Sometimes, the function that we have to integrate, is not that easy that we only have to use the basic rules previously mentioned. There are several techniques, or methods that we can use. Observe:

  Example 1.2 - Integration by Substitution

Find:
$\begin{align*} &\text{A. }\int \frac{x-1}{\sqrt{x^2-2x+10}}\;dx\\ &\text{B. }\int \frac{1}{\sqrt{1-4x^2}} \; dx\\ &\text{C. }\int x\sec{x^2} \; dx\\ \end{align*}$

Solution:

1. Let us define a new function u, such that:
   $\begin{align*} u&=x^2-2x+10\\ du&=(2x-2)\;dx\\ \frac{1}{2}du &= (x-1)\;dx\\ \end{align*}$
   
   So, we can substitute the above to the original integral:
   $\begin{align*} &\int \frac{x-1}{\sqrt{x^2-2x+10}}\;dx\\ &=\int \frac{1}{2\sqrt{u}} \; du\\ &= \sqrt{u} + C\\ &= \sqrt{x^2-2x+10} + C \end{align*}$


2. Let us define a new function u, such that:
   $\begin{align*} u&=2x\\ du&=2\;dx\\ \frac{du}{2}&=\;dx \end{align*}$
   
   So, we can substitute the above to the original integral:
   $\begin{align*} &\int \frac{1}{\sqrt{1-4x^2}}\;dx\\ &=\int \frac{1}{\sqrt{1-(2x)^2}}\;dx\\ &=\frac{1}{2} \int \frac{1}{\sqrt{1-u^2}}\;du\\ &=\frac{1}{2} \sin^{-1}{2x}+C\\ \end{align*}$


3. Let us define a new function u, such that:
   $\begin{align*} u&=x^2\\ du&=2x\;dx\\ \frac{du}{2}&=x\;dx \end{align*}$
   
   So, we can substitute the above to the original integral:
   $\begin{align*} &\int x\sec{x^2}\;dx\\ &=\frac{1}{2} \int \sec{u} \; du\\ &=\frac{1}{2} |\sec{u}+\tan{u}|+C\\ &=\frac{1}{2} |\sec{x^2}+\tan{x^2}|+C \end{align*}$

  Example 1.3 - Integration by Parts

Find:
$\begin{align*} &\text{A. }\int x \cos{2x} \; dx\\ &\text{B. }\int x^2 \ln{x} \; dx\\ &\text{C. }\int x^5 \sin{2x^3} \; dx\\ \end{align*}$

Solution:
Basically, Integration by Parts can be done with $\int u\;dv = uv - \int v\;du$. We must select u and dv appropriately, so we can arrive at the solution easier.

1. In this example, we can select u = x, because its derivative will eventually become 0:
   $\begin{align*} u&=x\\ dv&=\cos{2x} \; dx\\ du&=1\\ v&=\frac{1}{2}\sin{2x}\\ \end{align*}$
   
   So, we can do substitute the above to the partial integral form:
   $\begin{align*} &\int x \cos{2x} \; dx\\ &=\int u\;dv \\ &=uv - \int v\;du \\ &=\frac{1}{2}x \sin{2x} - \int \frac{1}{2}\sin{2x} \; dx\\ &=\frac{1}{2}x \sin{2x} + \frac{1}{4}\cos{2x} + C \\ \end{align*}$


2. Initially, we may want to use u = x². But instead, we should use u = ln x. This is because: If we use u = x², we have to find the integral of ln(x). On top of that, it most likely requires us to do partial integration again. If we use u = ln x, the integral is directly doable. So, for this problem:
   $\begin{align*} u&=\ln{x}\\ du&=\frac{1}{x} \; dx\\ dv&=x^2 \; dx\\ v&=\frac{1}{3}x^3\\ \end{align*}$
   
   So, we can do substitute the above to the partial integral form:
   $\begin{align*} &\int x^2 \ln{x} \; dx\\ &=\int u\;dv \\ &=uv - \int v\;du \\ &=\frac{1}{3}x^3 \ln{x}- \int \frac{1}{3}x^2 \; dx\\ &=\frac{1}{3}x^3 \ln{x}- \frac{1}{9}x^3 + C\\ \end{align*}$


3. In this problem, we cannot use both u = x⁵ and v = sin 2x³. Using either of them, will require us to do another partial integration. However, we can do this:
   $\begin{align*} \int x^5 \sin{2x^3} \; dx = \int x^3 x^2 \sin{2x^3} \; dx\\ \end{align*}$
   
   In this problem, we will use u = x³. Integrating dv involves Integration by Substitution, which we have already learned previously. We can always check if we did the integral correctly by finding the derivative of the result:
   $\begin{align*} u&=x^3\\ du&=3x^2\; dx\\ dv&=x^2 \sin{2x^3} \; dx\\ v&=-\frac{1}{6} \cos{2x^3}\\ \end{align*}$
   
   So, we can do substitute the above to the partial integral form. We are only required to do another Integration by Subsitution:
   $\begin{align*} &\int x^5 \sin{2x^3} \; dx\\ &=\int u\;dv \\ &=uv - \int v\;du \\ &=-\frac{1}{6}x^3 \cos {2x^3} + \frac{1}{6}\int (3x^2)\cos{2x^3} \; dx\\ &=-\frac{1}{6}x^3 \cos {2x^3} + \frac{1}{12}\sin{2x^3} + C\\ \end{align*}$

  Example 1.4 - Trigonometric Sub. I

Find:
$\begin{align*} &\text{A. }\int \sin^{7}{x} \cos^{2}{x}\;dx\\ &\text{B. }\int \sin^{3}{x} \cos^{5}{x}\;dx\\ &\text{C. }\int \sin^{2}{x} \cos^{2}{x}\;dx\\ \end{align*}$

Solution:
Recall the following identity:
$\begin{align*} &\sin^2{x}+\cos^2{x}=1\\ &\sin{2x}=2\sin{x}\cos{x}\\ &\cos{2x}=\cos^2{x}-\sin^2{x}\\ &\cos{2x}=1-2\sin^2{x}\\ &\cos{2x}=2\cos^2{x}-1\\ \end{align*}$

1. Let us manipulate the original function, such that:
   
   $\begin{align*} &\int \sin^{7}{x} \cos^{2}{x}\;dx\\ &= \int \sin^{6}{x} \cos^{2}{x} \sin{x}\;dx\\ &= \int (\sin^{2}{x})^3 \cos^{2}{x} \sin{x}\;dx\\ &= \int (1-\cos^{2}{x})^3 \cos^{2}{x} \sin{x}\;dx\\ &= \int (1-3\cos^{2}{x}+3\cos^4{x}-\cos^6{x}) \cos^{2}{x} \sin{x}\;dx\\ &= \int (\cos^{2}{x}-3\cos^{4}{x}+3\cos^{6}{x}-\cos^{8}{x}) \sin{x}\;dx\\ \end{align*}$
   
   Now, we can perform Integral by Substitution. So, let's define:
   $\begin{align*} u&=\cos{x}\\ du&=-\sin{x}\;dx \end{align*}$
   
   Lastly:
   
   $\begin{align*} &\int \sin^{7}{x} \cos^{2}{x}\;dx\\ &= \int (\cos^{2}{x}-3\cos^{4}{x}+3\cos^{6}{x}-\cos^{8}{x}) \sin{x}\;dx\\ &= -\int (u^2-3u^4+3u^6-u^8) \;du\\ &= -\left (\frac{1}{3}u^3-\frac{3}{5}u^5+\frac{3}{7}u^7 - \frac{1}{9}u^9\right ) + C\\ &= -\left (\frac{1}{3}\cos^3{x}-\frac{3}{5}\cos^5{x}+\frac{3}{7}\cos^7{x} - \frac{1}{9}\cos^9{x}\right ) + C\\ \end{align*}$


2. Let us manipulate the original function, such that:
   
   $\begin{align*} &\int \sin^{3}{x} \cos^{5}{x}\;dx\\ &= \int \sin^{3}{x} \cos^{4}{x} \cos{x}\;dx\\ &= \int \sin^{3}{x} (\cos^2{x})^2 \cos{x}\;dx\\ &= \int \sin^{3}{x} (1-\sin^2{x})^2 \cos{x}\;dx\\ &= \int \sin^{3}{x} (1-2\sin^2{x}+\sin^4{x}) \cos{x}\;dx\\ &= \int (\sin^{3}{x}-2\sin^5{x}+\sin^7{x}) \cos{x}\;dx\\ \end{align*}$
   
   Now, we can perform Integral by Substitution. So, let's define:
   $\begin{align*} u&=\sin{x}\\ du&=\cos{x}\;dx \end{align*}$
   
   Lastly:
   
   $\begin{align*} &\int \sin^{3}{x} \cos^{5}{x}\;dx\\ &= \int (\sin^{3}{x}-2\sin^5{x}+\sin^7{x}) \cos{x}\;dx\\ &= \int (u^3-2u^5+u^7) \;du\\ &= \frac{1}{4}u^4-\frac{1}{3}u^6+\frac{1}{8}u^8+C\\ &= \frac{1}{4}\sin^4{x}-\frac{1}{3}\sin^6{x}+\frac{1}{8}\sin^8{x}+C \end{align*}$


