Antiderivative via FTC and Intergration by parts

**Lupus7874** · December 7th 2009, 01:13 AM

Could anyone please help me with this?

Define the function g(x) = anti derivative of t^2 * cos(t) dt
top value = 1/x^2 bottom value = 0

a) solve using FTC
b) solve using intergration by parts

**mr fantastic** · December 7th 2009, 01:21 AM

Originally Posted by Lupus7874

Could anyone please help me with this?

Define the function g(x) = anti derivative of t^2 * cos(t) dt
top value = 1/x^2 bottom value = 0

a) solve using FTC
b) solve using intergration by parts

a) You want $\frac{dg}{dx} = \frac{d}{dx} \int_0^{1/x^2} t^2 \cos t \, dt$ . Let $u = \frac{1}{x^2}$ . Then from the chain rule:

$\frac{dg}{dx} = \frac{dg}{du} \cdot \frac{du}{dx} = \frac{d}{du} \int_0^{u} t^2 \cos t \, dt \cdot \frac{du}{dx}$

and you should be able to do the necessary calculations.

b) Use integration by parts twice. It should be clear what choices to make (Hint: You want to end up integrating only a trig function).

**Lupus7874** · December 7th 2009, 06:03 AM

I'm rather confused on the whole methodology of this problem, any chance someone could walk me through it?

I mean assuming I substitute what you said, I'm still lost on how to take the antiderivative of a product

**mr fantastic** · December 8th 2009, 12:30 AM

Originally Posted by Lupus7874

I'm rather confused on the whole methodology of this problem, any chance someone could walk me through it?

I mean assuming I substitute what you said, I'm still lost on how to take the antiderivative of a product

I told you exactly how to get the solution: $\frac{dg}{dx} = \frac{dg}{du} \cdot \frac{du}{dx} = \frac{d}{du} \int_0^{u} t^2 \cos t \, dt \cdot \frac{du}{dx}$ .

$\frac{d}{du} \int_0^{u} t^2 \cos t \, dt$ is found using the Fundamental Theorem of Calculus (then substitute back $u = \frac{1}{x^2}$ ).

Getting $\frac{du}{dx}$ from $u = \frac{1}{x^2}$ should be trivial at the level you're apparently studying.

I don't see where the trouble can be in this ....

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