Question

help integrating exp functions. medal and fan

Answer 1

You're not actually doing any integrating here. This is an application of the fundamental theorem of calculus, which says
\[\large \frac{d}{dx}\int_c^{g(x)}f(t)~dt=f(g(x))\cdot g'(x)\]
where \(c\) is a constant.

Answer 2

So if
\[\large y(x)=\int_0^{\ln3x}\sin\left(2e^t\right)~dt\]
then you have \(c=0\), \(g(x)=\ln3x\), and \(f(t)=\sin\left(2e^t\right)\).

Answer 3

ok

Answer 4

where does y'(x) com into play?

Answer 5

\(y'(x)\) is the derivative of \(y(x)\) with respect to \(x\). You're asked to find the rate of change of a definite integral, which represents area, so you're finding the rate of change of the area under a curve.

Answer 6

ok i found the definite integral sin(2 e^t) log(3 x) the i tried to find a derivative (sin(2 e^t))/x

it doesn't work

Answer 7

You found the definite integral....?
No no that's not what you're supposed to do :O

I'll repeat what Sith said.

You assume some anti-derivative exists, call it F(t):\[\Large\rm \int\limits_0^{\ln3} \sin(2e^t)dt=F(t)_0^{\ln3x}=F(\ln 3x)-F(0)\]Then after plugging in the limits as we've done here ^
you take a derivative.

\[\Large\rm f(\ln3x)\frac{d}{dx}(\ln 3x)-\cancel{\frac{d}{dx}F(0)}\]That second term is just a constant, so it's derivative is zero.
The first term, differentiating brings us back to little f,
but now we have to apply the chain rule because of our inner function.

Answer 8

i see

Answer 9

\[\Large\rm f(\color{orangered}{t})=\sin(2e^{\color{orangered}{t}})\]\[\Large\rm f(\color{orangered}{\ln3x})=\sin(2e^{\color{orangered}{\ln3x}})\]^ Which can be simplified a little bit.
And also the chain rule is giving us a \(\Large\rm \frac{3}{x}\) right?
Errr 1/x I think, like you had kind of posted.

Answer 10

thank you

Answer 11

ok so i have the answer but when do i know when use a definite integral or not?

Answer 12

When should you look for an actual anti-derivative? Do the integration?
When you're not taking the `derivative` of an `integral`.

When you integrate, then take a derivative,
you're doing something, then you're undoing it.
So it brings you back where you started.
(But something happens in the middle, sticking in limits).

So whenever you're taking the derivative of an integral, you can fall back on this identity.
The FTC 1 (Fundamental Theorem of Calculus, Part 1).

Answer 13

yeah that makes sense. i kind of get lost in this calc 2 material so basic theorems start to become unclear

Answer 14

Yah FTC 1 is a weird one at first.
Takes a little while to get that spark of intuitive with what's going on.