3. Let us manipulate the original function, such that:
   
   $\begin{align*} &\int \sin^{2}{x} \cos^{2}{x}\;dx\\ &= \int (\sin{x} \cos{x})^2\;dx\\ &= \int \left (\frac{\sin{2x}}{2} \right ) ^2\;dx\\ &= \frac{1}{4} \int \sin^2(2x)\;dx\\ &= \frac{1}{4} \int (1-\cos^2(2x))\;dx\\ &= \frac{1}{4} \int (1-(\cos(4x)+2\sin^2(2x)))\;dx\\ &= \frac{1}{4} \left ( \int 1 \;dx - \int \cos(4x) \;dx - 2 \int (\sin(2x))^2 \;dx\right )\\ &= \frac{1}{4} \left ( x - \frac{1}{4}\sin(4x) - 2 \int (2\sin(x)\cos(x))^2 \;dx\right )\\ &= \frac{1}{4} \left ( x - \frac{1}{4}\sin(4x) - 4 \int (\sin(x)\cos(x))^2 \;dx\right )\\ \end{align*}$
   
   
   Now, let us define:
   $A=\int \sin^{2}{x} \cos^{2}{x}\;dx$
   
   Lastly, we can do this: (C is an arbitrary constant)
   
   $\begin{align*} A&=\int \sin^{2}{x} \cos^{2}{x}\;dx\\ &= \frac{1}{4} \left ( x - \frac{1}{4}\sin(4x) - 4 \int (\sin(x)\cos(x))^2 \;dx\right ) + C\\ &= \frac{1}{4} \left ( x - \frac{1}{4}\sin(4x) - 4A\right ) + C\\ &= \frac{1}{4}x - \frac{1}{16}\sin(4x) - A + C\\ 2A&= \frac{1}{4}x - \frac{1}{16}\sin(4x) + C\\ A&= \frac{1}{8}x - \frac{1}{32}\sin(4x) + C\\ \end{align*}$
   
   The A we defined previously, is the answer we are looking for.

  Example 1.5 - Trigonometric Sub. II

Find:
$\begin{align*} &\text{A. }\int \tan^{3}{x} \sec^{5}{x}\;dx\\ &\text{B. }\int \tan^{3}{x} \sec^{4}{x}\;dx\\ &\text{C. }\int \cot^{3}{x} \csc^{3}{x}\;dx\\ \end{align*}$

Solution:
Recall the following identity:
$\begin{align*} &\tan^2{x}+1=\sec^2{x}\\ &1+\cot^2{x}=\csc^2{x} \end{align*}$

1. Let us manipulate the original function, such that:
   
   $\begin{align*} &\int \tan^{3}{x} \sec^{5}{x}\;dx\\ &= \int \tan{x} \tan^{2}{x} \sec^{5}{x}\;dx\\ &= \int \tan{x} (\sec^{2}{x}-1) \sec^{5}{x}\;dx\\ \end{align*}$
   
   Now, we can perform Integral by Substitution. So, let's define:
   $\begin{align*} u&=\sec{x}\\ du&=\sec{x} \tan{x}\;dx \end{align*}$
   
   Lastly:
   
   $\begin{align*} &\int \tan^{3}{x} \sec^{5}{x}\;dx\\ &= \int \tan{x} (\sec^{2}{x}-1) \sec^{5}{x}\;dx\\ &= \int \tan{x} \sec{x} (\sec^{6}{x}-\sec^{4}{x})\;dx\\ &= \int (u^6 - u^4) \;du\\ &= \frac{1}{7}u^7 - \frac{1}{5}u^5 + C\\ &= \frac{1}{7}\sec^7{x} - \frac{1}{5}\sec^5{x} + C\\ \end{align*}$


2. Let us manipulate the original function, such that:
   
   $\begin{align*} &\int \tan^{3}{x} \sec^{4}{x}\;dx\\ &= \int \tan^{3}{x} (\sec^{2}{x})^2 \; dx\\ &= \int \tan^{3}{x} (\tan^2{x}+1)(\sec^{2}{x}) \; dx\\ \end{align*}$
   
   Now, we can perform Integral by Substitution. So, let's define:
   $\begin{align*} u&=\tan{x}\\ du&=\sec^{2}{x}\;dx \end{align*}$
   
   Lastly:
   
   $\begin{align*} &\int \tan^{3}{x} \sec^{4}{x}\;dx\\ &= \int \tan^{3}{x} (\tan^2{x}+1)(\sec^{2}{x}) \; dx\\ &= \int u^3 (u^2+1) \;du\\ &= \int (u^5 + u^3) \;du\\ &= \frac{1}{6}u^6 + \frac{1}{4}u^4 + C\\ &= \frac{1}{6}\tan^6{x} + \frac{1}{4}\tan^4{x} + C\\ \end{align*}$


3. Let us manipulate the original function, such that:
   
   $\begin{align*} &\int \cot^{3}{x} \csc^{3}{x}\;dx\\ &= \int (\cot^2{x}\csc^2{x}) (\cot{x}\csc{x}) \;dx\\ &= \int (\csc^2{x}-1)(\csc^2{x}) (\cot{x}\csc{x}) \;dx\\ \end{align*}$
   
   Now, we can perform Integral by Substitution. So, let's define:
   $\begin{align*} u&=\csc{x}\\ du&=-\csc{x}\cot{x}\;dx \end{align*}$
   
   Lastly:
   
   $\begin{align*} &\int \cot^{3}{x} \csc^{3}{x}\;dx\\ &= \int (\csc^2{x}-1)(\csc^2{x}) (\cot{x}\csc{x}) \;dx\\ &= \int (u^4-u^2) \;du\\ &= \frac{1}{5}u^5 - \frac{1}{3}u^3 + C\\ &= \frac{1}{5}\csc^5{x} - \frac{1}{3}\csc^3{x}+C \end{align*}$
   
   

  Example 1.6 - Trigonometric Sub. III

Find:
$\begin{align*} &\text{A. }\int \frac{x^2}{\sqrt{4-x^2}}\;dx\\ &\text{B. }\int \frac{1}{x^2\sqrt{x^2+4}}\;dx\\ &\text{C. }\int \frac{1}{\sqrt{x^2-1}} \;dx\\ \end{align*}$

Solution:

1. Let us substitute:
   
   $\begin{align*} x &= 2 \sin {\theta}\\ dx &= 2 \cos {\theta}\; d\theta \end{align*}$
   
   Next:
   
   $\begin{align*} &\int \frac{x^2}{\sqrt{4-x^2}}\;dx\\ &=\int \frac{4 \sin^2 {\theta}}{\sqrt{4-4\sin^2{\theta}}} 2 \ cos{\theta} \; d\theta\\ &=\int \frac{4 \sin^2 {\theta}}{2\cos{\theta}} 2 \ cos{\theta} \; d\theta\\ &=\int 4 \sin^2{\theta} \; d\theta\\ &= 2\int \left ( 1-\cos{2\theta} \right ) \; d\theta\\ &= 2\theta - \sin{2\theta} + C\\ \end{align*}$
   
   
   With the help of the visualization above, we can express the integration result we have previously obtained in terms of x (not theta anymore). In this case, a = 2:
   
   $\begin{align*} &\int \frac{x^2}{\sqrt{4-x^2}}\;dx\\ &= 2\theta - \sin{2\theta} + C\\ &= 2\theta - 2\sin{\theta}\cos{\theta} + C\\ &= 2\sin^{-1}\left ( \frac{x}{2} \right ) - 2 \left ( \frac{x}{2} \right )\left ( \frac{\sqrt{4-x^2}}{2} \right ) + C\\ &= 2\sin^{-1}\left ( \frac{x}{2} \right ) - \left ( \frac{x\sqrt{4-x^2}}{2} \right ) + C \end{align*}$


2. Let us substitute:
   
   $\begin{align*} x &= 2 \tan {\theta}\\ dx &= 2 \sec^2 {\theta}\; d\theta \end{align*}$
   
   Next:
   
   $\begin{align*} &\int \frac{1}{x^2\sqrt{x^2+4}}\;dx\\ &=\int \frac{1}{(4\tan^2{\theta})\sqrt{4\tan^2{\theta}+4}}2 \sec^2 {\theta} \; d\theta\\ &=\int \frac{1}{(4\tan^2{\theta})(2\sec{\theta})}2 \sec^2 {\theta} \; d\theta\\ &=\frac{1}{4} \int \frac{1}{(\tan^2{\theta})(\sec{\theta})} \sec^2 {\theta} \; d\theta\\ &=\frac{1}{4} \int \frac{\sec{\theta}}{\tan^2{\theta}}\;d\theta\\ &=\frac{1}{4} \int \frac{\cos{\theta}}{\sin^2{\theta}}\;d\theta\\ &=\frac{1}{4} \int \cot{\theta}\csc{\theta}\;d\theta\\ &=-\frac{1}{4}\csc{\theta} + C\\ \end{align*}$
   
   
   With the help of the visualization above, we can express the integration result we have previously obtained in terms of x (not theta anymore). In this case, a = 2:
   
   $\begin{align*} &\int \frac{1}{x^2\sqrt{x^2+4}}\;dx\\ &=-\frac{1}{4}\csc{\theta} + C\\ &=-\frac{1}{4}\left (\frac{\sqrt{x^2+4}}{x} \right ) + C\\ \end{align*}$


3. Let us substitute:
   
   $\begin{align*} x &= \sec {\theta}\\ dx &= \sec{\theta} \tan{\theta}\; d\theta \end{align*}$
   
   Next:
   
