Question

Evaluate the integral.

Answer 1

\[\int\limits_{1}^{9}(\frac{ 8 }{ x ^{2} }+3x)dx\]

Answer 2

Start with the integration

Answer 3

It just might be simpler to understand the solution of this problem if you separate it into two different definite integrals: (1) the integral from 1 to 0 of 8/(x^2) and (2) the integral from 1 to 0 of 3x, in both cases integrating with respect to x.

Answer 4

not sure how to find the anti derivative of 8/x^2 @Luigi0210

Answer 5

The first integral could be re-written as 8 times the integral from 1 to 0 of x^(-2). Note that the negative exponent here stems from moving the x^2 in the denominator to the numerator.

Answer 6

So, arshia93, can you integrate x^(-2) with respect to x? Try it.

Answer 7

Hint: Use the power rule for integration.

Answer 8

@mathmale you realize its 9 and 1 as the intefrals right? how did you get 1 to 0. And im not sure how you got x^(-2)..

Answer 9

Thanks for the clarification. Integrating from 1 to 0, though possible, is odd. We will integrate from 1 to 9, as you point out.

Answer 10

I know how to do this problem the thing thats tricky for me is finding the anti derivatives

Answer 11

i think lol

Answer 12

Your integrand involves 8/(x^2). That x^2 is in the denominator. To apply the power rule for integration, that x^n must be moved to the numerator, and in doing so we must put a negative sign in front of the 2: 8/(x^2) becomes 8x^(-2).

Answer 13

That's one of the rules of exponents from algebra. Clear on this?

Answer 14

Thanks, Euler271. Your formula is correct (although it'd be better if you included the "+C).
And you are correct in rewriting 8/x^2 as 8*x^-2.

Answer 15

Arshia, can you now apply the power rule for integration to 8*x^-2?

Answer 16

@mathmale yes i get it, so 8(9)^-2 minus the antideriv of 3x with (1) plugged in?

Answer 17

wait thats not the full anti deriv? -16x^-3??

Answer 18

arshia93, you need to integrate 8*x^-2 with respect to x BEFORE you substitute the x values 1 and 9.
Pleaes try again.

Answer 19

@mathmale dude honetly the best way for me to learn math is by examples, i appreciate the help but please go a little further

Answer 20

By the power rule for integration, the integral of 8*x^-2 is -8*x^-1. (If necessary, ask for clarification on this.)
Evaluating this result at the limits 9 and 1 results in:
-8*[9^-1 - 1^-1], which simplifies to -8*[(1/9) - (1/1)].
Hope this helps. I need to get offline now, but am sure others can help you finish this problem if need be. Good luck!

Answer 21

@mathmale thanks for your help mathmale

Answer 22

\[\Large \bf\int\limits\limits_{1}^{9}\left(\frac{ 8 }{ x ^{2} }+3x\right)\;dx\quad=\quad \left(\frac{-8}{x}+\frac{3}{2}x^2\right)_1^9\]
What part are you confused on arshia, do you understand how to find the anti-derivative of those two terms as shown?

Answer 23

got it @zepdrix using the equation to show it on openstudy helps alot, i just didnt know how to get \[\frac{ 8 }{ x ^{2} }\to \frac{ -8 }{ x }\]

Answer 24

We can bring the x up from the denominator using rules of exponents,
and then apply the power rule for integration as we normally would.
\[\Large\bf \int\limits \frac{1}{x^n}\;dx \quad=\quad \int\limits x^{-n}\;dx\quad=\quad \frac{1}{-n+1}x^{-n+1}\]

Answer 25

pluggin in 9 to -8/x etc it getting really messy, is it supposed to be -8/9+121.5?

Answer 26

Hmm let's see.\[\large \bf\left(\frac{-8}{x}+\frac{3}{2}x^2\right)_{\color{royalblue}{1}}^{\color{orangered}{9}}\quad=\quad \left(\frac{-8}{\color{orangered}{9}}+\frac{3}{2}\color{orangered}{9}^2\right)-\left(\frac{-8}{\color{royalblue}{1}}+\frac{3}{2}\color{royalblue}{1}^2\right)\]

Answer 27

Does the plugging in make sense?
I have we have some simplification after that.

Answer 28

I guess we have* blah typo

Answer 29

yeah it makes sense but its soo messy lol i get 1085.5/9 on the left and -13/2 on the left..

Answer 30

I wouldn't simplify left and right separately.
See how in the second slots, each set of brackets has a common denominator of a 2?
Might be easier to combine them that way.

\[\Large\bf \frac{3\cdot81}{2}-\frac{3}{2}-\frac{8}{9}+8\quad=\quad \frac{240}{2}-\frac{8}{9}+8\]\[\Large\bf 120-\frac{8}{9}+8\quad=\quad 128-\frac{8}{9}\quad=\quad \frac{1072}{9}\]

Answer 31

Hmm it's supposed to be 1144/9 I made a booboo somewhere.

Answer 32

yeah this one was ugly for sure

Answer 33

So your answer worked out to 114.1 and here a link to the correct answer.
Very close, probably just some little fraction math to watch for.
http://www.wolframalpha.com/input/?i=integral+%288%2Fx%5E2%29%2B3x+from+x%3D1+to+x%3D9
If using decimals and all that nonsense makes more sense to you, then by all means take that approach hehe ^^