Question

Help, calculus

Answer 1

What would this simplify to?\[\Large \frac{ d }{ dx }( \int\limits_{\cos(66x)}^{1223}t^2+\ln t~dt )\]

Answer 2

This is sort of an amazing result. Think about how you evaluate ANY definite integral. What if those weird limits wire jsut simple numbers? What would you do?

1) Find the antiderivative.
2) Evaluate at the limits.

3) Deal with the pesky additional derivative.

If \(\dfrac{d}{dt}H(t) = h(t)\), then

Fundamental theorem: \(\dfrac{d}{dx}\left(\int_{f(x)}^{g(x)}h(t)\;dt\right) = \dfrac{d}{dx}\left(H(t)\mid_{f(x)}^{g(x)}\right) = \)

\( = \dfrac{d}{dx}\left[H(g(x)) - H(f(x))\right]= h(g(x)) - h(f(x))\)

Look at it until it soaks in. It's pretty much what you have been doing with integrals all along.

Answer 3

Small correction on that last line:\[= \dfrac{d}{dx}\left[H(g(x)) - H(f(x))\right]= h(g(x))g'(x) - h(f(x))f'(x)\]
What do you think Steve?
Still confused on this one?

Answer 4

Whoops, I did forget the Chain Rule, didn't I?! I was tired of coding.

The REAL point of the demonstration was H(t). We don't have to find it. We don't really even care if it exists. Just play like it's there and move right to the end! (as correctly noted by zepdrix).

Answer 5

Suppose that \(\color{black}{\displaystyle \int f(x)\mathrm{d}x=F(x)+C }\).
Then, \(\color{black}{\displaystyle \left.\int_{b}^{g(t)} f(x)\mathrm{d}x=F(x)\right|_{x=b}^{x=g(t)}=F(g(t))-F(b) }\).

What is the derivative of \(\color{black}{\displaystyle F(g(t))-F(b) }\)?
By chain rule, knowing that \(\color{black}{\displaystyle F'(x)=f(x)}\), it should be \(\color{black}{\displaystyle f'(g(t))\times g'(t) }\).

So, you have \(\color{black}{\displaystyle \frac{d}{dt}\left(\int_{b}^{g(t)} f(x)\mathrm{d}x\right)=f'(g(t))\times g'(t) }\).

Answer 6

Now, suppose that both limits were functions of t, then by the same logic you have:

\(\color{black}{\displaystyle \frac{d}{dt}\left(\int_{k(t)}^{p(t)} f(x)\mathrm{d}x\right)=f'(p(t))\times p'(t) -f'(k(t))\times k'(t) }\)

Answer 7

Or, if your limits of integration are flipped as follows:

\(\color{black}{\displaystyle \frac{d}{dt}\left(\int_{g(t)}^{b} f(x)\mathrm{d}x\right)=\frac{d}{dt}\left(-\int_{b}^{g(t)} f(x)\mathrm{d}x\right) =-f'(g(t))g'(t)}\)

Answer 8

this is really not that complicated

Answer 9

first change \[( \int\limits_{\cos(66x)}^{1223}t^2+\ln t~dt )\] to \[-\int\limits_{1223)}^{\cos(66x)}t^2+\ln t~dt \]

Answer 10

then, where you see a \(t\) put \(\cos(66x)\) and finally multiply by the derivative of \(\cos(66x)\)

Answer 11

as we say in math english, the derivative of the integral is the integrand
aka the fundamental theorem of calculus

Answer 12

Thank you all who replied! I think I understand it a little more now. I have a question: so the b value of the integral can be completely ignored since it's a constant?

Answer 13

yes, its a constant so if you plug in and take derivative it turns to 0

Answer 14

Yes, because with b in one of the limits, you are differentiating F(b) (which is a constant) with respect to t, and derivative of a constant is ....

Answer 15

zero! Ah thank you.

Answer 16

yw

Answer 17

Basically, the point here (as other people expressed before me) is (1) you really don't care what the integral of this function would be, (2) the chain rule for whatever function is in the limits of the integral.