   $\begin{align*} &\int \frac{1}{\sqrt{x^2-1}}\;dx\\ &=\int \frac{1}{\sqrt{\sec^2{\theta}-1}} (\sec{\theta} \tan{\theta})\;d\theta\\ &=\int \frac{1}{\tan{\theta}} (\sec{\theta} \tan{\theta})\;d\theta\\ &=\int \sec{\theta} \;d\theta\\ &=\ln{|\sec{\theta}+\tan{\theta}|}+C \end{align*}$
   
   
   With the help of the visualization above, we can express the integration result we have previously obtained in terms of x (not theta anymore). In this case, a = 1:
   
   $\begin{align*} &\int \frac{1}{\sqrt{x^2-1}}\;dx\\ &=\ln{|\sec{\theta}+\tan{\theta}|}+C\\ &=\ln{|x+{\sqrt{x^2-1}}|}+C\\ \end{align*}$

  Example 1.7 - Partial Fractions

Find:
$\begin{align*} &\text{A. }\int \frac{2x+3}{x^2+4x+3}\;dx\\ &\text{B. }\int \frac{x+2}{x^2-4x+4}\;dx\\ &\text{C. }\int \frac{x^2+x+1}{(x-1)(x-2)(x-3)}\;dx\\ &\text{D. }\int \frac{x^2+6x+5}{(x+4)^2(x+2)}\;dx\\ &\text{E. }\int \frac{1}{(x-1)(x^2+x+1)}\;dx\\ \end{align*}$

Solution:

1. Partial fractions:
   $\begin{align*} \frac{2x+3}{x^2+4x+3} &= \frac{A}{(x+1)} + \frac{B}{(x+3)}\\ 2x+3 &= A(x+3) + B(x+1)\\ 2x+3 &= (A+B)x + (3A+B) \end{align*}$
   
   By doing elimination, or substitution to the following system of equation:
   $\begin{align*} 2&=A+B\\ 3&=3A+B \end{align*}$
   
   We will obtain A=1/2 and B=3/2. Then, we can integrate:
   
   $\begin{align*} \int \frac{2x+3}{x^2+4x+3} \;dx &= \int \left (\frac{A}{(x+1)}+ \frac{B}{x+3} \right ) \; dx\\ &= \int \left (\frac{1}{2(x+1)}+ \frac{3}{2(x+3)} \right ) \; dx\\ &= \frac{1}{2}\ln{|x+1|} + \frac{3}{2}\ln{|x+3|} + C \end{align*}$


2. Partial fractions:
   $\begin{align*} \frac{x+2}{x^2-4x+4} &= \frac{A}{(x-2)} + \frac{B}{(x-2)^2}\\ x+2 &= A(x-2) + B\\ x+2 &= Ax + (B-2A) \end{align*}$
   
   By doing elimination, or substitution to the following system of equation:
   $\begin{align*} 1&=A\\ 2&=B-2A \end{align*}$
   
   We will obtain A=1 and B=4. Then, we can integrate:
   
   $\begin{align*} \int \frac{x+2}{x^2-4x+4} \;dx &= \int \left (\frac{A}{(x-2)}+ \frac{B}{(x-2)^2} \right ) \; dx\\ &= \int \left (\frac{1}{x-2}+ \frac{4}{(x-2)^2} \right ) \; dx\\ &= \ln{|x-2|} - \frac{4}{x-2} + C \end{align*}$


3. Partial fractions:
   $\begin{align*} \frac{x^2+x+1}{(x-1)(x-2)(x-3)} &= \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}\\ x^2+x+1 &= A(x^2-5x+6) + B(x^2-4x+3) + C(x^2-3x+2) \end{align*}$
   
   By doing elimination, or substitution to the following system of equation:
   $\begin{align*} 1&=A+B+C\\ 1&=-5A - 4B - 3C\\ 1&=6A + 3B + 2C \end{align*}$
   
   We will obtain A=3/2, B=-7 and C=13/2. Then, we can integrate:
   
   $\begin{align*} \int \frac{x^2x+1}{(x-1)(x-2)(x-3)} \;dx &= \int \left (\frac{A}{x-1}+ \frac{B}{x-2} + \frac{C}{x-3} \right ) \; dx\\ &= \int \left (\frac{3}{2(x-1)} - \frac{7}{x-2} + \frac{13}{2(x-3)} \right ) \; dx\\ &= \frac{3}{2}\ln{|x-1|} -7 \ln{|x-2|} + \frac{13}{2} \ln{|x-3|} + C \end{align*}$


4. Partial fractions:
   
   $\begin{align*} \frac{x^2+6x+5}{(x+4)^2(x+2)} &= \frac{A}{(x+4)}+ \frac{B}{(x+4)^2} + \frac{C}{(x+2)}\\ (x^2+6x+5) &= A(x+4)(x+2) + B(x+2) + C(x+4)^2\\ (x^2+6x+5) &= A(x^2+6x+8) + B(x+2) + C(x^2+8x+16)\\ (x^2+6x+5) &= (A+C)x^2 + (6A+B+8C)x + (8A + 2B + 16C)\\ \end{align*}$
   
   
   By doing elimination, or substitution to the following system of equation:
   $\begin{align*} A+C&=1\\ 6A+B+8C&=6\\ 8A+2B+16C&=5 \end{align*}$
   
   We will obtain A=7/4, B=3/2 and C=-3/4. Then, we can integrate:
   
   $\begin{align*} \int \frac{x^2+6x+5}{(x+4)^2(x+2)}\;dx &= \int \left (\frac{A}{(x+4)}+ \frac{B}{(x+4)^2} + \frac{C}{(x+2)} \right ) \; dx\\ &= \int \left (\frac{7}{4(x+4)}+ \frac{3}{2(x+4)^2} - \frac{3}{4(x+2)} \right ) \; dx\\ &= \frac{7}{4}\ln{|x+4|} - \frac{3}{2(x+4)} - \frac{3}{4}\ln{|x+2|}+C \end{align*}$


5. Partial fractions:
   
   $\begin{align*} \frac{1}{(x-1)(x^2+x+1)} &= \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}\\ 1 &= A(x^2+x+1) + (Bx+C)(x-1)\\ &= A(x^2+x+1) + (Bx^2+(C-B)x-C)\\ \end{align*}$
   
   
   By doing elimination, or substitution to the following system of equation:
   $\begin{align*} 0 &= A+B\\ 0 &= A+C-B\\ 1 = A-C \end{align*}$
   
   We will obtain A=1/3, B=-1/3 and C=-2/3. Then, we can integrate:
   
   $\begin{align*} \int \frac{1}{(x-1)(x^2+x+1)} \;dx &= \int \left (\frac{A}{x-1}+ \frac{Bx+C}{x^2+x+1}\right ) \; dx\\ &= \int \left (\frac{1}{3(x-1)}+ \frac{-\frac{1}{3}x-\frac{2}{3}}{x^2+x+1}\right ) \; dx\\ &= \frac{1}{3}\ln{|x-1|} - \frac{1}{3} \int \left ( \frac{x+2}{x^2+x+1} \right )\; dx \end{align*}$
   
   
   Now, we have another integral to solve. (We will call this integral F) We can manipulate x+2 to $\frac{1}{2}(2x+1)+\frac{3}{2}$. We can see in the third line below that we have to do more integrals. The first integral can be done using the substitution method (won't be shown), and the second integral can be done using the completing the square method:
   
   $\begin{align*} F&=\int \left ( \frac{x+2}{x^2+x+1} \right )\; dx\\ &= \int \left ( \frac{\frac{1}{2}(2x+1)}{x^2+x+1} + \frac{\frac{3}{2}}{x^2+x+1} \right )\; dx\\ &= \frac{1}{2} \int \frac{2x+1}{x^2+x+1}\; dx + \frac{3}{2} \int \frac{1}{x^2+x+1} \; dx\\ &= \frac{1}{2} \ln{|x^2+x+1|} + \frac{3}{2} \int \frac{1}{x^2+x+1} \; dx\\ \end{align*}$
   
   
   Doing the second integral that requires completing the square method:
   $\begin{align*} &\int \frac{1}{x^2+x+1} \; dx\\ &\int \frac{1}{\left ( x+\frac{1}{2} \right )^2 + \frac{3}{4}} \; dx\\ &= \int \frac{\frac{4}{3}}{ \frac{4}{3}\left (x+\frac{1}{2} \right )^2 + 1} \; dx\\ &= \frac{4}{3}\int \frac{1}{\frac{4}{3}\left (x+\frac{1}{2} \right )^2 + 1} \; dx\\ \end{align*}$
   
   This form is similar to $\frac{1}{u^2+1}$. Integrating that would be easier:
   $\begin{align*} u^2&= \frac{4}{3} \left ( x+\frac{1}{2} \right )^2\\ u&= \frac{2}{\sqrt{3}} \left (x+\frac{1}{2}\right )\\ &= \frac{2}{\sqrt{3}} \left (x+\frac{1}{2}\right )\\ &= \frac{2x+1}{\sqrt{3}}\\ \\ du&= \frac{2}{\sqrt{3}} \;dx\\ dx&=\frac{\sqrt{3}}{2} \;du \end{align*}$
   
   $\begin{align*} &\int \frac{1}{\left (x+\frac{1}{2} \right )^2 + \frac{3}{4}} \; dx\\ &= \frac{4}{3}\int \frac{1}{ \frac{4}{3} \left ( x+\frac{1}{2} \right )^2 + 1} \; dx\\ &= \frac{2}{\sqrt{3}}\int \frac{1}{u^2 + 1} \; du\\ &= \frac{2}{\sqrt{3}} \tan^{-1}{\left ( \frac{2x+1}{\sqrt{3}} \right )} + C \end{align*}$
   
