Symmetrical Areas
Key Questions

Well, we need to be careful when you say "under the graph" since
#f(x)=x^3# goes below the xaxis when#x<0# , but you meant the region between the graph and the xaxis, then the area of the region is 1/2.Since there are two regions: one from
#x=1# to#x=0# and the other from#x=0# to#x=1# , the area#A# can be found by
#A=int_{1}^0(0x^3)dx+int_0^1(x^30)dx#
by Power Rule,
#=[x^4/4]_{1}^0+[x^4/4]_0^1=1/4+1/4=1/2# 
If
#f# is an even function (symmetric about the yaxis), then#int_{a}^a f(x) dx=2int_0^a f(x) dx# .If
#f# is an odd function (symmetric about the origin), then#int_{a}^a f(x) dx=0# .Symmetries can be used to simplify computation of definite integrals. Let us look at the following examples.
Example 1 (Even Function)
#int_{1}^1(3x^2+1) dx =2int_0^1(3x^2+1) dx=2[x^3+x]_0^1=2(20)=4# Example 2 (Odd Function)
#int_{pi/3}^{pi/3}{sin theta}/{sqrt{cos^2 theta+1}} d theta=0# I hope that this was helpful.