If $f\left( {a + x} \right) = f\left( x \right)$, then $\mathop \smallint \limits_0^{na} \,f\,\,\left( x \right)\,\,dx$ is equal to $\left( {n \in N} \right)$

$n\,\mathop \smallint \limits_0^a \,f\left( x \right)\,\,dx$

$\left( {n - 1} \right)\,\,\mathop \smallint \limits_0^a f\left( x \right)\,\,dx$

$\mathop \smallint \limits_0^{\left( {n - 1} \right)a} \,f\left( x \right)\,\,dx$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Understanding +4 | -1

f\left( {a + x} \right) = f\left( x \right)

, then

\int \limits_0^{na} \,f\,\,\left( x \right)\,\,dx

is equal to

\left( {n \in N} \right)

View answer and explanation

Correct option: B

n\,\int \limits_0^a \,f\left( x \right)\,\,dx

Option B (

n\int \limits_0^a \\,f\left( x \right)\\,,dx

) is correct because the integral over

na

can be split into

n

integrals over the interval

[0, a]

due to the periodicity of

f(x)