Finding F(x) (INTEGRAL)

**Archduke01** · February 22nd 2010, 07:36 PM

Consider the function

.
Let

be the antiderivative of

with

.
Then

?

I don't understand what it's asking. Do I simply find the antiderivative?

**Prove It** · February 22nd 2010, 07:38 PM

Originally Posted by Archduke01

Consider the function

.
Let

be the antiderivative of

with

.
Then

?

I don't understand what it's asking. Do I simply find the antiderivative?

Yes, $F(x)$ is the antiderivative of $f(x)$ .

And then you use the boundary condition to evaluate the integration constant.

**Archduke01** · February 22nd 2010, 07:41 PM

I'm doing something wrong. I get $-6 x^{-2} /2 + 2 x^{-6} /6$ .

Procedure;

.

I separated the 6 and 2, then brought the denominators up getting $6 \int x^{-3} - 2 \int x^{-7}$ . Do I apply the power rule at this point?

**Prove It** · February 22nd 2010, 07:44 PM

Originally Posted by Archduke01

I'm doing something wrong. I get $-6 x^{-2} /2 + 2 x^{-6} /6$ .

Procedure;

.

I separated the 6 and 2, then brought the denominators up getting $6 \int x^{-3} - 2 \int x^{-7}$ . Do I apply the power rule at this point?

Your writing of integrals needs work. You can't leave off the $dx$ 's.

It should read $6\int{x^{-3}\,dx} - 2\int{x^{-7}\,dx}$ .

But yes, now apply the power rule.

**Archduke01** · February 22nd 2010, 07:48 PM

I did apply the power rule, and the original answer is the answer I got. It's not correct though.

**Prove It** · February 22nd 2010, 07:56 PM

Originally Posted by Prove It

Your writing of integrals needs work. You can't leave off the $dx$ 's.

It should read $6\int{x^{-3}\,dx} - 2\int{x^{-7}\,dx}$ .

But yes, now apply the power rule.

$6\int{x^{-3}\,dx} - 2\int{x^{-7}\,dx} = 6\left(-\frac{1}{2}x^{-2}\right) - 2\left(-\frac{1}{6}x^{-6}\right) + C$

$= -\frac{3}{x^2} - \frac{1}{3x^6} + C$ .

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