Question

If f(x) is a continuous function defined on [-a, a], then ${\int }_{-a}^{a}f\left(x\right)\mathrm{dx}$

• A. ${\int }_0^a\left(f\left(x\right)+f(-x)\right\}\mathrm{dx}$
• B. ${\int }_0^a\left(f\left(x\right)+f(a-x)\right\}\mathrm{dx}$
• C. 0, if f is an odd function
• D. $2{\int }_0^{a}f\left(x\right)\mathrm{dx}$, if f is an even function.

Answer 1

Answer: (A), (C), (D)

The correct options are
A ${\int }_0^a\left(f\left(x\right)+f(-x)\right\}\mathrm{dx}$
C 0, if f is an odd function
D $2{\int }_0^{a}f\left(x\right)\mathrm{dx}$, if f is an even function.

We have,
${\int }_{-a}^{a}f\left(x\right)\mathrm{dx}={\int }_{-a}^{0}f\left(x\right)\mathrm{dx}+{\int }_0^{a}f\left(x\right)\mathrm{dx}...\left(1\right]\mathrm{Let}x=-t.\mathrm{Then}\mathrm{dx}=-\mathrm{dt}\mathrm{Also},x=-a\Rightarrow t=a\mathrm{and},x=0\Rightarrow t=0\therefore {\int }_{-a}^{0}f\left(x\right)\mathrm{dx}={\int }_a^{0}f\left(-t\right)\left(-\mathrm{dt}\right)=-{\int }_a^{0}f\left(-t\right)\mathrm{dt}={\int }_0^{a}f\left(-x\right)\mathrm{dx}\mathrm{Now},\left(1\right]\Rightarrow {\int }_{-a}^{a}f\left(x\right)\mathrm{dx}={\int }_0^{a}f\left(x\right)\mathrm{dx}+{\int }_0^{a}f\left(-x\right)\mathrm{dx}\Rightarrow {\int }_{-a}^{a}f\left(x\right)\mathrm{dx}={\int }_0^a\left(f\left(x\right)+f\left(-x\right)\right\}\mathrm{dx}\mathrm{If}f\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{function},f\left(-x\right)=f\left(x\right)\mathrm{and},\mathrm{If}f\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function},f\left(-x\right)=-f\left(x\right)\therefore {\int }_{-a}^{a}f\left(x\right)\mathrm{dx}=(\begin{array}{c}2{\int }_0^{a}f\left(x\right)\mathrm{dx},\mathrm{if}f\mathrm{is}\mathrm{even}\\ 0,\mathrm{if}f\mathrm{is}\mathrm{odd}\end{array}$