   So, we can use this result to solve the integral F:
   
   $\begin{align*} F&=\int \left ( \frac{x+2}{x^2+x+1} \right )\; dx\\ &= \int \left ( \frac{\frac{1}{2}(2x+1)}{x^2+x+1} + \frac{\frac{3}{2}}{x^2+x+1} \right )\; dx\\ &= \frac{1}{2} \int \frac{2x+1}{x^2+x+1} + \frac{3}{2} \int \frac{1}{x^2+x+1} \; dx\\ &= \frac{1}{2} \ln{|x^2+x+1|} + \frac{3}{2} \int \frac{1}{\left (x+\frac{1}{2} \right )^2 + \frac{3}{4}} \; dx\\ &= \frac{1}{2} \ln{|x^2+x+1|} + \sqrt{3} \tan^{-1}{\left ( \frac{2x+1}{\sqrt{3}} \right )} + C\\ \end{align*}$
   
   
   Lastly, we return to the main integral of this problem:
   
   $\begin{align*} \int \frac{1}{(x-1)(x^2+x+1)} \;dx &= \int \left (\frac{A}{x-1}+ \frac{Bx+C}{x^2+x+1}\right ) \; dx\\ &= \int \left (\frac{1}{3(x-1)}+ \frac{-\frac{1}{3}x-\frac{2}{3}}{x^2+x+1}\right ) \; dx\\ &= \frac{1}{3}\ln{|x-1|} - \frac{1}{3} \int \left ( \frac{x+2}{x^2+x+1} \right )\; dx\\ &= \frac{1}{3}\ln{|x-1|} - \frac{1}{3} F\\ &= \frac{1}{3}\ln{|x-1|} - \frac{1}{6} \ln{|x^2+x+1|} - \frac{ \tan^{-1}{\left ( \frac{2x+1}{\sqrt{3}} \right )}}{\sqrt{3}} + C \\ \end{align*}$

Area Problem

Given a function f(x) as shown above, how would we determine the area of the purple region?
We could approximate by using the area of the rectangles as shown above. We could make smaller rectangles so that the area of the rectangles approaches the actual area of the purple region.

Area of the Region Under the Graph of the Function Let f be a continuous, nonnegative function defined on an interval [a,b]. Suppose that [a,b] is divided into n subintervals of equal length $\Delta x = \frac{b-a}{n}$ by means of (n+1) equally spaced points. Then, the area of the region that lies under the graph of f on [a,b] is: $A= \lim_{n\to \infty}A_n = \lim_{n\to \infty} \sum_{k=1}^{n} f(c_k) \Delta x$ where ck lies in the k-th subinterval [xk-1,xk]

The formula above is also known as Riemann Sum. There are three types of Riemann Sum. Observe:

Left Riemann Sum

Middle-Point Riemann Sum

Right Riemann Sum

There is another sum, which is Trapezoid Sum. But, it is not a part of the Riemann Sum. Just like the previous ones, observe the following:

The green area does not entirely cover the area under the curve, because it consists of rectangles and triangles. However, approximating the area using the Trapezoid Sum can yield a much accurate result.

  Example 2.1 - Riemann Sum

The function shown as an example to demonstrate the types of Riemann Sum above is $f(x)=-\frac{x^2}{2}+2$. Approximate the area between f(x) and the x-axis from x = 0 and x = 2 (or, the purple region) using Left Riemann Sum, Middle-Point Riemann Sum, Right Riemann Sum and the Trapezoid Sum with 4 equal subdivisions.

Solution:
By looking at the visualizations of the left, middle-point and right riemann sums, we can directly calculate the area of the green area. For the Riemann sums, we only need to find the area of the rectangles formed. For the Trapezoid rule, we can either find the area of the trapezoid, or find the area of both the rectangles and triangles (then add it).
$\begin{align*} \Delta x &= \frac{b-a}{n}\\ &= \frac{2-0}{4}\\ &=\frac{1}{2} \end{align*}$

Approximation of the area of the purple region, using Left Riemann Sum:

$\begin{align*} A&=\left ( f(c_1)+f(c_2)+f(c_3)+f(c_4) \right ) \Delta x\\ &= \left ( f(0)+ f\left (\frac{1}{2} \right )+f(1)+f\left ( \frac{3}{2}\right ) \right ) \left ( \frac{1}{2} \right )\\ &= \left ( 2+\frac{15}{8}+\frac{3}{2}+\frac{7}{8} \right ) \left ( \frac{1}{2} \right )\\ &= \frac{25}{8} \end{align*}$

Approximation of the area of the purple region, using Middle Point Riemann Sum:

$\begin{align*} A&=\left ( f(c_1)+f(c_2)+f(c_3)+f(c_4) \right ) \Delta x\\ &= \left ( f\left (\frac{1}{4} \right)+ f\left (\frac{1}{2} \right )+ f\left (\frac{3}{4} \right ) +f\left ( 1\right ) \right ) \left ( \frac{1}{2} \right )\\ &= \left ( \frac{63}{32} + \frac{55}{32} + \frac{39}{32} + \frac{15}{32} \right ) \left ( \frac{1}{2} \right )\\ &= \frac{43}{16} \end{align*}$

Approximation of the area of the purple region, using Right Point Riemann Sum:

$\begin{align*} A&=\left ( f(c_1)+f(c_2)+f(c_3)+f(c_4) \right ) \Delta x\\ &= \left ( f\left (\frac{1}{2} \right )+f(1)+f\left ( \frac{3}{2}\right ) +f(2) \right ) \left ( \frac{1}{2} \right )\\ &= \left ( \frac{15}{8}+\frac{3}{2}+\frac{7}{8}+0 \right ) \left ( \frac{1}{2} \right )\\ &= \frac{17}{8} \end{align*}$

Approximation of the area of the purple region, using Trapezoid Sum:

$\begin{align*} A&=\left ( f(c_1)+f(c_2)+f(c_2)+f(c_3)+f(c_3)+f(c_4)+f(c_4)+f(c_5) \right ) \frac{\Delta x}{2}\\ &= \left ( f(0)+2f\left (\frac{1}{2} \right )+2f(1)+2\left (\frac{3}{2} \right )+f(2) \right ) \left ( \frac{1}{4} \right )\\ &= \left ( 2 + \frac{15}{4} + 3 + \frac{7}{4} \right ) \left ( \frac{1}{4} \right )\\ &= \frac{21}{8} \end{align*}$

Definite Integral

Definition
Let f be a continuous, nonnegative function defined on an interval [a,b]. Suppose that [a,b] is divided into n subintervals of equal length $\Delta x = \frac{b-a}{n}$ by means of (n+1) equally spaced points. Let c₁,...,c_n be arbitrary points in the respective subintervals with c_k lying in the k-th subinterval [x_k-1,x_k]. Then, the definite integral of f on [a,b] or denoted as such, is:

$\intop\nolimits_{a}^{b}f(x)\;dx = \lim_{n\to \infty} \sum_{k=1}^{n} f(c_k) \Delta x$

Properties of Definite Integral

$\begin{align*} \text{1. }&\intop\nolimits_{a}^{a} f(x) \; dx = 0\\ \text{2. } &\intop\nolimits_{a}^{b} f(x) \; dx = -\intop\nolimits_{b}^{a} f(x)\;dx,\;\text{if }a>b\\ \text{3. } &\intop\nolimits_{a}^{b} c \; dx = c(b-a),\;\text{if }c \in \mathbb{R} \\ \text{4. } &\intop\nolimits_{a}^{b} \left [ f(x) \pm g(x) \right ] \; dx = \intop\nolimits_{a}^{b} f(x)\;dx \pm \intop\nolimits_{a}^{b} g(x)\;dx\\ \text{5. } &\intop\nolimits_{a}^{b} c.f(x) \; dx = c\intop\nolimits_{a}^{b} f(x)\;dx \\ \text{6. } &\intop\nolimits_{a}^{b} f(x) \; dx = \intop\nolimits_{a}^{c} f(x)\;dx \pm \intop\nolimits_{c}^{b} g(x)\;dx \;,\text{if }a \leq c \leq b\\ \text{7. } &\intop\nolimits_{a}^{b} f(x) \; dx \geq 0,\text{if }f(x) \geq 0\text{ on }[a,b]\\ \text{8. } &\intop\nolimits_{a}^{b} f(x) \; dx \geq \intop\nolimits_{a}^{b} g(x) \; dx,\text{if }f(x) \geq g(x)\text{ on }[a,b]\\ \end{align*}$

Theorems

Average Value of a Function
If f is integrable on [a,b], then the average value of f over [a,b] is:
$f_{av}=\frac{1}{b-a}\intop\nolimits_{a}^{b} f(x)\;dx$

Mean Value Theorem
If f is continuous on [a,b], then there exists a number c in [a,b] such that:
$f(c)=\frac{1}{b-a}\intop\nolimits_{a}^{b} f(x)\;dx$

First Fundamental Theorem of Calculus
If f is continuous on [a,b], then the function F defined by:
$F(x)=\intop\nolimits_{a}^{x} f(t)\;dt\;\text{ }a\leq x \leq b$
is differentiable on (a,b) and:
$F'(x)=\frac{d}{dx} \intop\nolimits_{a}^{x} f(t)\;dt = f(x)$

Second Fundamental Theorem of Calculus
If f is continuous on [a,b], then:
$\intop\nolimits_{a}^{b} f(x)\;dx=F(b)-F(a)$
where F is any antiderivative of f, that is F' = f.

