Question

Suponha que F(t) = ∫_1^x▒〖f(t)dt=x^2-2x+1〗. Determine f(x)

Answer 1

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Solução!

Esse exercício é uma demonstração do TFC.

$\boxed{\dfrac{d}{dx}[\int_{a}^{x} f(t)dt]=f(x)}$



$\displaystyle \int_{1}^{x} x^{2} -2x+1 dx\\\\\\\ I= \frac{ x^{3} }{3}+ \frac{2 x^{2} }{2}+x \bigg| _{1}^{x}\\\\\\\\\\\\ I= \left (\frac{ x^{3} }{3}+ \frac{2 x^{2} }{2}+x \right )- \left (\frac{ ^{3} }{3}+ \frac{2 x^{2} }{2}+x \right )\\\\\\\\\\\ I= \left (\frac{ x^{3} }{3}+ \frac{2 x^{2} }{2}+x \right )- \left (\frac{ 1^{3} }{3}+ \frac{1^{2} }{2}+1 \right )\\\\\\\\\\ I= \left (\frac{ x^{3} }{3}+ \frac{2 x^{2} }{2}+x \right )- \left (\frac{ 1 }{3}+ \frac{1}{2}+1 \right )$


$Fazendo~~a~~derivada!\\\\\\\ I= \left (\dfrac{ x^{3} }{3}+ \dfrac{2 x^{2} }{2}+x \right )- \left (\dfrac{ 1 }{3}+ \dfrac{1}{2}+1 \right )\\\\\\\\\\ I'= \left (\dfrac{ 3.x^{3-1} }{3}+ \dfrac{2.2 x^{2-1} }{2}+ x^{1-1} \right )- \left (0+0+0\right )\\\\\\\\\\ I'= \left ( x^{2} +2 x+ 1 \right )$


$Logo,~~pelo~~TFC~~temos\\\\\\\ F'(t)= \dfrac{ x^{3} }{3}+ \dfrac{2 x^{2} }{2}+x\\\\\\\ f(x)= x^{2} +2 x+ 1\\\\\\\ F'(t)=f(x)\\\\\\\\\ \boxed{\dfrac{ x^{3} }{3}+ \dfrac{2 x^{2} }{2}+x=x^{2} +2 x+ 1 }$


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