Without integrating, explain why ∫−22x(x2+1)2dx=0

Question

Without integrating, explain why  ∫−22x(x2+1)2dx=0

Alannej · Accepted Answer

Step 1
Odd Function
A function f(x) is a odd function if and only if
$f (- x) = - f (x)$
Even Function
A function f(x) is an even function if and only if
$f (- x) = f (x)$
Now, if we have $I = \int_{- a}^{a} f (x) d x$ , then
$I = {\begin{cases} 0 & if f(x) is an odd function \\ 2 \int_{0}^{2} f (x) d x & if f(x) is an even function \end{cases}$
With the help of this result we will solve our given integration.
Step 2
Given integration is $\int_{- 2}^{2} x {(x^{2} + 1)}^{2} d x$
Here, $f (x) = x {(x^{2} + 1)}^{2}$ and $a = 2$
Now, to check the function, that is function is odd or even, consider
$f (- x)$ , replace x by -x, we get
$f (- x) = - x {({(- x)}^{2} + 1)}^{2}$
$= - x {(x^{2} + 1)}^{2}$
$= - (x {(x^{2} + 1)}^{2})$
$= - f (x)$
$\Rightarrow f (- x) = - f (x)$
hence function $x {(x^{2} + 1)}^{2}$ is an odd function.
So, by the result of integration,
$\int_{- 2}^{2} x {(x^{2} = 1)}^{2} d x = 0$
Hence proved

Without integrating, explain why \int_{-2}^{2}x(x^{2}+1)^{2}dx=0

Answered question

Answer & Explanation

New Questions in Calculus 1