If $f\left( x \right) = f\left( {a - x} \right),\;\,\,{\rm{then}}\,\,\mathop \smallint \nolimits_0^a \,\,f\left( x \right)\,\,dx$ is equal to

$\frac{a}{2}\,\,\mathop \smallint \nolimits_0^a \,\,f\left( x \right)\,\,dx$

$\mathop \smallint \nolimits_0^a \,\,f\left( x \right)\,\,dx$

$\frac{{{a^2}}}{2}\,\,\mathop \smallint \nolimits_0^a \,\,f\left( x \right)dx$

$ - \frac{a}{2}\,\,\mathop \smallint \nolimits_0^a \,\,f\left( x \right)\,\,dx$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

f\left( x \right) = f\left( {a - x} \right),\;\,\,\mathrm{then}\,\,\int \nolimits_0^a \,\,f\left( x \right)\,\,dx

is equal to

View answer and explanation

Correct option: C

\frac{a}{2}\,\,\int \nolimits_0^a \,\,f\left( x \right)\,\,dx

Option C (

\frac{a}{2}\,\,\int \nolimits_0^a \,\,f(x)\,\,\;dx

) is correct because using the property

f(x) = f(a - x)

simplifies the integral to

\frac{a}{2}\,\,\int \nolimits_0^a \,\,f(x)\,\,\;dx