Question

If $f\left(x\right)$ is an even function in $[-a,a]$ then ${\int }_{-a}^{+a}f\left(x\right)dx=?$

• A. $0$
• B. ${\int }_0^{2a}\left[f\right(x)+f(a-x\left)\right]dx$
• C. $2{\int }_0^{a}f\left(x\right)dx$
• D. $2$

Answer 1

The correct option is C $2{\int }_0^{a}f\left(x\right)dx$

$={\int }_{-a}^{a}f\left(x\right)dx$

$={\int }_{-a}^{0}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

$I_1={\int }_{-a}^{0}f\left(x\right)dx$ let $x=-t$

$-{\int }_{-a}^{0}f(-t)\cdot dt=-{\int }_a^{0}f\left(t\right)dt={\int }_0^{a}f\left(t\right)dt$

($\because$ it is even)

$I_1={\int }_0^{a}f\left(x\right)dx$

So, $I_1=I_2={\int }_0^{a}f\left(x\right)dx$

$I=2{\int }_0^{a}f\left(x\right)dx$