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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: 6,823 Thanks: 723 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: 20,978 Thanks: 2229 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 $\displaystyle \frac{dg}{dx}= 2xf(x^2)- f(x)$. Tags integration, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Spelcy.z Real Analysis 1 May 29th, 2013 04:38 AM steve93 Calculus 2 June 3rd, 2010 02:44 AM khyratmath123 Calculus 5 January 24th, 2010 06:35 PM khyratmath123 Calculus 4 January 16th, 2010 01:48 AM Spelcy.z Calculus 0 December 31st, 1969 04:00 PM

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