If $f\left( x \right)$ be continuous function such that the area bounded by the curve $y = f\left( x \right)$, the $x$-axis and the lines $x = a{\rm{\;and}}\;x = 0\;{\rm{is}}\frac{{{a^2}}}{2} + \frac{a}{2}\sin a + \frac{{\rm{\pi }}}{2}\cos a.\,{\rm{Value\;of\;}}f\left( {\frac{{\rm{\pi }}}{2}} \right)\;{\rm{is}}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

f\left( x \right)

be continuous function such that the area bounded by the curve

y = f\left( x \right)

, the

x

-axis and the lines

x = a\mathrm{\;and}\;x = 0\;\mathrm{is}\frac{{{a^2}}}{2} + \frac{a}{2}\sin a + \frac{\mathrm{\pi }}{2}\cos a.\,\mathrm{Value\;of\;}f\left( {\frac{\mathrm{\pi }}{2}} \right)\;\mathrm{is}

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because it is the result of differentiating the area function and substituting

\frac{\pi}{2}

for

a