Question

Evaluate integral of $\frac{x^2}{\left(x^2+{\left(9-x\right)}^2\right)}$ with limits from $4$ to $5$ . The result is

• A. $\frac{1}{2}$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

Let $I=\underset{4}{\overset{5}{\int }}\frac{x^2}{x^2+{\left(9-x\right)}^2}dx\dots \left(i\right)$
$\Rightarrow I=\underset{4}{\overset{5}{\int }}\frac{{\left(9-x\right)}^2}{{\left(9-x\right)}^2+x^2}dx\dots \left(ii\right)$
$\left[\because \underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx\right]$
Adding $\left(i\right)$ & $\left(ii\right)$, we get
$2I=\underset{4}{\overset{5}{\int }}\frac{x^2+{\left(9-x\right)}^2}{x^2+{\left(9-x\right)}^2}dx$
$=\underset{4}{\overset{5}{\int }}1dx={\left[x\right]}_4^5$
$\Rightarrow 2I=5-4=1$
$\Rightarrow I=\frac{1}{2}$