Question

Help please! ((:
Integral problem below.

Answer 1

The easiest way is to write the powers of x as negative powers.

\[\frac{1}{x^a} = x^{-a} \]

Then you can use your normal rules for the integral.

Answer 2

After you actually take the integral, you can apply the condition given to you to find the value of the integration constant (the + C at the end).

Answer 3

\[\frac{ -15(x^2-2) }{ x^6 } +C\]

Answer 4

@Vandreigan

Answer 5

I'm not sure what you did there, but be careful with negative exponents.

Answer 6

I integrated the function

Answer 7

\[\int\limits 5x^{-3}-6x^{-5}dx = -\frac{5}{2}x^{-2}+\frac{3}{2}x^-4 = \frac{1}{x^2}(\frac{3}{2x^2}-\frac{5}{2})+C\]

Answer 8

or should I write it like this
\[5x ^{-3}-6x ^{-5}\]

Answer 9

ope. you wrote out what i was about to write out haha

Answer 10

So, once we have that, we know that F(1)=0. So we plug in 1 for x, set it equal to zero, and solve for C.

Then we're done. We know the whole equation.

Answer 11

c=2.5
\[(\frac{ 1 }{ x^2 })((\frac{ 2 }{ 2x^2 })-(\frac{ 5 }{ 2 }))+2.5\]
@Vandreigan

Answer 12

wait that's wrong

Answer 13

C does not equal 2.5

Answer 14

Nope :)

Answer 15

C=1.5?

Answer 16

In your fraction, you have:
\[\frac{2}{2x^2}\]

This should be

\[\frac{3}{2x^2}\]

Answer 17

C=1 !

Answer 18

There we go :)

Answer 19

so it would be
\[\frac{ 1 }{ x^2 }((\frac{ 3 }{ 2x^2})-(\frac{ 5 }{ ?2 }))+1\]

Answer 20

ignore the "?"

Answer 21

That's what I get. Well done :)

Answer 22

It appears to be the wrong answer whn I plug it in :(

Answer 23

Ugh, programs that grade...

Try:

\[-\frac{5}{2x^2}+\frac{3}{2x^4}+1\]

Answer 24

Note that it's the same thing we found, but without re-ordering or factoring.

Answer 25

Oh, I see what you did there... Okay. It worked this time!
Thank you soooo much!

Answer 26

My pleasure :)

Answer 27

Do you work at Chick-fil-A?! haha @Vandreigan

Answer 28

Sorry if that offended you. It's just what all the workers say.. haha