Transwiki:Integral (examples)

Example 1

Consider the function f(x)=7. Suppose we search for $\underset{3}{\overset{11}{\int }}f(x)dx$.

$\int f(x)dx$	$=\int 7dx$
	$=\int 7x^{0}dx$
	$=7x+C,C\in ℝ$

C is the arbitrary constant of integration. While evaluating the definite integral below, I (arbitrarily) choose the antiderivative where C = 0.

$\underset{3}{\overset{11}{\int }}7dx$	$={\left[7x\right]}_3^{11}$
	$=77-21=56$

Example 2

Consider the function $f(x)=x^4+1+2sin(x)$. Suppose we search for $\underset{6}{\overset{8}{\int }}f(x)dx$.

$\int f(x)dx=\int x^{4}dx+\int 1dx+\int 2sin(x)dx$

$=\frac{x^5}{5}+x+C-2\int -sin(x)dx$

$=\frac{x^5}{5}+x+C-2cos(x)$

$\underset{6}{\overset{8}{\int }}f(x)dx={[\frac{x^5}{5}+x-2cos(x)]}_6^8$

$=(\frac{8^5}{5}+8-2\cos (8))-(\frac{6^5}{5}+6-2\cos (6))$

$=+5000.408.$

Example of integrals of an irrational function

$f(x)=\frac{1}{{(x^2+a^2)}^{\frac{3}{2}}}\int f(x)dx=\frac{x}{a^2\sqrt{x^2+a^2}}+C$

First part of the solution:

$\int f(x)dx=\int {(x^2+a^2)}^{-\frac{3}{2}}dx$

$=\int {\left(x^2(1+\frac{a^2}{x^2})\right)}^{-\frac{3}{2}}dx$

$=\int x^{-3}{(1+\frac{a^2}{x^2})}^{-\frac{3}{2}}dx$

$=\int {(1+\frac{a^2}{x^2})}^{-\frac{3}{2}}\left(x^{-3}dx\right)$

Perform the following substitution:

$u=1+\frac{a^2}{x^2}$

$\frac{du}{dx}=-2a^2{x}^{-3}⟹x^{-3}dx=-\frac{du}{2a^2}$

Second part of the solution:

$\int {(1+\frac{a^2}{x^2})}^{-\frac{3}{2}}\left(x^{-3}dx\right)=$

$=\int u^{-\frac{3}{2}}(-\frac{du}{2a^2})$

$=-\frac{1}{2a^2}\int u^{-\frac{3}{2}}du$

$=-\frac{1}{2a^2}(-\frac{2}{\sqrt{u}})+C(C\in ℝ)$

$=\frac{1}{a^2\sqrt{1+\frac{a^2}{x^2}}}+C$

$=\left(\frac{x}{x}\right)\frac{1}{a^2\sqrt{1+\frac{a^2}{x^2}}}+C$

$=\frac{x}{a^2\sqrt{x^2+a^2}}+C$

Alternate method

Using the trigonometric substitution $x=a\tan \theta$, then $dx=a{\sec }^2\theta d\theta$ and $\sqrt{x^2+a^2}=a\sec \theta$ (when $-\pi /2<\theta <\pi /2$).

$\int f(x)dx=\int \frac{1}{{(x^2+a^2)}^{\frac{3}{2}}}dx$

$=\int \frac{1}{a^3{\sec }^3\theta }a{\sec }^2\theta d\theta$

$=\frac{1}{a^2}\int \cos \theta d\theta$

$=\frac{1}{a^2}\sin \theta +C$

$=\frac{1}{a^2}\frac{a\tan \theta }{a\sec \theta }+C$

$=\frac{x}{a^2\sqrt{x^2+a^2}}+C$