# simple derivative but confusing me

• Nov 10th 2008, 10:30 PM
tsal15
simple derivative but confusing me
without evaluating the integral, is the derivative of the function $\int_{2}^{x} e^{t^{2}-2} dt$ just $e^{t^{2}-2}$?

Thanks in advance for any assistance recieved :)
• Nov 10th 2008, 10:37 PM
ajj86
Response
Quote:

Originally Posted by tsal15
without evaluating the integral, is the derivative of the function $\int_{2}^{x} e^{t^{2}-2} dt$ just $e^{t^{2}-2}$?

Thanks in advance for any assistance recieved :)

You have the general form correct, but you would have to evaluate the function as follows:

e^(x^2 - 2)
• Nov 10th 2008, 10:47 PM
tsal15
Quote:

Originally Posted by ajj86
You have the general form correct, but you would have to evaluate the function at its endpoints.

So, it would be F(x) - F(2), or

e^(x^2 - 2) - e^(2^2 - 2)
e^(x^2 - 2) - e^2

Hey ajj86,

First I gotta say, you've been very helpful within the period of 20mins, more than some others on this forum have been in 20days.

MMM, I believe that you are implying the fundamental theorm of calculus whilst using antidifferentiation techniques aka finding the area under graph aka evaluating the integral...? I'm aware it would be a more perfect answer (can u get more perfect than perfect?), but the question strictly states that i must: "Without evaluating the integral, find F'(x) where F(x) = $\int_{2}^{x} e^{t^2-2} dt$" so having said that would $e^{t^2-2}$ be the answer?

Thank you

tsal15
• Nov 10th 2008, 10:52 PM
ajj86
Response
I just found this:

But, I'm not really sure as to whether the answer I gave is correct. Let me do a little more investigating.
• Nov 10th 2008, 11:01 PM
tsal15
Quote:

Originally Posted by ajj86
I just found this:

But, I'm not really sure as to whether the answer I gave is correct. Let me do a little more investigating.

ok. no problems. I'll wait anxiously (Nod)
• Nov 10th 2008, 11:03 PM
ajj86
Response
According to the form given, I'm thinking the answer would be:

e^(x^2 - 2)
• Nov 10th 2008, 11:06 PM
tsal15
Quote:

Originally Posted by ajj86
According to the form given, I'm thinking the answer would be:

e^(x^2 - 2)

ok. 2 things

1. Did u evaluate the integral or how did u do this?

2. why is it not $e^{t^2-2}$
• Nov 11th 2008, 07:05 AM
ajj86
Response
Tsal15, I apologize for not replying last night. I'm not trying to take the easy way out, but I think this might give a better explanation than I can give:

Calculus Facts: Derivative of an Integral

Let me know what you think.
• Nov 11th 2008, 07:25 AM
Mathstud28
This is just a simple application of the Second Fundamental Theorem of Calculus

$\frac{d}{dx}\int_a^{g(x)}f(t)dt=f(g(x))\cdot{g'(x) }$

The proof of this is most likely beyond the scope of your class.
• Nov 12th 2008, 05:04 AM
tsal15
Quote:

Originally Posted by Mathstud28
This is just a simple application of the Second Fundamental Theorem of Calculus

$\frac{d}{dx}\int_a^{g(x)}f(t)dt=f(g(x))\cdot{g'(x) }$

The proof of this is most likely beyond the scope of your class.

I'm sure of it :) but could you show me anyways... I'd like to learn something outside the class rather than just from the tutor... thanks
• Nov 12th 2008, 05:19 AM
tsal15
Quote:

Originally Posted by ajj86
Tsal15, I apologize for not replying last night. I'm not trying to take the easy way out, but I think this might give a better explanation than I can give:

Calculus Facts: Derivative of an Integral

Let me know what you think.

Thanks that was quite helpful. I'll be sure to recommend you to my girlfriends...i think they need more help than i do hehehehe

well, thanks again, ajj86 :)
• Nov 12th 2008, 02:18 PM
Mathstud28
Quote:

Originally Posted by tsal15
I'm sure of it :) but could you show me anyways... I'd like to learn something outside the class rather than just from the tutor... thanks

Wikipedia actually has a pretty non-analysis friendly version. Look here

Fundamental theorem of calculus - Wikipedia, the free encyclopedia