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