Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 5

y=e^x,y=x^2-1 , x=-1 , x=1
Refer the attached image. Graph of e^x is plotted in red color and y=x^2-1 is plotted in blue color.
From the graph ,the region of y=e^x lies above the region of y=x^2-1,
Area of the region enclosed by the given curves A=int_(-1)^1((e^x-(x^2-1))dx
A=int_(-1)^1(e^x-x^2+1)dx
A=[e^x-x^3/3+x]_(-1)^1
A=(e^1-1^3/3+1)-(e^-1-(-1)^3/3-1)
A=(e-1/3+1)-(1/e+1/3-1)
A=e+2/3-1/e+2/3
A=e-1/e+4/3
A~~3.68374