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

1.	If f(x) =lower lt 0, upper lt x integration e^2tsin3t dt, then f'(x)= –
Answer» $\large\underline{\sf{Given- }}$ $\rm :\longmapsto\: \sf \: f(x) = \sf \displaystyle\int^{x} _{0} \sf \: {e}^{2t} \: sin3t \: dt$ $\large\underline{\sf{To\:Find - }}$ $\rm :\longmapsto\:f'(x)$ $\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$ $\sf \: Differentiation \: under \: the \: integral \: <klux>SIGN</klux> \:$ $\bf \: The \: Leibniz \: Rule$ $\sf \: \dfrac{d}{dx}\sf \displaystyle\int^{b} _{a} \sf \:f(x,t)dt = \sf \displaystyle\int^{b} _{a} \sf \: \dfrac{ \partial}{ \partial \: x} f(x,t)dt + \dfrac{db}{dx}f(x,b) - \dfrac{da}{dx}f(x,a)$ $\large\underline{\sf{Solution-}}$ 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.

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

Answer»

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

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

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

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

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

$\sf \: Differentiation \: under \: the \: integral \: <klux>SIGN</klux> \:$

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

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

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

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.