If $f$ is an odd function, then $\displaystyle\int_{-a}^{a} f(x)\,dx$ equals
A$2\displaystyle\int_0^a f(x)\,dx$
B$\displaystyle\int_0^a f(x)\,dx$
C$0$
D$2f(a)$
Answer & Solution
Correct answer: C. $0$
1. For an odd function $f(-x)=-f(x)$.
2. By property $\mathbf{P_7}$(ii), the contributions over $[-a,0]$ and $[0,a]$ cancel.
3. Hence $\int_{-a}^{a} f(x)\,dx=0$.
4. Option A is the even-function case.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.343_
Related questions
$\displaystyle\int_2^3 \dfrac{x\,dx}{x^2+1}$ equals$\displaystyle\int_0^{2a} f(x)\,dx$ equals $2\displaystyle\int_0^{a} f(x)\,dx$ precisely wIf $f(a+b-x)=f(x)$, then $\displaystyle\int_a^b x\,f(x)\,dx$ is equal to$\displaystyle\int \frac{dx}{e^{x}+e^{-x}}$ is equal toA rational function $\dfrac{P(x)}{Q(x)}$ is called proper when$\displaystyle\int_0^{\pi/4} \tan x\,dx$ equals$\displaystyle\int_0^{1} x e^{x^2}\,dx$ equalsThe value of $\displaystyle\int_0^{\pi/2} \log\!\left(\dfrac{4+3 in x}{4+3\cos x}\right)dx