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December 13th, 2017, 10:59 AM   #1
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Integration question

Calculate G'(x) given that:

\[g(x)=\int_{x}^{x^2}f(t)dt\] where f(t) is differentiable.
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December 13th, 2017, 12:09 PM   #2
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Quote:
Originally Posted by Jaket1 View Post
Calculate G'(x) given that:

\[g(x)=\int_{x}^{x^2}f(t)dt\] where f(t) is differentiable.
I assume you mean $g'(x)$ ...

$\displaystyle \dfrac{d}{dx} \bigg[\int_v^u f(t) \, dt \bigg] = f(u) \cdot \dfrac{du}{dx} - f(v) \cdot \dfrac{dv}{dx}$
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December 13th, 2017, 01:10 PM   #3
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Let F(t) be the indefinite integral of f(t). $\displaystyle g(x)=F(x^2)-F(x).\ g'(x)=(2x-1)f(x)$.
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December 13th, 2017, 02:51 PM   #4
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Not quite... $\displaystyle g'(x) = 2x\hspace{1px}f\!\left(x^2\right) - f(x)$.
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December 14th, 2017, 01:40 PM   #5
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Quote:
Originally Posted by skipjack View Post
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.
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December 16th, 2017, 02:50 AM   #6
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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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