  Example 3.1 - First Fundamental Theorem of Calculus

Find the derivative of the function $F(x)= \intop\nolimits_{x}^{1} -\frac{1}{t^2+1}\;dt$.

Solution:
The integrand is continuous everywhere:

$\begin{align*} F'(x)&=\frac{d}{dx} \intop\nolimits_{x}^{1} -\frac{1}{t^2+1}\;dt\\ \end{align*}$

Using Property 2:

$\begin{align*} F'(x)&=-\frac{d}{dx} \left [\intop\nolimits_{1}^{x} -\frac{1}{t^2+1}\;dt \right ]\\ &=\frac{d}{dx} \intop\nolimits_{1}^{x} \frac{1}{t^2+1}\;dt \\ &=\frac{1}{x^2+1} \\ \end{align*}$

  Example 3.2 - Second Fundamental Theorem of Calculus

Find the value of:
$\int_{0}^{2} (3x^2 + 2x) \; dx$

Solution:

$\begin{align*} \int_{0}^{2} (3x^2 + 2x) \; dx&= \int_{0}^{2} 3x^2\;dx + \int_{0}^{2} 2x\;dx \\ &= \left (\frac{3}{2+1}x^{2+1} \right ]_{0}^{2} + \left (\frac{2}{1+1}x^{1+1} \right ]_{0}^{2}\\ &= \left (x^3 \right ]_{0}^{2} + \left (x^2 \right ]_{0}^{2}\\ &= (2^3-0^3)+(2^2-0^2)\\ &=12 \end{align*}$

  Example 3.3 - Definite Integral to Determine Area I

The function shown as an example to demonstrate the types of Riemann Sum above is $f(x)=-\frac{x^2}{2}+2$. Find the exact area between f(x) and the x-axis from x = 0 and x = 2 (or, the purple region).

Solution:
The area of the region can be calculated through a definite integral:
$\begin{align*} \intop\nolimits_{0}^{2}\left (-\frac{x^2}{2}+2 \right )\;dx &= \left (-\frac{x^3}{6}+2x \right] _{0}^{2} \\ &= \left ( -\frac{2^3}{6} + 2(2) \right ) - \left ( 0 \right )\\ &= \frac{8}{3} \end{align*}$

  Example 3.4 - Definite Integral to Determine Area II

The function shown above, is $f(x)=x^{3}-8x^{2}+19x-12$. Find the exact shaded region.

Solution:
We cannot directly use definite integral to find the exact area. However, we can do a separate integral (one for the function when it's above the x-axis and another for the function when it's below the x-axis):

$\begin{align*} A&=\intop\nolimits_{1}^{3}\left (x^{3}-8x^{2}+19x-12 \right )\;dx - \intop\nolimits_{3}^{4}\left (x^{3}-8x^{2}+19x-12 \right )\;dx \\ &= \left (\frac{1}{4}x^4-\frac{8}{3}x^3+\frac{19}{2}x^2-12x\right] _{1}^{3} - \left (\frac{1}{4}x^4-\frac{8}{3}x^3+\frac{19}{2}x^2-12x\right] _{3}^{4} \\ &=\left ( \left ( \frac{81}{4} - 72 + \frac{171}{2} - 36 \right ) - \left ( \frac{1}{4} - \frac{8}{3} + \frac{19}{2} - 12 \right ) \right ) - \left ( \left ( 64 - \frac{512}{3} + 152 - 48 \right ) - \left ( \frac{81}{4} - 72 + \frac{171}{2} - 36 \right ) \right )\\ &= \frac{8}{3} - (-\frac{5}{12})\\ &= \frac{37}{12} \end{align*}$

Notice how the result of the second integral is negative? It's because the function is under the x-axis at those interval. Had we directly integrated directly using the interval 1 to 4, like the following:

$\begin{align*} A&=\intop\nolimits_{1}^{4}\left (x^{3}-8x^{2}+19x-12 \right )\;dx \\ &= \left (\frac{1}{4}x^4-\frac{8}{3}x^3+\frac{19}{2}x^2-12x\right] _{1}^{4}\\ &=\left ( \left ( 64 - \frac{512}{3} + 152 - 48 \right ) - \left ( \frac{1}{4} - \frac{8}{3} + \frac{19}{2} - 12 \right ) \right ) \\ &= - \frac{8}{3} - \left (-\frac{59}{12} \right ) \\ &= \frac{9}{4} \end{align*}$

We would have gotten the wrong result.

  Example 3.5 - Average Value of a Function

The function shown above, is $f(x)=x^{3}-8x^{2}+19x-12$. Find the average value of f(x) over the interval [1,3].

Solution:

$\begin{align*} f_{av}&=\frac{1}{b-a} \intop\nolimits_{a}^{b} f(x)\;dx\\ &=\frac{1}{3-1} \intop\nolimits_{1}^{3} (x^{3}-8x^{2}+19x-12)\;dx\\ &=\frac{1}{2} \left (\frac{1}{4}x^4-\frac{8}{3}x^3+\frac{19}{2}x^2-12x \right ]_{1}^{3}\\ &=\frac{1}{2} \left ( \left ( \frac{81}{4} - 72 + \frac{171}{2} - 36 \right ) - \left ( \frac{1}{4} - \frac{8}{3} + \frac{19}{2} - 12 \right ) \right )\\ &=\frac{1}{2} \left ( \frac{8}{3} \right )\\ &= \frac{4}{3} \end{align*}$

  Example 3.6 - Mean Value Theorem

The function shown above, is $f(x)=-\frac{x^2}{2}+2$. Using the Mean Value Theorem, find the value of c for f(x) on the interval [0,2].

Solution:

$\begin{align*} f(c)&=\frac{1}{b-a} \intop\nolimits_{a}^{b} f(x)\;dx\\ &=\frac{1}{2} \intop\nolimits_{0}^{2} \left ( -\frac{x^2}{2}+2 \right ) \;dx\\ &=\frac{1}{2} \left ( -\frac{x^3}{6}+2x \right ]_{0}^{2}\\ &=\frac{1}{2} \left ( -\frac{8}{6}+4 \right )\\ &= \frac{4}{3}\\ -\frac{c^2}{2}+2 &= \frac{4}{3}\\ -c^2&= -\frac{4}{3}\\ c&=\frac{2\sqrt{3}}{3} \end{align*}$

Area Between Curves
Let f(x) and g(x) be functions, where f(x) ≥ g(x) on the interval [a, b]. The area between those functions are:

$\begin{align*} \text{9. }&A=\intop\nolimits_{a}^{b} [f(x)-g(x)] \; dx\\ \end{align*}$

We can also look it at another point of view. Let f(y) and g(y) be functions, where f(y) ≥ g(y) on the interval [c, d]. The area between those functions are:

$\begin{align*} \text{10. }&A=\intop\nolimits_{c}^{d} [f(y)-g(y)] \; dy\\ \end{align*}$

The area calculated from both formula are equal.

  Example 3.7 - Area between Curves I

Calculate the area bounded by the given functions:

1. y = x² and y = x + 2.
2. y = (x+2)^-1, y = (x+2)², x = -³/₂ and x = 1.
3. y = ⁸/_x, y = 2x, and x = 4.

Solution:

1. First, we can try sketching the entire graph:
   
   Function colors:
   • y = x²
   • y = x + 2
   
   
   So, we know that the x-coordinate of the intersection points are the bounds for definite integral. These x-coordinates are found from:
   
   $\begin{align*} y_1&=y_2\\ x^2&=x+2\\ x^2-x-2&=0\\ (x-2)(x+1)&=0\\ x&=2\\ x&=-1 \end{align*}$
   
   These x-coordinates are the bounds for the definite integral:
   
   $\begin{align*} A&=\intop\nolimits_{-1}^{2} [y_2-y_1] \; dx\\ &=\intop\nolimits_{-1}^{2} [(x+2)-x^2] \; dx\\ &=\left (\frac{1}{2}x^2 + 2x - \frac{1}{3}x^3 \right ]_{-1}^{2}\\ &=\left ( 2 + 4 - \frac{8}{3} \right ) - \left ( \frac{1}{2} - 2 + \frac{1}{3} \right )\\ &=\left ( 5 - \frac{1}{2} \right )\\ &=\frac{9}{2} \end{align*}$


2. First, we can try sketching the entire graph:
   
   Function colors:
   • y = (x+2)^-1
   • y = (x+2)²
   
   
   
   We are already given the bounds for the definite integral. However, we have to find the exact x-coordinate of the intersection point of both function:
   
   $\begin{align*} y_1&=y_2\\ \frac{1}{x+2}&=(x+2)^2\\ 1&=(x+2)^3\\ 1&=x+2\\ x&=-1\\ \end{align*}$
   
   Lastly:
   
   $\begin{align*} A&=\intop\nolimits_{-\frac{3}{2}}^{-1} \left (\frac{1}{x+2}-(x+2)^2\right ) \; dx + \intop\nolimits_{-1}^{1} \left ((x+2)^2 - \frac{1}{x+2} \right ) \; dx \\ &=\left (\ln{|x+2|}-\frac{1}{3}(x+2)^3\right ]_{-\frac{3}{2}}^{-1}+ \left (\frac{1}{3}(x+2)^3 - \ln{|x+2|} \right ]_{-1}^{1} \\ &=\left ( -\frac{1}{3} - \left [ \ln{ \left |-\frac{1}{2} \right |}-\frac{1}{24} \right ] \right )+ \left ( \left [9 - \ln{|3|} \right ] - \left [ \frac{1}{3} \right ] \right ) \\ &=\frac{67}{8}-\ln{\left ( \frac{1}{2} \right )} - \ln{3} \end{align*}$


3. First, we can try sketching the entire graph:
   
   Function colors:
   • y = ⁸/_x
   • y = 2x
   
   
   
   So, we know that the x-coordinate of the intersection points is the lower bound for definite integral. This x-coordinate is found from:
   
   $\begin{align*} y_1&=y_2\\ \frac{8}{x}&=2x\\ x&=2 \end{align*}$
   
   These x-coordinates are the bounds for the definite integral:
   
   $\begin{align*} A&=\intop\nolimits_{2}^{4} \left (2x - \frac{8}{x} \right ) \; dx\\ &=\left (x^2 - 8\ln{|x|} \right ]_{2}^{4} \\ &= \left ( 16 - 8\ln{|4|}\right ) - \left ( 4 - 8\ln{|2|}\right ) \\ &=12 - 8 \ln{|2|} \end{align*}$

  Example 3.8 - Area between Curves II

Calculate the area bounded by the given functions:

1. x = sin(y), x = cos(y), y = 0 and y = π/4.
2. x = 10-x², x = (y-2)².
3. x = e^1+2y, x = e^1-y, y = -2 dan y = 1.

