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Question Number 137378 by benjo_mathlover last updated on 02/Apr/21

Given ∫^( x) _0 f(t) dt = x^2  sin (πx)  find f(2).

$${Given}\:\underset{\mathrm{0}} {\int}^{\:{x}} {f}\left({t}\right)\:{dt}\:=\:{x}^{\mathrm{2}} \:\mathrm{sin}\:\left(\pi{x}\right) \\ $$$${find}\:{f}\left(\mathrm{2}\right). \\ $$

Answered by EDWIN88 last updated on 02/Apr/21

 (d/dx) [ ∫_0 ^( x) f(t) dt ] = (d/dx) [ x^2  sin (πx) ]  ⇒ f(x) = 2x sin (πx)+πx^2  cos (πx)  ⇒f(2)= 4sin (2π) + 4π cos (2π)= 4π

$$\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\:\int_{\mathrm{0}} ^{\:\mathrm{x}} \mathrm{f}\left(\mathrm{t}\right)\:\mathrm{dt}\:\right]\:=\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\:\mathrm{x}^{\mathrm{2}} \:\mathrm{sin}\:\left(\pi\mathrm{x}\right)\:\right] \\ $$$$\Rightarrow\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{2x}\:\mathrm{sin}\:\left(\pi\mathrm{x}\right)+\pi\mathrm{x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\pi\mathrm{x}\right) \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{2}\right)=\:\mathrm{4sin}\:\left(\mathrm{2}\pi\right)\:+\:\mathrm{4}\pi\:\mathrm{cos}\:\left(\mathrm{2}\pi\right)=\:\mathrm{4}\pi \\ $$

Answered by mathmax by abdo last updated on 02/Apr/21

∫_0 ^x  f(t)dt =x^2 sin(πx) ⇒f(x)=2xsin(πx)+πx^2 cos(πx) ⇒  f(2)=4sin(2π)+4πcos(2π) =4π

$$\int_{\mathrm{0}} ^{\mathrm{x}} \:\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\:=\mathrm{x}^{\mathrm{2}} \mathrm{sin}\left(\pi\mathrm{x}\right)\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2xsin}\left(\pi\mathrm{x}\right)+\pi\mathrm{x}^{\mathrm{2}} \mathrm{cos}\left(\pi\mathrm{x}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{2}\right)=\mathrm{4sin}\left(\mathrm{2}\pi\right)+\mathrm{4}\pi\mathrm{cos}\left(\mathrm{2}\pi\right)\:=\mathrm{4}\pi \\ $$

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