posted by Anonymous .
Solve for x: S(0 to x) (t^3-2t+3)dt=4
I got a different answer from the back of the book. The back of the book said that x=1.63052 or x=3.09131. How am I suppose to solve this using a calculator?
I integrated, substituted the x, and simplified to get the following equation
x^4 - 4x^2 + 12x - 4 = 0
I then used Newton's Method and after 5 iterations I get x= 1.63052
So it appears the book is right.
I am not familar with that method. What is it that I do?
- calculus - drwls