Solution:

1. First, we can try sketching the entire graph:
   
   Function colors:
   • x = cos y
   • x = sin y
   
   
   
   We are already given the bounds for the definite integral. So:
   $\begin{align*} A&=\intop\nolimits_{0}^{\frac{\pi}{4}} [x_2-x_1] \; dy\\ &=\intop\nolimits_{0}^{\frac{\pi}{4}} [\cos{y}-\sin{y}] \; dy\\ &=\left (\sin{y}+\cos{y} \right ]_{0}^{\frac{\pi}{4}}\\ &=\left (\sin{\frac{\pi}{4}}+\cos{\frac{\pi}{4}} \right ) - \left (\sin{0}+\cos{0} \right ) \\ &=\sqrt{2} - 1\\ \end{align*}$


2. First, we can try sketching the entire graph:
   
   Function colors:
   • x = 10 - y²
   • x = (y - 2)²
   
   
   
   So, we know that the y-coordinate of the intersection points are the bounds for definite integral. These y-coordinates are found from:
   $\begin{align*} x_1&=x_2\\ 10-y^2&=(y-2)^2\\ 0&=2y^2-4y-6\\ &=y^2-2y-3\\ &=(y-3)(y+1)\\ \\ y&=-1\\ y&=3 \end{align*}$
   
   These y-coordinates are the bounds for the definite integral:
   $\begin{align*} A&=\intop\nolimits_{-1}^{3} [x_2-x_1] \; dy\\ &=\intop\nolimits_{-1}^{3} [-2y^2+4y+6] \; dy\\ &=\left (-\frac{2}{3}y^3 + 2y^2 + 6y \right ]_{-1}^{3}\\ &=\left (-\frac{2}{3}(27) + 2(9) + 6(3) \right ) - \left (\frac{2}{3} + 2 - 6 \right ) \\ &=\frac{64}{3}\\ \end{align*}$


3. First, we can try sketching the entire graph:
   
   Function colors:
   • x = e^1+2y
   • x = e^1-y
   
   
   We are already given the bounds for the definite integral. So:
   $\begin{align*} A&=\intop\nolimits_{-2}^{0} (e^{1-y}-e^{1+2y}) \; dy + \intop\nolimits_{0}^{1} (e^{1+2y}-e^{1-y}) \; dy \\ &=\left (-e^{1-y} - \frac{1}{2}e^{1+2y} \right]_{-2}^{0} + \left (\frac{1}{2}e^{1+2y} + e^{1-y} \right]_{0}^{1}\\ &=\left ( \left [ -e - \frac{1}{2}e \right ] - \left [ -e^{3} - \frac{1}{2}e^{-3} \right ] \right ) - \left ( \left [ \frac{1}{2}e^3 + 1 \right ] - \left [ \frac{1}{2}e + e \right ]\right )\\ &=\frac{3}{2}e^{3}-3e+\frac{1}{2}e^{-3}+1\\ \end{align*}$

Volume of Solid: Disc Method
Let f(x) be a function. The volume formed by rotating the area between f(x) on interval [a, b] along the x-axis is:

$\begin{align*} \text{11. }&V=\pi \intop\nolimits_{a}^{b} (f(x))^2 \; dx\\ \end{align*}$

If it were to be rotated along y-axis:

$\begin{align*} \text{12. }&V=\pi \intop\nolimits_{c}^{d} (f(y))^2 \; dy\\ \end{align*}$

Volume of Solid: Ring Method
Let f(x) and g(x) be functions, where f(x) ≥ g(x) on the interval [a, b]. The volume formed by rotating the area between those functions along the x-axis is:

$\begin{align*} \text{13. }&V=\pi \intop\nolimits_{a}^{b} [(f(x))^2-(g(x))^2] \; dx\\ \end{align*}$

If it were to rotated along y-axis:

$\begin{align*} \text{14. }&V=\pi \intop\nolimits_{c}^{d} [(f(y))^2-(g(y))^2] \; dy\\ \end{align*}$

  Example 3.9 - Disc & Ring Method I

1. Find the formula of a sphere with radius r. Use the Disc Method along x-axis.
2. Calculate the volume obtained by the rotating the area bounded by the given functions: y = 7 - x², x = -2 and x = 2 along the x-axis.
3. Calculate the volume obtained by the rotating the area bounded by the given functions: y = 2x², and y = x³ along the x-axis.

Solution:

1. First, let us recall the general function of circle, which is x² + y² = r². If we rewrite it in explicit way, we will obtain the $y=\sqrt{r^2-x^2}$. When drawn:
   
   
   
   Rotating the shaded area along x-axis would certainly yield a sphere. So:
   $\begin{align*} V&=\pi \intop\nolimits_{-r}^{r} \left ( \sqrt{r^2-x^2} \right)^2 \;dx\\ &=\pi \intop\nolimits_{-r}^{r} \left ( r^2-x^2 \right ) \;dx\\ &=\pi \left ( r^2x-\frac{1}{3}x^3 \right ]_{-r}^{r} \\ &=\pi \left ( \left [ r^3-\frac{1}{3}r^3 \right ] - \left [ -r^3 + \frac{1}{3}r^3 \right ] \right )\\ &=\pi \left ( 2r^3-\frac{2}{3}r^3 \right )\\ &=\frac{4}{3}\pi r^3 \end{align*}$


2. First, we can try sketching the entire graph:
   
   
   Using the formula directly:
   $\begin{align*} V&=\pi \intop\nolimits_{-2}^{2} \left ( 7-x^2 \right )^2 \; dx\\ &=\pi \intop\nolimits_{-2}^{2} \left ( 49-14x^2+x^4 \right ) \; dx\\ &= \left ( 49x-\frac{14}{3}x^3+\frac{1}{5}x^5 \right ]_{-2}^{2}\\ &= \frac{2012}{15}\pi \end{align*}$


3. First, we can try sketching the entire graph:
   
   Function colors:
   • y = 2x²
   • y = x³
   
   
   Finding the upper and lower bounds:
   $\begin{align*} y_1&=y_2\\ 2x^2&=x^3\\ x&=0\\ x&=2 \end{align*}$
   
   Now, we can calculate:
   $\begin{align*} V&=\pi \intop\nolimits_{0}^{2} \left ( \left ( 2x^2 \right )^2 - \left ( x^3 \right )^2 \right ) \; dx\\ &=\pi \intop\nolimits_{0}^{2} \left ( 4x^4 - x^6 \right ) \; dx\\ &= \left ( \frac{4}{5}x^5 - \frac{1}{7}x^7 \right ]_{0}^{2}\\ &= \frac{256}{35}\pi \end{align*}$

  Example 3.10 - Disc & Ring Method II

1. Find the formula of a cylinder with radius r and height h. Use the Disc Method along y-axis.
2. Calculate the volume obtained by the rotating the area bounded by the given function: x = y², and x = y + 2 along the y-axis.
3. Calculate the volume obtained by the rotating the area bounded by the given function: y = 2x + 1, x = 4 and y = 3 along the line x = -4.

