$\displaystyle \int \frac{{{2x}^2+2}}{x^2}dx $ There's an upper limit of 4 and lower limit of 1.
Follow Math Help Forum on Facebook and Google+
Originally Posted by Lolcats $\displaystyle \int \frac{{{2x}^2+2}}{x^2}dx $ There's an upper limit of 4 and lower limit of 1. Note that $\displaystyle \frac{2x^2+2}{x^2}=2+2x^{-2}$ So, can you find $\displaystyle \int_1^4(2+2x^{-2})dx$ ?
Originally Posted by Lolcats $\displaystyle \int \frac{{{2x}^2+2}}{x^2} $ There's an upper limit of 4 and lower limit of 1. Hello . $\displaystyle \frac{{{2x}^2+2}}{x^2}=\frac{2x^2}{x^2}+\frac{2}{x ^2}$ $\displaystyle =2+2x^{-2}$ Now, $\displaystyle \int_{1}^{4} \frac{{{2x}^2+2}}{x^2} dx$ $\displaystyle =2\int_{1}^{4}dx+2\int_{1}^{4}x^{-2}dx$. Can you complete it ?
Jeez these are easy problems I'm screwing up. Thanks !
Sorry to be a bother, the answer should be 4 shouldn't it?
Originally Posted by Lolcats Sorry to be a bother, the answer should be 4 shouldn't it? No. See this: integrate [ 2 + 2 x^(-2) ] from x=1 to x=4 - Wolfram|Alpha Please show us your work, to see your mistakes.
Originally Posted by Lolcats Sorry to be a bother, the answer should be 4 shouldn't it? no $\displaystyle 15/2$
K well i'd integrate and get 2x-(2/x), then insert the upper limit so: 2(4) + 2/4 = 4 then minus that from 2(1) + 2/1 = 0 Sorry i'm replying slow I have a seperated shoulder atm and can only type with 1 hand.
Originally Posted by Lolcats K well i'd integrate and get 2x-(2/x), then insert the upper limit so: 2(4) + 2/4 = 4 then minus that from 2(1) + 2/1 = 0 Sorry i'm replying slow I have a seperated shoulder atm and can only type with 1 hand. See the red.
Wops.. thats embarrasing. Thanks again.
View Tag Cloud