If $F(x)=\int_\limits{x^2}^{x^3} \log t \,d t,(x > 0)$, then $F^{\prime}(x)=$

$\left(4 x-9 x^2\right) \log x$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

F(x)=\int\limits_{{x^2}}^{x^3} \log t \,d t,(x > 0)

, then

F^{\prime}(x)=

View answer and explanation

Correct option: A

\left(9 x^2-4 x\right) \log x

Option A ((9x^2-4x)log x) is correct because it is the result of applying Leibniz's rule for differentiating an integral with variable limits.

Practice All Chapter MCQs →

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

f\left( x \right)

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

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

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

[*]

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

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

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

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

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