JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

The smallest interval

[a, b]

such that

\int\limits_{0}^1 \frac{d x}{\sqrt{1+x^4}} \in[a, b]

is given by

View answer and explanation

Correct option: A

\left[\frac{1}{\sqrt{2}}, 1\right]

Option A (

\left[\frac{1}{\sqrt{2}}, 1\right]

) is correct because it provides the smallest interval that contains the value of the definite integral, which is bounded by the minimum and maximum values of the integrand.

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)