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 December 13th, 2017, 10:59 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 Integration question Calculate G'(x) given that: $g(x)=\int_{x}^{x^2}f(t)dt$ where f(t) is differentiable.
December 13th, 2017, 12:09 PM   #2
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 Originally Posted by Jaket1 Calculate G'(x) given that: $g(x)=\int_{x}^{x^2}f(t)dt$ where f(t) is differentiable.
I assume you mean $g'(x)$ ...

$\displaystyle \dfrac{d}{dx} \bigg[\int_v^u f(t) \, dt \bigg] = f(u) \cdot \dfrac{du}{dx} - f(v) \cdot \dfrac{dv}{dx}$

 December 13th, 2017, 01:10 PM #3 Global Moderator   Joined: May 2007 Posts: 6,786 Thanks: 708 Let F(t) be the indefinite integral of f(t). $\displaystyle g(x)=F(x^2)-F(x).\ g'(x)=(2x-1)f(x)$. Thanks from greg1313
 December 13th, 2017, 02:51 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,823 Thanks: 2159 Not quite... $\displaystyle g'(x) = 2x\hspace{1px}f\!\left(x^2\right) - f(x)$. Thanks from greg1313 and Jaket1
December 14th, 2017, 01:40 PM   #5
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 Originally Posted by skipjack Not quite... $\displaystyle g'(x) = 2x\hspace{1px}f\!\left(x^2\right) - f(x)$.
You are right - I stand corrected. Getting sloppy in my old age.

 December 16th, 2017, 02:50 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Leibniz's rule: $\displaystyle \frac{\d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x, t)dt= f(x, \beta(x)}\frac{\d\beta}{dx}- f(x, \alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f(x, y)}{\partial y}dy$. Here $\displaystyle \beta(x)= x^2$, $\displaystyle \alpha(x)= x$ f is independent of x so $\displaystyle \frac{dg}{dx}= 2xf(x^2)- f(x)$.

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