Question

let h(x)= the integral of -3 to 4x^2-7 e^(t^2) dt
find H'(2)
I'm not sure what to do here, should I take the derivative of the function first and then integrate or vice versa or do something completely different?

Answer 1

The `Fundamental Theorem of Calculus, Part 1`:\[\large \frac{d}{dx}\int\limits_c^x f(t)\;dt \qquad = \qquad f(x)\]

What happens is, you're first taking the anti-derivative. Then you're taking the derivative of the result. So you're effectively doing something, then "undoing" what you did.

We have to pay attention to what happens `in between` these step.
See how the `limits of integration` involve x?
It will change the variable of our function.

Answer 2

\[\large h(x)=\int\limits_{-3}^{4x^2-7}e^{t^2}\;dt\]

Let's simplify this down a second so we can see what's happening.\[\large h(x)=\int\limits\limits_{-3}^{4x^2-7}f(t)\;dt\]Taking the anti-derivative,\[\large h(x)=F(t)|_{-3}^{4x^2-7}\]\[\large h(x)=F(4x^2-7)-F(-3)\]

Taking the derivative,\[\large h'(x)=f(4x^2-7)\color{royalblue}{(4x^2-7)'}\]

Understand where the blue part comes from?
We have to apply the chain rule when differentiating.

Answer 3

The process is a little confusing if you haven't seen it before. Maybe a simpler example is needed? :O

Answer 4

I see what you're saying up until the very end, it looks like f(-3) just goes away? and is the x value of the funtion both the black and blue equations

Answer 5

F(-3) is a constant. S it's derivative gives us zero.

Example, if we integrated some function and got \(\large t^2\),
Evaluating that function at the lower limit, t=-3, gives us, \(\large (-3)^2\).
It's just a constant right? :) It's derivative will be zero.

I should have written -0 in that last step perhaps. heh

Answer 6

The new value of the function is just the black part. The blue is being multiplied on the outside, due to the chain rule.

Answer 7

Err I should be clear what I'm saying, the new value of \(\large f\) is the black part.

Answer 8

\[\huge f(\color{orangered}{t})\qquad = \qquad e^{\color{orangered}{t}^2}\]

We now have, \[\huge f(\color{#CC0033}{4x^2-7}) \qquad = \qquad e^{\color{#CC0033}{4x^2-7}^2}\]

Understand how the \(\large f\) function is changing?
This problem really deserves an example :) lol ... tough one to explain.

Answer 9

woops, i should have put brackets around the new Red exponent. My bad.

Answer 10

okay and after you put in your x value (2 in this case) you would have your answer? or do you multiply what you get by the blue function? or does that just go away?

Answer 11

I was kinda all over the place, lemme simplify it down a sec.

Answer 12

\[\large h(x)=\int\limits\limits_{-3}^{4x^2-7}e^{t^2}\;dt\]
\[\large h'(x)=e^{(4x^2-7)^2}(4x^2-7)'\]

The variable changes due to the process we took, anti-differentiating, then plugging in our limits, and finally undoing the anti-differentiation by taking the derivative.
So we end up with the same function we started with, but with respect to x now.

When we took the derivative, we have to apply the chain rule. Multiplying by the derivative of the exponent. That's why that prime is there, we still need the derivative of that part.

Answer 13

So if I'm understanding you correctly, the solution of this problem would be 16e^81?

Answer 14

Yes that sounds right! :)
Do you have an answer key we can check that against or no? :D

Answer 15

I don't but I'm going to hope that's correct! Thank you so much that was a wonderful explanation!

Answer 16

There is one part in the middle I'm a little unsure about.
But I'm pretty sure it's correct.
Imma see if I can throw it into wolfram or something maybe :o

Answer 17

Ok good same answer :) Yay team!
http://www.wolframalpha.com/input/?i=%28d%2Fdx%29+integral_%7B-3%7D%5E%7B4x%5E2-7%7D%28e%5E%28%28t%29%5E2%29%29dt

Answer 18

great, thanks again!