since the integrand is an even function, i.e. ...
I have a real problem here. It's from Stewart's Early Transcendental's 5th Edition. Section 8.3 problem 33. I'm studying from this book as a hobby. I'm just trying to solve for M sub x. The formula is density times the integral from -1 to +1 of the function 1/2(2-2x)^2 dx. The solutions manual acts like it's an even symmetrical function and converts it to
2*density integral from 0 to +1 of the same function. I cant see how that function is an even function. But even if I try to solve it without calling it an even function I cant match the book's answer. I cant find this problem anywhere on the web and I dont know how to draw it here. I'm hoping that someone out there has this book and understands this problem. Thanks
Thanks Skeeter so it's the absolute value sign that causes that function to become even? I dont think I've ever seen an absolute value sign in the integrand. The student guide never mentions the abolute value sign but I remember from the discussion on cusps that you need them. Thanks a lot I feel better now!