What would be the rule for something like this? $\displaystyle \[\int_4^6 \left(\frac{d}{dt}\sqrt{4 + 4 t^4}\right)\, dt\]$
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You can take the derivative of $\displaystyle \sqrt{4 + 4 t^4}$ and integrate the result?
$\displaystyle \displaystyle \int_a^bf'(x) dx = f(b)-f(a).$
wait..I got it. At least I think I got it. $\displaystyle \sqrt{4}\int_{4}^{6}\frac{1}{(1+0)}t^{1+0}+\sqrt{4 }\int_{4}^{6}\frac{1}{(1+4)}t^{4+1}$ Nevermind I got it from the above post.
Originally Posted by TheCoffeeMachine $\displaystyle \displaystyle \int_a^bf'(x) dx = f(b)-f(a).$ To make it crystal clear for the OP, I'd state it as $\displaystyle \displaystyle \int_a^b \frac{df}{dx} \, dx = f(b)-f(a)$.
$\displaystyle \int_{ 2 } ^ { 7 } (2 x + 8) dx$ I think something like this is what I take the anti-derivative of. Yeah I got it now, I just now have to deal with a long tedious task of entering in the answers. This is a lot easier than reimann sums
The "chain rule for integration" (i.e. the inverse of the chain rule) is "substitution".