# derivative of an integral

• Sep 19th 2009, 03:54 PM
calcbeg
derivative of an integral
Hi

The goal is the find f'(x) if

f(x) = Integral sign (a= sec x; b= x^2) (1 + 2u^2)^ (1/2) <- that is the square root of (1 + 2u^2)

Hopefully that makes sense in how I am communicating the question.

So I started with letting t= sec x so dx/dt = sec x tan x and I am thinking I need to do the same with the x^2 but I don't know if it should be a different letter and then what do I do???

Any help or direction you can offer is greatly appreciated.

This calculus beginner
• Sep 19th 2009, 04:05 PM
Jester
Quote:

Originally Posted by calcbeg
Hi

The goal is the find f'(x) if

f(x) = Integral sign (a= sec x; b= x^2) (1 + 2u^2)^ (1/2) <- that is the square root of (1 + 2u^2)

Hopefully that makes sense in how I am communicating the question.

So I started with letting t= sec x so dx/dt = sec x tan x and I am thinking I need to do the same with the x^2 but I don't know if it should be a different letter and then what do I do???

Any help or direction you can offer is greatly appreciated.

This calculus beginner

The second fundemental thm of calculus is

if $\displaystyle F(x) = \int_a^x f(t)\, dt$ then $\displaystyle F'(x) = f(x).$

For your problem, split the integral up. i.e.

$\displaystyle \int_{\sec x}^{x^2} \sqrt{1+ 2u^2}\, du = \int_{\sec x}^{a} \sqrt{1+ 2u^2}\, du + \int_{a}^{x^2} \sqrt{1+ 2u^2}\, du$
$\displaystyle = - \int_a^{\sec x} \sqrt{1+ 2u^2}\, du + \int_{a}^{x^2} \sqrt{1+ 2u^2}\, du,$

where $\displaystyle a$ is a constant. Then use the chain rule on each piece.
• Sep 19th 2009, 04:12 PM
redsoxfan325
Quote:

Originally Posted by Danny
The second fundemental thm of calculus is

if $\displaystyle F(x) = \int_a^x f(t)\, dt$ then $\displaystyle F'(x) = f(x).$

For your problem, split the integral up. i.e.

$\displaystyle \int_{\sec x}^{x^2} \sqrt{1+ 2u^2}\, du = \int_{\sec x}^{a} \sqrt{1+ 2u^2}\, du + \int_{a}^{x^2} \sqrt{1+ 2u^2}\, du$
$\displaystyle = - \int_a^{\sec x} \sqrt{1+ 2u^2}\, du + \int_{a}^{x^2} \sqrt{1+ 2u^2}\, du,$

where $\displaystyle a$ is a constant. Then use the chain rule on each piece.

It seems unnecessary to split up the integral. Assume that $\displaystyle \int\sqrt{1+2u^2}\,du = F(u)$. That is to say, $\displaystyle F'(u)=\sqrt{1+2u^2}$.

Thus $\displaystyle \int_{\sec x}^{x^2}\sqrt{1+2u^2}\,du = F(x^2)-F(\sec x)$.

Taking the derivative of this expression (using the chain rule) gives you $\displaystyle \sqrt{1+2(x^2)^2}\cdot2x-\sqrt{1+2(\sec x)^2}\cdot\sec x\tan x$.

Then you can clean that up a bit if you want.
• Sep 19th 2009, 04:17 PM
Jester
Quote:

Originally Posted by redsoxfan325
It seems unnecessary to split up the integral. Assume that $\displaystyle \int\sqrt{1+2u^2}\,du = F(u)$. That is to say, $\displaystyle F'(u)=\sqrt{1+2u^2}$.

Thus $\displaystyle \int_{\sec x}^{x^2}\sqrt{1+2u^2}\,du = F(x^2)-F(\sec x)$.

Taking the derivative of this expression (using the chain rule) gives you $\displaystyle \sqrt{1+2(x^2)^2}\cdot2x-\sqrt{1+2(\sec x)^2}\cdot\sec x\tan x$.

Then you can clean that up a bit if you want.

I believe what you did and what I'm suggesting is equivalent!
• Sep 19th 2009, 04:18 PM
redsoxfan325
There's no question that they give the same answer. It's just that your method involves an extra step - just one more chance to make a mistake.