Question

FTC question

Answer 1

\[\frac{ d(\int\limits_{3sinx}^{2015}t^2\cos(t) dt) }{ dx }\]
how do i deal with the 3sinx?
i know that FTC can just turn the function into a function of x but how do i apporach this question?

Answer 2

ok apparently i really use substitution but why is there a need to mutiply the integral by the derivative of the substitution, e.g. 3sinx

Answer 3

ummmmmm.......................
idk
im sorrrrrryyyyyyyyyyyyyy

Answer 4

What I like to do is pretend I've already evaluated the integral like this:
\[F(t) = \int t^2\cos(t) dt\]
Then I plug in my bounds:
\[F(2015)-F(3 \sin x) = \int\limits_{3 \sin x}^{2015}t^2\cos(t) dt\]

Now if we throw that derivative back in there, it's a little clearer how to use the chain rule:

\[\frac{d}{dx} \left( \int\limits_{3 \sin x}^{2015}t^2\cos(t) dt \right) = \frac{d}{dx}( F(2015)-F(3 \sin x) )\]

When we evaluate this: \[\frac{d}{dx}( F(2015)-F(3 \sin x) )\] we know F(2015) is a constant so it goes away, and so we are just differentiating the other term really, so by the chain rule we have:
\[\frac{d}{dx} -F(3 \sin x ) = - F'(3 \sin x) * 3 \cos x \]

But now remember F was an integral so F' is the derivative of an integral, so we can replace this with \(F'(t) = t^2 \cos t \)
\[- F'(3 \sin x) * 3 \cos x = -(3 \sin x)^2 \cos( 3 \sin x) *3 \cos x \]

Answer 5

i think that answers my question. thanks

Answer 6

Cool, yeah there are multiple ways to do these problems but I find this is the easiest way.