Solution:

1. So, imagine a rectangle:
   
   
   
   The graph above is just a representation of a line y = h, x = r. When the shaded area is rotated, it'll form a cylinder:
   $\begin{align*} V&=\pi \intop\nolimits_{0}^{h} \left ( r \right )^2 \; dy\\ &=\pi \intop\nolimits_{0}^{h} r^2 \; dy\\ &=\pi \left ( r^2y \right ]_{0}^{h}\\ &=\pi \left ( r^2 (h) - r^2 (0)\right )\\ &=\pi r^2 h \end{align*}$


2. First, we can try sketching the entire graph:
   
   Function colors:
   • x = y²
   • x = y + 2
   
   
   
   Finding the upper and lower bounds:
   $\begin{align*} x_1&=x_2\\ y+2&=y^2\\ 0&=y^2-y-2\\ &=(y-2)(y+1)\\ y&=2\\ y&=-1 \end{align*}$
   
   Now, we can calculate:
   $\begin{align*} V&=\pi \intop\nolimits_{-1}^{2} \left ( (y+2)^2 - \left (y^2 \right )^2 \right ) \; dy\\ &=\pi \intop\nolimits_{-1}^{2} (y^2+4y+4 - y^4) \; dy\\ &=\pi \left (-\frac{1}{5}y^5 + \frac{1}{3}y^3 + 2y^2 + 4y \right ]_{-1}^{2}\\ &=\frac{72}{5}\pi \end{align*}$


3. First, we can try sketching the entire graph:
   
   Function colors:
   • x = (y-1)/2
   • y = 3
   • x = 4
   
   
   
   The triangle area will be rotated around the line x = -4. We only used the formula for solids that are rotated along the x, or y axis. So, to make this triangle area rotates around the y-axis, we can shift the functions by 4 (add 4 to all x). It will look like this:
   
   
   So, our current graph consists of the following functions:
   
   • x = (y-1)/2 + 4 = (y+7)/2
   • y = 3
   • x = 8
   
   Substituting x = 8 to the line equation will yield the upper bound for the integral, which is y = 9. Now, we can calculate:
   $\begin{align*} V&=\pi \intop\nolimits_{3}^{9} \left ( (8)^2 - \left (\frac{y+7}{2} \right )^2 \right ) \; dy\\ &=\pi \intop\nolimits_{3}^{9} \left ( (8)^2 - \left (\frac{y+7}{2} \right )^2 \right ) \; dy\\ &=\pi \intop\nolimits_{3}^{9} \left (64 - \frac{(y+7)^2}{4} \right ) \; dy\\ &=\pi \left ( 64y - \frac{1}{12}(y+7)^3 \right ]_{3}^{9}\\ &=126\pi \end{align*}$

Volume of Solid: Cylinder (Shell) Method

  Example 3.11 - Cylinder Method

Using the Cylinder Method, calculate the volume obtained by the rotating the area bounded by the given function: y = 2x + 7, x = 8 and y = 3 along the y-axis.

Solution:
Basically, we use one of the following formulas according to the axis we use for the rotation. If we rotate along the x-axis, we will use the second formula. If we rotate along the y-axis, we will use the first formula:
$\begin{align*} V&=2\pi\intop\nolimits_{a}^{b} \left ( Radius \right ) \left ( Height \right ) \; dx\\ V&=2\pi\intop\nolimits_{c}^{d} \left ( Radius \right ) \left ( Height \right ) \; dy\\ \end{align*}$

Since we are rotating along the y-axis, we will use the first formula.

Original desmos project used for the 3d simulation: desmos.com/calculator/8qdpedyypl

These graphs show the result of rotating the shaded area along the y-axis. This is the same as Example 3.10.C, so that we can compare the results since both methods should yield the same result. Using the Ring Method, we obtained the result 126π.



So, according to that, the integral would be:
$\begin{align*} V&=2\pi\intop\nolimits_{5}^{8} \left ( Radius\right )\left (Height \right ) \; dx\\ &=2\pi\intop\nolimits_{5}^{8} \left ( x\right )\left (y-3 \right ) \; dx\\ &=2\pi\intop\nolimits_{5}^{8} \left ( x\right )\left ((2x-7)-3 \right ) \; dx\\ &=2\pi\intop\nolimits_{5}^{8} \left ( x\right )\left (2x -10 \right ) \; dx\\ &=2\pi\intop\nolimits_{5}^{8} \left ( 2x^2 - 10x \right ) \; dx\\ &=2\pi \left (\frac{2}{3}x^3 - 5x^2 \right ]_{5}^{8}\\ &=2\pi \left (\frac{1024}{3} - 320 - \frac{250}{3} + 125 \right )\\ &=2\pi \left (63\right )\\ &=126\pi \end{align*}$

See more examples of using the Cylinder Method: click me

Arc Length

$\begin{align*} \text{13. }&L=\intop\nolimits_{x_1}^{x_2} \sqrt{ 1+(y')^2} \; dx\\ \text{14. }&L=\intop\nolimits_{y_1}^{y_2} \sqrt{ 1+(x')^2} \; dy\\ \text{15. }&L=\intop\nolimits_{t_1}^{t_2} \sqrt{ \left ( \frac{dy}{dt} \right )^2 + \left ( \frac{dx}{dt} \right )^2 } \; dt\\ \end{align*}$

  Example 3.12 - Arc Length of a Function

Given a function along with points, determine the arc length formed!
$\begin{align*} \text{A. }&y=\ln{\sec{x}}\\ &x=0 \text{ to }x=\frac{\pi}{4}\\ \text{B. }&x=\frac{2}{3}(y-1)^{\frac{3}{2}}\\ &y=1 \text{ to }y=4\\ \text{C. }&x=3\cos{t}\\ &y=3\sin{t}\\ &t=0 \text{ to }t=\frac{\pi}{2} \end{align*}$

Solution:

1. Deriving the function:
   $\begin{align*} y&=\ln{\sec{x}}\\ y'&=\frac{\sec{x}\tan{x}}{\sec{x}}\\ &=\tan{x}\\ \\ L&=\intop\nolimits_{a}^{b} \sqrt{ 1+(y')^2} \; dx\\ &=\intop\nolimits_{0}^{\frac{\pi}{4}} \sqrt{ 1+\left ( \tan{x} \right )^2} \; dx\\ &=\intop\nolimits_{0}^{\frac{\pi}{4}} \sec{x} \; dx\\ &= \left( \ln{|\sec{x} + \tan{x}|} \right ]_{0}^{\frac{\pi}{4}}\\ &= ( \ln{|\sec{\frac{\pi}{4}} + \tan{\frac{\pi}{4}}|} ) - ( \ln{|\sec{0} + \tan{0}|} )\\ &= ( \ln{|\sqrt{2} + 1|} ) - \ln{|1|}\\ &= \ln{|\sqrt{2} + 1|} \end{align*}$


2. Deriving the function:
   $\begin{align*} x&=\frac{2}{3}(y-1)^{\frac{3}{2}}\\ x'&=\sqrt{y-1})\\ \\ L&=\intop\nolimits_{a}^{b} \sqrt{ 1+(x')^2} \; dy\\ &=\intop\nolimits_{1}^{4} y \; dy\\ &=\intop\nolimits_{1}^{4} \frac{2}{3}y^{\frac{3}{2}} \; dx\\ &= \frac{2}{3} \left ( \sqrt{y^3} \right ]_{1}^{4} \\ &= \frac{14}{3} \end{align*}$


3. Deriving the function:
   $\begin{align*} x&=3\cos{t}\\ x'&=-3\sin{t}\\ y&=3\sin{t}\\ y'&=3\cos{t} \\ L&=\intop\nolimits_{t_1}^{t_2} \sqrt{ \left ( \frac{dy}{dt} \right )^2 + \left ( \frac{dx}{dt} \right )^2 } \; dt\\ &=\intop\nolimits_{0}^{\frac{\pi}{2}} \sqrt{ 9\cos^2{t} + 9\sin^2{t} } \; dt\\ &=\intop\nolimits_{0}^{\frac{\pi}{2}} 3\sqrt{ \cos^2{t} + \sin^2{t} } \; dt\\ &=\intop\nolimits_{0}^{\frac{\pi}{2}} 3 \; dt\\ &= \left ( 3t \right ]_{0}^{\frac{\pi}{2}}\\ &= \frac{3}{2}\pi \end{align*}$

Surface Area of Solid

$\begin{align*} \text{16. }&S=2 \pi \intop\nolimits_{a}^{b} y \sqrt{ 1+(y')^2} \; dx\\ \text{17. }&S=2 \pi \intop\nolimits_{a}^{b} x \sqrt{ 1+(x')^2} \; dy\\ \end{align*}$

  Example 3.13 - Surface Area of Solid

1. Find the area of the surface obtained by revolving the graph of $f(x)=\sqrt{x}$ on the interval [0,2] about the x-axis.
2. Find the area of the surface obtained by revolving the graph of x = y³ on the interval [0,1] about the y-axis.

Solution:

1. Determining y and y':
   $\begin{align*} y&=\sqrt{x}\\ y'&=\frac{1}{2\sqrt{x}}\\ \end{align*}$
   
   Inputting y and y' into formula (Integral by Substitution will be done):
   $\begin{align*} A&=2 \pi \intop\nolimits_{a}^{b} y \sqrt{ 1+f(y')^2} \; dx\\ &=2 \pi \intop\nolimits_{0}^{2} \sqrt{x} \sqrt{ 1+ \frac{1}{4x}} \; dx\\ &=2 \pi \intop\nolimits_{0}^{2} \sqrt{x+\frac{1}{4}} \; dx\\ \\ u&=\frac{1}{4}+x\\ du&=dx\\ \\ A &=2 \pi \intop\nolimits_{0}^{2} \sqrt{u} \; du\\ &=2 \pi \left (\frac{2}{3}\sqrt{u^3} \right ]_{0}^{2} \\ &=2 \pi \left (\frac{2}{3}\sqrt{\left ( \frac{1}{4}+x \right ) ^3} \right ]_{0}^{2} \\ &=\frac{4}{3}\pi \left( \frac{26}{\sqrt{4^3}}\right ) \end{align*}$


2. Using the formula:
   $\begin{align*} x&=y^3\\ x'&=3y^2\\ \end{align*}$
   
   Inputting x and x' into formula (Integral by Substitution will be done):
   $\begin{align*} A&=2 \pi \intop\nolimits_{a}^{b} x \sqrt{ 1+(x')^2} \; dy\\ &=2 \pi \intop\nolimits_{0}^{1} y^3 \sqrt{ 1+ 9y^4} \; dy\\ \\ u&=1+9y^4\\ \frac{1}{36} du&= y^3 dy\\ \\ A&=\frac{1}{18} \pi \intop\nolimits_{1}^{10} \sqrt{u} \; du\\ &=\frac{1}{27} \pi \left ( u^{\frac{3}{2}} \right ]_{1}^{10} \\ &=\frac{1}{27} \pi \left ( 10\sqrt{10} - 1 \right )\\ \end{align*}$
   
   Since the integrand is changed to u, we have to switch the bounds aswell. Just substitute the bounds of y into u, and we'll obtain 1 and 10 as the new bounds.

