Question

If f(x) =lower lt 0, upper lt x integration e^2tsin3t dt, then f'(x)= –
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Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\: \sf \: f(x) = \sf \displaystyle\int^{x} _{0} \sf \: {e}^{2t} \: sin3t \: dt$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:f'(x)$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\sf \: Differentiation \: under \: the \: integral \: sign \:$

$\bf \: The \: Leibniz \: Rule$

$\sf \: \dfrac{d}{dx}\sf \displaystyle\int^{b} _{a} \sf \:f(x,t)dt = \sf \displaystyle\int^{b} _{a} \sf \: \dfrac{ \partial}{ \partial \: x} f(x,t)dt + \dfrac{db}{dx}f(x,b) - \dfrac{da}{dx}f(x,a)$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: \sf \: f(x) = \sf \displaystyle\int^{x} _{0} \sf \: {e}^{2t} \: sin3t \: dt$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \sf \: \dfrac{d}{dx}f(x) = \dfrac{d}{dx}\sf \displaystyle\int^{x} _{0} \sf \: {e}^{2t} \: sin3t \: dt$

$\rm :\longmapsto\: \sf \: f'(x) = \sf \displaystyle\int^{x} _{0} \sf \:\dfrac{ \partial}{ \partial \: x} \: {e}^{2t} \: sin3t \: dt + \dfrac{d}{dx}(x) \: {e}^{2x} sin3x - 0$

$\rm :\longmapsto\:f'(x) = 0 + 1 \times {e}^{2x} sin3x$

$\bf\implies \:f'(x) = {e}^{2x} sin3x$

Additional Information :-

• The Leibniz rule for differentiating under the integral sign is also known as Feynman's trick or technique for integration.