Derivative of the product of 2 definite integrals

[Dba18]

Homework Statement

Find f'(x) for (integral from 0 to x of cos^5(t)dt)* (integral from x^2 to 1 of e^t^2 dt). No differentiation allowed in the answer

Homework Equations

The Attempt at a Solution

. I used the product rule and integrated then differentiated the first term --> cos^5x* integral of e, etc. but I'm stumped on what to do with the these one term. Obviously I leave it as is for the first part of the product rule. So it's the second half where I'm looking for the differential of the integral of e^t^2

 

[LCKurtz]

Homework Statement

Find f'(x) for (integral from 0 to x of cos^5(t)dt)* (integral from x^2 to 1 of e^t^2 dt). No differentiation allowed in the answer

Homework Equations

The Attempt at a Solution

. I used the product rule and integrated then differentiated the first term --> cos^5x* integral of e, etc. but I'm stumped on what to do with the these one term. Obviously I leave it as is for the first part of the product rule. So it's the second half where I'm looking for the differential of the integral of e^t^2

You don't have to integrate then differentiate. Use the fundamental theorem of calculus:$$
\frac d {dx}\int_a^x f(t)\, dt = f(x)$$

 

[Dba18]

I apologize for being so dense, but how does that apply to the stated answer for the quiz of cos^5(x) * integral with e + integral with cos^5(t) * (-2x)e^x^4 it's the answer for the last of the four terms in the result that's throwing me. To me, it seems the derivative of the integral of e^t^2 should just be e^x^2.

Thanks,

 

[LCKurtz]

Homework Statement

Find f'(x) for (integral from 0 to x of cos^5(t)dt)* (integral from x^2 to 1 of e^t^2 dt). No differentiation allowed in the answer

Homework Equations

The Attempt at a Solution

. I used the product rule and integrated then differentiated the first term --> cos^5x* integral of e, etc. but I'm stumped on what to do with the these one term. Obviously I leave it as is for the first part of the product rule. So it's the second half where I'm looking for the differential of the integral of e^t^2

I apologize for being so dense, but how does that apply to the stated answer for the quiz of cos^5(x) * integral with e + integral with cos^5(t) * (-2x)e^x^4 it's the answer for the last of the four terms in the result that's throwing me. To me, it seems the derivative of the integral of e^t^2 should just be e^x^2.

Thanks,

You have given me more information now, so I can see what your problem is. You need to use the chain rule:$$
\frac d {dx}\int_{x^2}^1 e^{t^2}\ dt = -\frac d {dx}\int_{1}^{x^2} e^{t^2}\ dt$$
You plug in the $x^2$ for the $t$ in $e^{t^2}$ but you must multiply by the derivative of $x^2$ using the chain rule. Some versions of Leibnitz rule show this.

 

[Dba18]

$$( \int {cos^t dt})$$ *($$\int{ e^t^2 dt) }$$. I am trying out the text editor I just found on the site to see if I can more clearly write oute the problem. So I'll post to see if this comes out ok

 

[Dba18]

Well my first attempts at using the editor aren't so good. I'm going to go take the laundry out of the dryer and sit and play with your answer for a While to make sure I know what's going on. And the reason you inverted the integral is because it went from a higher number to a lower one, right?

 

[eumyang]

$$( \int {cos^t dt})$$ *($$\int{ e^t^2 dt) }$$. I am trying out the text editor I just found on the site to see if I can more clearly write oute the problem. So I'll post to see if this comes out ok

You're mixing non-LaTeX symbols with the LaTeX commands. Just use one set of tex tags.
What you want to type between the tex tags is this:
\left( \int_{0}^{x} \cos^5 t\ dt \right) \cdot \left( \int_{x^2}^{1} e^{t^2}\ dt \right)
Which gives you this:
$$\left( \int_{0}^{x} \cos^5 t\ dt \right) \cdot \left( \int_{x^2}^{1} e^{t^2}\ dt \right)$$

 

[SammyS]

Dba18,

Here's how I go about remembering what to do with something like $\displaystyle \frac{d}{dx} \left( \int_{x^2}^{1} e^{t^2}\ dt \right)\ .$

Let G(t) be the anti derivative of $\displaystyle e^{t^2}\,,$ so that $\displaystyle G\,'(t)=e^{t^2}\,,$

Then $\displaystyle \int_{x^2}^{1} e^{t^2}\ dt =<br /> G(1)-G(x^2)\ .$

Therefore, $\displaystyle \frac{d}{dx} \left( \int_{x^2}^{1} e^{t^2}\ dt \right)=<br /> \frac{d}{dx}\left(G(1)-G(x^2)\right)=\underline{\ \ \ ?\ \ \ } \ .$