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

• Feb 12th 2009, 04:17 PM
fattydq
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.
• Feb 12th 2009, 04:37 PM
HallsofIvy
Quote:

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?

Quote:

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.
• Feb 12th 2009, 04:42 PM
fattydq
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?
• Feb 12th 2009, 04:51 PM
mr fantastic
Quote:

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 $\int 3^t \, dt = \int e^{(\ln 3) t} \, dt \, ....$
• Feb 12th 2009, 04:57 PM
fattydq
Quote:

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?
• Feb 12th 2009, 05:00 PM
mr fantastic
Quote:

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 ....
• Feb 12th 2009, 05:01 PM
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)
• Feb 12th 2009, 05:21 PM
mr fantastic
Quote:

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.
• Feb 12th 2009, 05:24 PM
fattydq
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.
• Feb 12th 2009, 05:30 PM
mr fantastic
Quote:

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.
• Feb 12th 2009, 05:40 PM
skeeter
Quote:

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 ...

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

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

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

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

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

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

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

$h'(x) = 0 - \left[\frac{3^{\sin{x}}}{\ln{3}} \cdot \cos{x} \cdot \ln{3} + 7\cos{x}\right] = -3^{\sin{x}}\cos{x} - 7\cos{x}$
• Feb 12th 2009, 05:42 PM
fattydq
Quote:

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 (Talking)

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...

Quote:

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
• Feb 12th 2009, 05:55 PM
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).
• Feb 13th 2009, 04:26 AM
fattydq
Quote:

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...(Angry)...calc II is ridiculous lol
• Feb 13th 2009, 04:27 AM
mr fantastic
Quote:

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

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).