Position, Velocity and Acceleration
From the previous topic, we know that when we derive a function of position, we obtain a function of velocity. If we derive it again, we obtain a function of acceleration. Likewise, if we integrate a function of acceleration, we will obtain a function of velocity. If we integrate it again, we obtain a function of position.

  Example 3.14 - Position, Velocity and Acceleration

Suppose the velocity of a car is depicted by the velocity function $v(t)=3x^2-2x+2$. It is also known that when t=2, the car's position is s(2)=4.

1. Find the position function s(x).
2. Find the distance traveled by the car from t=0 to t=2.

Solution:

1. We can integrate the given function v(t):
   $\begin{align*} s(t)&= \intop\nolimits v(t)\;dt\\ &= \intop\nolimits (3t^2-2t+2) \; dt\\ &= t^3-t^2+2t+C \end{align*}$
   
   We are not done yet, since we haven't found C. We also know that when t=2, the car's position is s(2)=4. This information can be utilized to find the exact value of C. So:
   $\begin{align*} s(t) &= t^3-t^2+2t+C\\ s(2) &= 8 - 4 + 4 + C\\ 4 &= 8 + C\\ C &= -4 \end{align*}$
   
   We have our final answer: s(t) = t³ - t² + 2t - 4.


2. Definite integral:
   $\begin{align*} D&=\intop\nolimits_{0}^{2} v(t) \; dt\\ &=\left (s(t) \right ]_{0}^{2}\\ &=s(2)-s(0)\\ &=4 \end{align*}$

Work
The work W done by a constant force F in moving a body a distance d in the direction of the force is:
$W = Fd$

Suppose the force F is a function that is continuous on [a, b], acts on a body moving it along the x-axis. The work done by the force in moving the body from x = a to x = b is:
$W = \int_{a}^{b} F(x)\;dx$

According to Hooke's Law, the force required to stretch the spring meters beyond its natural length is:
$F(x)=kx$

  Example 3.15 - Work

A force of 250 N stretches a spring 30 cm long. How much work is done in stretching the string from 20 cm to 50 cm?

Solution:
The cm unit must be converted to m.

Using the Hooke's Law:
$\begin{align*} F(x)&=kx\\ 250&=\frac{3}{10}k\\ k&=\frac{2500}{3} \end{align*}$

Now, we can integrate F(x) to determine the work:
$\begin{align*} \int_{\frac{2}{10}}^{\frac{5}{10}} F(x)\;dx &= \int_{\frac{2}{10}}^{\frac{5}{10}} \frac{2500}{3}x \; dx\\ &= \left ( \frac{1250}{3}x^2 \right ]_{\frac{2}{10}}^{\frac{5}{10}}\\ &= \frac{1250}{3} \left ( \frac{25}{100} - \frac{4}{100} \right ) \\ &= \frac{1250}{3} \left ( \frac{21}{100} \right ) \\ &= \frac{175}{2} \\ \end{align*}$

So, the work done is 87.5 N.

Fluid Force
The force exerted by a fluid of constant weight-density: &ro; against a submerged vertical plane region from y = c to y = d is:
$F=\rho g \int_{c}^{d} h(y) L(y) \; dy$

where h(y) is the depth of the fluid at y and L(y) is the horizontal length of the region at y.

  Example 3.16 - Fluid Force

A vertical gate in a dam has the shape of an isosceles trapezoid 8 m across the top and 6 m across the bottom, with a height of 5 m. What is the fluid force on the gate when the top of the gate is 4 m below the surface of the water? Density of water is 1000 km.m^-3 and g = 9.81 m.s^-2

Solution:
First, let us find the line L(y):
$\begin{align*} L(-4)&=8\\ L(-9)&=6\\ \\ L(y)&=\frac{2}{5}y+\frac{48}{5}\\ \end{align*}$

We define the function H(y) as (-y), since we used negative sign to indicate depth.

Now, we can integrate:
$\begin{align*} F&=\rho g \int_{c}^{d} h(y) L(y) \; dy\\ &=(1000)(9.81)\int_{-9}^{-4} \left ( -\frac{2}{5}y^2-\frac{48}{5}y \right ) \; dy\\ &=(1000)(9.81)\left ( -\frac{2}{15}y^3-\frac{24}{5}y^2 \right ]_{-9}^{-4}\\ &=(1000)(9.81)\left ( \frac{670}{3} \right )\\ &=(1000)(9.81)\left ( \frac{670}{3} \right )\\ &= 2190900\\ &= 2.1909 \times 10^6\\ \end{align*}$

So, the force is 2,1909 × 10⁶ N.

Area's Mass Center
The center of mass of an area:
$\begin{align*} x &= \frac{1}{A} \int_{a}^{b} x[f(x)-g(x)] \; dx\\ y &= \frac{1}{A} \int_{a}^{b} \frac{1}{2}\left ( [f(x)]^2 - [g(x)]^2 \right ) \; dx \end{align*}$

where:
$\begin{align*} A &= \int_{a}^{b} [f(x)-g(x)] \; dx\\ \end{align*}$

  Example 3.17 - Area's Mass Center

Determine the center of mass for the region bounded by y = x² and y = x.

Solution:
First, let us find the A:
$\begin{align*} A &= \int_{a}^{b} [f(x)-g(x)] \; dx\\ &= \int_{0}^{1} [x-x^2] \; dx\\ &= \left ( \frac{1}{2}x^2 - \frac{1}{3}x^3 \right ] _{0}^{1}\\ &= \left ( \frac{1}{2} - \frac{1}{3} \right )\\ &= \frac{1}{6} \end{align*}$

Finding the x-coordinate of the mass center:
$\begin{align*} x&=\frac{1}{A} \int_{a}^{b} x(f(x)-g(x)) \; dx\\ &=6 \int_{0}^{1} x(x-x^2) \; dx\\ &=6 \int_{0}^{1} (x^2-x^3) \; dx\\ &=6 \left ( \frac{1}{3}x^3 - \frac{1}{4}x^4 \right ] _{0}^{1}\\ &= \frac{1}{2} \end{align*}$

Finding the y-coordinate of the mass center:
$\begin{align*} y&=\frac{1}{A} \int_{a}^{b} \frac{1}{2}([f(x)]^2-[g(x)]^2) \; dx\\ &=3 \int_{0}^{1} (x^2-x^4) \; dx\\ &=3 \left ( \frac{1}{3}x^3 - \frac{1}{5}x^5 \right ] _{0}^{1}\\ &=\frac{2}{5} \end{align*}$

So, the center of mass of the bounded area is $\left (\frac{1}{2},\frac{2}{5}\right )$

Improper Integral

Improper integral is convergent if the limit exists. It is divergent if the limit does not exist.

  Example 4.1 - Improper Integral

Evaluate:
$\begin{align*} &\text{A. } \intop\nolimits_{-1}^{\infty} e^{-x} \; dx\\ &\text{B. } \intop\nolimits_{-\infty}^{1} \frac{1}{x^2+1} \; dx\\ &\text{C. } \intop\nolimits_{-\infty}^{\infty} \frac{1}{x^2+1} \; dx\\ \end{align*}$

Solution:

1. So:
   $\begin{align*} A&= \intop\nolimits_{-1}^{\infty} e^{-x} \; dx\\ A_t&=\intop\nolimits_{-1}^{t} e^{-x} \; dx\\ &=\left ( -e^{-x} \right ]_{-1}^{t}\\ &=-e^{-t}+e\\ \\ \lim_{t\to \infty} A_t &= \lim_{t\to \infty} \left ( e - e^{-t} \right )\\ &= \lim_{t\to \infty} \left ( e - \frac{1}{e^t} \right )\\ &= e \end{align*}$


2. So:
   $\begin{align*} A &=\intop\nolimits_{-\infty}^{1} \frac{1}{x^2+1} \; dx\\ A_t&=\intop\nolimits_{t}^{1} \frac{1}{x^2+1} \; dx\\ &=\left ( \tan^{-1} {x} \right ]_{t}^{1}\\ &=\tan^{-1} {1} - \tan^{-1} {t} \\ \\ \lim_{t\to \infty} A_t &= \lim_{t\to \infty} \left ( \tan^{-1} {1} - \tan^{-1} {t} \right )\\ &= \left ( \frac{\pi}{2} + \frac{\pi}{4} \right )\\ &= \frac{3}{4}\pi \end{align*}$


3. So:
   $\begin{align*} A &=\intop\nolimits_{-\infty}^{\infty} \frac{1}{x^2+1} \; dx\\ A_t&=\intop\nolimits_{-t}^{t} \frac{1}{x^2+1} \; dx\\ &=\left ( \tan^{-1} {x} \right ]_{-t}^{t}\\ &=\tan^{-1} {t} - \tan^{-1} {-t} \\ \\ \lim_{t\to \infty} A_t &= \lim_{t\to \infty} \left ( \tan^{-1} {t} - \tan^{-1} {-t} \right )\\ &= \left ( \frac{\pi}{2} - \left ( -\frac{\pi}{2} \right ) \right )\\ &= \pi \end{align*}$