December 13th, 2017, 11:59 AM  #1 
Senior Member Joined: Jan 2016 From: Blackpool Posts: 100 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, 01:09 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449  
December 13th, 2017, 02:10 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,660 Thanks: 647 
Let F(t) be the indefinite integral of f(t). $\displaystyle g(x)=F(x^2)F(x).\ g'(x)=(2x1)f(x)$.

December 13th, 2017, 03:51 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,089 Thanks: 1902 
Not quite... $\displaystyle g'(x) = 2x\hspace{1px}f\!\left(x^2\right)  f(x)$.

December 14th, 2017, 02:40 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,660 Thanks: 647  
December 16th, 2017, 03:50 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
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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