Evaluate the integral ∫ |x|dx limit from -1 to 1?

1 Answer

11|x|dx=1

Explanation:

You define thef(x)=|x|, which is same as defining
f(x)=x,x<0 and f(x)=x,x0 at the same time
So you integral is 11f(x)dx=01xdx+10xdx
=[x22]01+[x22]10=0(12)+120=1