JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

Let

f(x)=\{x\}

, the fractional part of

x

then

\int\limits_{{-1}}^1 f(x) d x

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because the integral of the fractional part function over the interval [-1, 1] can be calculated by splitting the integral and using the definition of the fractional part function.

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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}

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

equals The value of

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

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[*]

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\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

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

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