# Math Help - Integral question

1. ## Integral question

Consider the function .
Let be the antiderivative of with .
Then ?

I'm guessing you have to find the antiderivative, then find C and then evaluate the antiderivative with 4 as your x input.

For the antiderivative I ended up with $-7/2x^2 + 6/7x^7 + 2.64$

However the webwork assignment doesn't accept my final answer of 2.42

Can anyone please tell me where I messed up? Or at least just tell me whether my antiderivative was correct or not.

2. Originally Posted by Archduke01
Or at least just tell me whether my antiderivative was correct or not.
Looks good to me.

3. Even the C value?

Because the webwork assignment considers my answer wrong.

4. I did not check the C value, you should double check that. Leave your answer as an improper fraction.

5. I don't get what I'm doing wrong.

So I plug in 4 for the x values in my antiderivative; . I get 2.42 as my answer - which is wrong.

6. $\int \frac{7}{x^3}-\frac{6}{x^8} ~dx = \frac{-7}{2x^{2}}+\frac{6}{7x^7} +C$

$F(1)= 0 \implies 0 = \frac{-7}{2(1)^{2}}+\frac{6}{7(1)^7} +C$

$0 = \frac{-7}{2}+\frac{6}{7} +C$

$0 = \frac{-49}{14}+\frac{12}{14} +C$

$0 = \frac{-49+12}{14}+ +C$

$0 = \frac{-37}{14}+C$

$C = \frac{37}{14}$

Did this rather quick, might need to check my arithmetic...

$F(x)= \frac{-7}{2x^{2}}+\frac{6}{7x^7} +\frac{37}{14}$

$F(4)= \frac{-7}{2\times 4^{2}}+\frac{6}{7\times 4^7} +\frac{37}{14}=\dots$

7. That's exactly what I did. The webwork doesn't recognize it as correct. Either we both made the same mistake or the question is bugged.

EDIT: Nevermind ... it just needed me to put the answer in 4 decimal places. Jeez, what a waste of time.