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I have the following practice problem that I'd like to know how to solve before taking my test, can someone explain what is necessary?

$D_x \int_{0}^{2x} \left[15 \sqrt{2t^2 + 3t + 4} \right] dt$

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    Do you see what to do if the $2x$ were replaced by an $x$ (use the fundamental theorem of calculus)? Do that, but use the chain rule to account for the $2x$.2012-12-08

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$\frac{d}{dx}\,\int_{a(x)}^{b(x)}f(t)\,dt = f(b(x))\,b'(x) - f(a(x))\,a'(x)$

In this case, $b(x)=2x$ and $a(x) = 0$.

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    Wouldn't that leave us with the answer: $15 \sqrt{2(2t)^2 + 3(2t) + 4} * 2 - 15 \sqrt{2(0)^2 + 3(0) + 4} * 0$?2012-12-09
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You know from the fundamental theorem of calculus that $\frac{d}{du}\int_0^uf(t)~dt=f(u)\;.\tag{1}$ If $u=2x$ and $f(t)=15\sqrt{2t^2+3t+4}$, $(1)$ becomes

$\frac{d}{du}\int_0^u15\sqrt{2t^2+3t+4}~dt=15\sqrt{2u^2+3u+4}=15\sqrt{8x^2+6x+4}\;.$

If $F(x)=\int_0^{2x}15\sqrt{2t^2+3t+4}~dt\;,$

you now know $\dfrac{dF}{du}$; how do you get $\dfrac{dF}{dx}$ from this?

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    I'm not sure I understand. I'm used to doing integrals like so: f(b) * f'(b) - f(a) * f'(a). I got really confused when$u$substitution came into play and what not. Why can't this equation be solved this way? (Every time I read more to understand, the more I get confused)2012-12-09