JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

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Analyzing +4 | -1

Area bounded by

y=f^{-1}(x)

and tangent and normal drawn to it at the points with abscissae

\pi

and

2 \pi

, where

f(x)=\sin x-x

View answer and explanation

Correct option: B

\frac{\pi^2}{2}-2

Option B (

\frac{{\pi^{2}}}{{2}}-2

) is correct because the area bounded by the inverse function and the tangent and normal lines is calculated by integrating the difference between the function and the lines.

More Questions from DEFINITE INTEGRATION

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\int \nolimits_0^\infty \,\,\log \,\,\left( {x + \frac{1}{x}} \right)\,\,\frac{{dx}}{{1 + {x^2}}}

is equal to If

f\left( x \right)

is differentiable and

\int \nolimits_0^{{t^2}} \,\,x\;\,f\left( x \right)\,\,dx = \frac{2}{5}\,\,{t^5}

, then

f\left( {\frac{4}{{25}}} \right)

equals The value of

\int\limits_{{0}}^{11}[x]^{3} . \mathrm{dx}

, where

[*]

denotes the greatest integer function, is The integral

\int\limits_{2}^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} d x

is equal to: The definite integral

\int\limits_{{-2}}^0\left(x^3+3 x^2+3 x+3+(x+1) \cos (x+1)\right) d x

equals If

I\,\,\left( {m,\,\,n} \right) = \int \nolimits_0^1 \,\,{t^m}\,\,{\left( {1 + t} \right)^n}\,\,dt

,then the expression for

I\left( {m,\,\,n} \right)

in terms of

I\,\,\left( {m + 1,\,\,n + 1} \right)