Asked by Hannah
evaluate the integral
integral of 3 to 2 x/(x^2-2)^2 dx
u=x^2-2
du=2x dx
1/2 du = x dx
integral of 1/u^2 du
-1/(x^2-2) Then I plug in 3 and 2 and subtract them form each other
-1/(3^2-2) - (-1/(2^2-2) Is this correct?
integral of 3 to 2 x/(x^2-2)^2 dx
u=x^2-2
du=2x dx
1/2 du = x dx
integral of 1/u^2 du
-1/(x^2-2) Then I plug in 3 and 2 and subtract them form each other
-1/(3^2-2) - (-1/(2^2-2) Is this correct?
Answers
Answered by
MathMate
The (1/2) has been missed out, should read:
∫ du/(<b>2</b>u^2)
=-1/(<b>2</b>(x^2-2))
After that you can plug in the limits of integration. I get 5/28 integrating from 2 to 3.
The bottom limit is the start value, and the top value is the end value.
It is usual to go from 2 to 3, and rarely from 3 to 2. Please check.
∫ du/(<b>2</b>u^2)
=-1/(<b>2</b>(x^2-2))
After that you can plug in the limits of integration. I get 5/28 integrating from 2 to 3.
The bottom limit is the start value, and the top value is the end value.
It is usual to go from 2 to 3, and rarely from 3 to 2. Please check.