Asked by Cindy
How to do this vertical slicing question: find the area under the curve for the function y=(2x+1)^2 of the interval -1<=x<=3. I am not sure how to do this, but it may involve breaking the integral up. How though?
Answers
Answered by
bobpursley
area= INT f(x)dx
- INT (2x+1)^2 dx
= 1/6 (2x+1)^3 over limits -1 to 3
= 1/6 [ (2*3+1)^3 - (-2+1)^3]
= 1/6 (7^3+1) =344/6
check that
- INT (2x+1)^2 dx
= 1/6 (2x+1)^3 over limits -1 to 3
= 1/6 [ (2*3+1)^3 - (-2+1)^3]
= 1/6 (7^3+1) =344/6
check that