Let $f\left( x \right)$ be a non-negative continuous function such that the area bounded by the curve $y = f\left( x \right),\;x$-axis and the ordinates $x = \frac{{\rm{\pi }}}{4}{\rm{\;}}$ and $x = {\rm{\beta }} > \frac{{\rm{\pi }}}{4}\,{\rm{is}}\,{\rm{\pi }}$ $\left( {{\rm{\beta }}\sin {\rm{\beta }} + \frac{{\rm{\pi }}}{4}\cos {\rm{\beta }} + \sqrt 2 {\rm{\beta }}} \right)\;{\rm{the\;}}f\left( {\frac{{\rm{\pi }}}{2}} \right){\rm{s}}$

$\left( {1 - \frac{{\rm{\pi }}}{4} + \sqrt 2 } \right)$

$\left( {1 - \frac{{\rm{\pi }}}{4} - \sqrt 2 } \right)$

$\left( {\frac{{\rm{\pi }}}{4} - \sqrt 2 + 1} \right)$

$\left( {\frac{{\rm{\pi }}}{4} + \sqrt 2 - 1} \right)$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Medium

1 of 1

Application +4 | -1

Let

f\left( x \right)

be a non-negative continuous function such that the area bounded by the curve

y = f\left( x \right),\;x

-axis and the ordinates

x = \frac{\mathrm{\pi }}{4}\mathrm{\;}

and

x = \mathrm{\beta } > \frac{\mathrm{\pi }}{4}\,\mathrm{is}\,\mathrm{\pi }

\left( {\mathrm{\beta }\sin \mathrm{\beta } + \frac{\mathrm{\pi }}{4}\cos \mathrm{\beta } + \sqrt 2 \mathrm{\beta }} \right)\;\mathrm{the\;}f\left( {\frac{\mathrm{\pi }}{2}} \right)\mathrm{s}

View answer and explanation

Correct option: A

\left( {1 - \frac{\mathrm{\pi }}{4} + \sqrt 2 } \right)

Option A (

\left( {1 - \frac{\mathrm{\pi }}{4} + \sqrt 2 } \right)

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

\beta = \frac{{\pi }}{2}