If f(x) =lower lt 0, upper lt x integration e^2tsin3t dt, then f'(x)= –
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1 thought on “<br />If f(x) =lower lt 0, upper lt x integration e^2tsin3t dt, then f'(x)= –<br />​”

1. [tex]\large\underline{\sf{Given- }}[/tex]

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

   [tex]\large\underline{\sf{To\:Find – }}[/tex]

   [tex]\rm :\longmapsto\:f'(x)[/tex]

   [tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used – }}}} \end{gathered}[/tex]

   [tex] \sf \: Differentiation \: under \: the \: integral \: sign \: [/tex]

   [tex] \bf \: The \: Leibniz \: Rule[/tex]

   [tex] \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) [/tex]

   [tex]\large\underline{\sf{Solution-}}[/tex]

   Given that

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

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

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

   [tex]\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[/tex]

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

   [tex]\bf\implies \:f'(x) = {e}^{2x} sin3x[/tex]

   Additional Information :-

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

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