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How can I differentiate this function - $3\int_0^te^uf(u)du$

Does differentiation cancel out the integration? And do I then have

$3(e^tf(t) - e^0f(0))$

I am not sure if that is right and I'm not sure of the step by step process for 'cancelling' the integration?

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    The [Fundamental Theorem of Calculus](http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part) states that $ \frac{d}{dt} \left(3 \int_0^t e^u f(u) du\right) = 3 e^t f(t)$2012-11-18

2 Answers 2

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In general $\frac{d}{{dt}}\int_{u(t)}^{v(t)}f(x,t)dx = \int_{u(t)}^{v(t)}{\frac{\partial}{{\partial t}}f(x,t)dx + f(v(t),t)v'(t)} - f(u(t),t)u'(t).$ When $u$ and $v$ are constants, this simplifies to $\frac{d}{{dt}}\int_{u}^{v}f(x,t)dx = \int_u^v{\frac{\partial}{{\partial t}}f(x,t)dx}.$ This very powerful rule is known as Leibniz rule. This is how use differentiate integrals. Hope this helps.

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$\dfrac{d}{dt} \left( \int_0^t F(x) dx\right) = F(t)$ Hence, the answer is just $3 e^t f(t)$

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    @css: No, because it does not depend upon $t$, so its derivative is zero.2012-12-03