Question

Evaluate:$\int\limits_{-2}^{2}{\dfrac{{{x}^{2}}}{1+{{5}^{x}}}}dx.$

Answer 1

Hint:
In the given question we have to find out the value of a definite integration. In order to solve such type of question we have to use the properties of definite integration. Here it is advisable to use the property $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(a+b-x)dx}}$, in this question we have the value of $a=-2,b=+2$
And the function here given is $f(x)=\dfrac{{{x}^{2}}}{1+{{5}^{x}}}$.

Complete step by step answer:
Let us assume that
$I=\int\limits_{-2}^{2}{\dfrac{{{x}^{2}}}{1+{{5}^{x}}}}dx------(1)$
Now we have to use the property of definite integration $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(a+b-x)dx}}$, we can write
$I=\int\limits_{-2}^{2}{\dfrac{{{\left( 2-2-x \right)}^{2}}}{1+5\left( ^{2-2-x} \right)}}dx.$
So, we can write the above integration as
$I=\int\limits_{-2}^{2}{\dfrac{{{\left( -x \right)}^{2}}}{1+{{5}^{-x}}}}dx.$
This can be further written as
$\begin{align}
  & I=\int\limits_{-2}^{2}{\dfrac{{{x}^{2}}}{1+\dfrac{1}{{{5}^{x}}}}}dx \\
 & \Rightarrow I=\int\limits_{-2}^{2}{\dfrac{{{5}^{x}}{{x}^{2}}}{{{5}^{x}}+1}}dx------(2) \\
\end{align}$
Now we have to add $(1)\text{ and }(2)$ we can write
$I+I=\int\limits_{-2}^{2}{\dfrac{{{x}^{2}}}{1+{{5}^{x}}}}dx+\int\limits_{-2}^{2}{\dfrac{{{5}^{x}}{{x}^{2}}}{1+{{5}^{x}}}}dx$
Hence, we can write the above sum of integral as
$2I=\int\limits_{-2}^{2}{\dfrac{{{x}^{2}}+{{5}^{x}}{{x}^{2}}}{1+{{5}^{x}}}}dx$
In the above step we use the property of definite integration
\[\int\limits_{a}^{b}{f(x)dx+}\int\limits_{a}^{b}{g(x)dx=}\int\limits_{a}^{b}{\left( f(x)+g(x) \right)dx}\]
Now we can take ${{x}^{2}}$as common we can write the above integral as
$2I=\int\limits_{-2}^{2}{\dfrac{{{x}^{2}}\left( 1+{{5}^{x}} \right)}{1+{{5}^{x}}}}dx$
As we see here $1+{{5}^{x}}$is in numerator as well as denominator, so both cancel each other so we can write further
$2I=\int\limits_{-2}^{2}{{{x}^{2}}}dx$
Now as we know that $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c$
So, we can write further
$\begin{align}
  & 2I={{\left[ \dfrac{{{x}^{2+1}}}{2+1} \right]}^{2}}_{-2} \\
 & 2I={{\left[ \dfrac{{{x}^{3}}}{3} \right]}^{2}}_{-2} \\
\end{align}$
Now we have to put the limit and upper limit we can write
\[2I=\left[ \dfrac{{{2}^{3}}}{3}-\dfrac{{{\left( -2 \right)}^{3}}}{3} \right]\]
After simplification we can write
$\begin{align}
  & I=\dfrac{1}{2}\left( \dfrac{8}{3}+\dfrac{8}{3} \right) \\
 & I=\dfrac{16}{6} \\
 & I=\dfrac{8}{3} \\
\end{align}$
Hence the value of given definite integration is$\dfrac{8}{3}$.

Note:
It should be important to note that in case of definite integration we cannot write integration constant as the value of definite integral is unique, for, if
$\int{f(x)dx=F(x)+c}$then we can write
$\int\limits_{a}^{b}{f(x)dx=\left[ F(x)+c \right]}_{a}^{b}$
This can we written further
$\begin{align}
  & \int\limits_{a}^{b}{f(x)dx=\left[ F(b)+c-F(a)-c \right]} \\
 & \int\limits_{a}^{b}{f(x)dx=\left[ F(b)-F(a) \right]} \\
\end{align}$
Hence, we see here there is no any integration constant exist in definite integration.
Also, when we have to find out the definite integration of a function the function must be continuous.