Thread: Finding the derivative of this integral using part I AND part II of FTC

1. Finding the derivative of this integral using part I AND part II of FTC

I have to find the derivative of the same integral twice, once using each part of the FTC. The integral is from sin(x) to 2 of the function (3^t+7). I think using part 1 it would just be 3^2+7 (but that seems way too simple?) and I don't really know how to do it using part two. I know part two deals with antiderivatives, but I don't know how that helps me here.

2. Originally Posted by fattydq
I have to find the derivative of the same integral twice, once using each part of the FTC. The integral is from sin(x) to 2 of the function (3^t+7).
$\int_{sin(x)}^2 3^t+ 7 dt$? That would be a function of x, not a number. Are you sure you have given the problem correctly?

I think using part 1 it would just be 3^2+7 (but that seems way too simple?) and I don't really know how to do it using part two. I know part two deals with antiderivatives, but I don't know how that helps me here.

3. Of course I gave the problem correctly. It says consider the given function h(x)=integral from sin(x) to 2 of (3^t+7)dt and then #1 is find h prime of x using part 1 of the FTC and #2 is the same just using part II

using part 1 I think it would be -(3^sin(x)+7) times cos(x). Does this sound correct? And how would I go about using part II to find the same answer?

4. Originally Posted by fattydq
using part 1 I think it would be -(3^sin(x)+7) times cos(x). Does this sound correct? Mr F says:Yes.

And how would I go about using part II to find the same answer?
Now calculate the integral and differentiate the result. Note that $\int 3^t \, dt = \int e^{(\ln 3) t} \, dt \, ....$

5. Originally Posted by mr fantastic
Now calculate the integral and differentiate the result. Note that $\int 3^t \, dt = \int e^{(\ln 3) t} \, dt \, ....$
what do you mean by calculate the integral? Just evaluate it?

6. Originally Posted by fattydq
what do you mean by calculate the integral? Just evaluate it?
Well yes. That's why I gave a result that will help you ....

7. I'm sorry I just don't understand. Could you go step by step?

I just don't see how you can evaluate a function with an exponent of sinx when one of the values you have to plug in is sin(x)

8. Originally Posted by fattydq
I'm sorry I just don't understand. Could you go step by step?

I just don't see how you can evaluate a function with an exponent of sinx when one of the values you have to plug in is sin(x)
$\left[ \frac{1}{\ln 3} \, e^{(\ln 3) t} + 7t \right]_{\sin x}^2 = \left[ \frac{1}{\ln 3} \, 3^t + 7t \right]_{\sin x}^2 = \, ....$

where I'm assuming you know what to do with integral terminals.

Now differentiate the result.

9. yes but I'm saying how can I put sin(x) in for t which is an exponent, when I don't know what x is. It's just sin(x) which doesn't help me at all.

It's like you're speaking a foreign language man. I know that in the end my answer will be the same as in part I I just don't understand how to arrive to that, and what you're saying is just jumbling tons of numbers into the mix, I don't understand how that will lead me to the original answer I got when using part I.

10. Originally Posted by mr fantastic
$\left[ \frac{1}{\ln 3} \, e^{(\ln 3) t} + 7t \right]_{\sin x}^2 = \left[ \frac{1}{\ln 3} \, 3^t + 7t \right]_{\sin x}^2 = \, ....$

where I'm assuming you know what to do with integral terminals.

Now differentiate the result.
*Sigh* This is what you should have been taught to do with integral terminals:

$\left( \frac{1}{\ln 3} \, 3^2 + 7(2) \right) - \left( \frac{1}{\ln 3} \, 3^{\sin x} + 7(\sin x) \right)$

Simplify this. Now differentiate with respect to x.

11. Originally Posted by fattydq
... consider the given function h(x)=integral from sin(x) to 2 of (3^t+7)dt and then #1 is find h prime of x using part 1 of the FTC and #2 is the same just using part II
here you go ...

$h(x) = \int_{\sin{x}}^2 3^t + 7 \, dt$

$h(x) = -\int_2^{\sin{x}} 3^t + 7 \, dt$

$h'(x) = -(3^{\sin{x}} + 7) \cdot \cos{x}$

$h'(x) = -3^{\sin{x}}\cos{x} - 7\cos{x}$

$h(x) = \int_{\sin{x}}^2 3^t + 7 \, dt$

$h(x) = \left[\frac{3^t}{\ln{3}} + 7t\right]_{\sin{x}}^2$

$h(x) = \left[\frac{3^2}{\ln{3}} + 7(2)\right] - \left[\frac{3^{\sin{x}}}{\ln{3}} + 7\sin{x}\right]$

$h'(x) = 0 - \left[\frac{3^{\sin{x}}}{\ln{3}} \cdot \cos{x} \cdot \ln{3} + 7\cos{x}\right] = -3^{\sin{x}}\cos{x} - 7\cos{x}$

12. Originally Posted by mr fantastic
*Sigh* This is what you should have been taught to do with integral terminals:

$\left( \frac{1}{\ln 3} \, 3^2 + 7(2) \right) - \left( \frac{1}{\ln 3} \, 3^{\sin x} + 7(\sin x) \right)$

Simplify this. Now differentiate with respect to x.
I appreciate you helping me out man, I'm sorry for being such a pain

I understand that solving that will get me the answer I want but how did you get those terms to begin with?

Imagine I'm starting from the beginning, with the original integral and solving it using part two...

Originally Posted by skeeter
here you go ...

$h(x) = \int_{\sin{x}}^2 3^t + 7 \, dt$

$h(x) = -\int_2^{\sin{x}} 3^t + 7 \, dt$

$h'(x) = -(3^{\sin{x}} + 7) \cdot \cos{x}$

$h'(x) = -3^{\sin{x}}\cos{x} - 7\cos{x}$

$h(x) = \int_{\sin{x}}^2 3^t + 7 \, dt$

This. In between these two steps what are you guys doing to reach this next step?

$h(x) = \left[\frac{3^t}{\ln{3}} + 7t\right]_{\sin{x}}^2$

$h(x) = \left[\frac{3^2}{\ln{3}} + 7(2)\right] - \left[\frac{3^{\sin{x}}}{\ln{3}} + 7\sin{x}\right]$

$h'(x) = 0 - \left[\frac{3^{\sin{x}}}{\ln{3}} \cdot \cos{x} \cdot \ln{3} + 7\cos{x}\right] = -3^{\sin{x}}\cos{x} - 7\cos{x}$
I wrote in your quote box exactly where I get stuck

13. You have remember that the derivative of 3^t= (ln3)3^t so in order to integrate 3^t you need a ln3, thats where they get the antiderivative of 3^t/(ln3). To cancel the ln3 you get when you derive 3^t/(ln3).

14. Originally Posted by aervid
You have remember that the derivative of 3^t= (ln3)3^t so in order to integrate 3^t you need a ln3, thats where they get the antiderivative of 3^t/(ln3). To cancel the ln3 you get when you derive 3^t/(ln3).
wow yet another thing I wasn't aware I should know......calc II is ridiculous lol

15. Originally Posted by fattydq
I appreciate you helping me out man, I'm sorry for being such a pain

I understand that solving that will get me the answer I want but how did you get those terms to begin with?

Imagine I'm starting from the beginning, with the original integral and solving it using part two...

I wrote in your quote box exactly where I get stuck
Both of us are doing nothing more than simple integration (the result I gave you way back is useful for doing this).