# Differentiating $3\int_0^te^uf(u)du$

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 states that $$\frac{d}{dt} \left(3 \int_0^t e^u f(u) du\right) = 3 e^t f(t)$$ – Pragabhava Nov 18 '12 at 22:06

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.
$$\dfrac{d}{dt} \left( \int_0^t F(x) dx\right) = F(t)$$ Hence, the answer is just $$3 e^t f(t)$$
What about evaluating it at 0? Are you not leaving out the term $e^0f(0)$? – csss Nov 18 '12 at 22:09
@css: No, because it does not depend upon $t$, so its derivative is zero. – Ross Millikan Dec 3 '12 at 17:50