Question

If $f\left(x\right)$ is a continuous function for all real values of $x$ and satisfises $\underset{n}{\overset{n+1}{\int }}f\left(x\right)dx=\frac{n^2}{2}\forall n\in I$, then $\underset{-3}{\overset{5}{\int }}f\left(|x|\right)dx$ is equal to

• A. $\frac{19}{2}$
• B. $\frac{35}{2}$
• C. $\frac{17}{2}$
• D. $\frac{30}{2}$

Answer 1

The correct option is B $\frac{35}{2}$

$I=\underset{-3}{\overset{5}{\int }}f\left(|x|\right)dx=\underset{-3}{\overset{3}{\int }}f\left(|x|\right)dx+\underset{3}{\overset{5}{\int }}f\left(|x|\right)dx=2\underset{0}{\overset{3}{\int }}f\left(x\right)dx+\underset{3}{\overset{5}{\int }}f\left(x\right)dx[\because \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{if}f\left(x\right)\text{is even}]=2\left(\underset{0}{\overset{1}{\int }}f\left(x\right)dx+\underset{1}{\overset{2}{\int }}f\left(x\right)dx+\underset{2}{\overset{3}{\int }}f\left(x\right)dx\right)+\left(\underset{3}{\overset{4}{\int }}f\left(x\right)dx+\underset{4}{\overset{5}{\int }}f\left(x\right)dx\right)=2(0+\frac{1}{2}+\frac{2^2}{2})+(\frac{9}{2}+\frac{16}{2})\therefore I=\frac{35}{2}$