How do you evaluate the definite integral #int x^3 dx# from #[-1,1]#?

Question

Cameron Alexander · Accepted Answer

We will need two ideas here:

So when we integrate #x^3#, we get #x^(3+1)/(3+1)=x^4/4#. Applying the bounds means we plug in #1# to this, then subtract what happens when we plug in #-1#.

This is written as:

#int_(-1)^1x^3dx=[x^4/4]_(-1)^1=(1^4/4)-((-1)^4/4)#

#color(white)(int_(-1)^1x^3dx)=1/4-1/4#

#color(white)(int_(-1)^1x^3dx)=0#

Cameron Alexander · Answer

We can also note that #x^3# is an odd function. An odd function #f# has the quality #f(-x)=-f(x)#. For #f(x)=x^3#, we see that #f(-x)=(-x)^3=-x^3=-f(x)#.

An odd function is one where it is reflected over the #x# and #y# axes. This means that its area on one side of the #y# axis will be mirrored on the other side of the #y# axis, only as the opposite area. Take a look at the graph of #x^5-16x# as a model odd function:

graph{x^5-16x [-4, 4, -30, 30]}

We can see that the area from the #y# axis to #x=2# will be the opposite of the area from the #y# axis to #x=-2#. If we were to add these areas, we would have a net area of #0#. This is integration—so we can formulate the following rule, where #f# is an odd function:

#int_(-a)^af(x)dx=0#

(Since the area from #x=-a# to #x=0# will have the opposite area as the area from #x=0# to #x=a#.)

We can also see this as:

#int_(-a)^0f(x)dx=-int_0^af(x)dx#

#=>int_(-a)^0f(x)dx+int_0^af(x)dx=0#

#=>int_(-a)^af(x)dx=0#

The same is applicable to the integral we have here! For #int_(-1)^1x^3dx#, we see that #x^3# is an odd function that goes from a negative value to its corresponding positive value the same distance away from the #y# axis, thus:

#int_(-1)^1x^3dx=0#

Luke Alvarez · Answer

undefined To evaluate the definite integral ∫x^3 dx from -1 to 1, we can use the fundamental theorem of calculus. First, we find the antiderivative of x^3, which is (1/4)x^4. Then, we evaluate this antiderivative at the upper and lower bounds of integration (-1 and 1) and subtract the results.

(1/4)(1^4) - (1/4)(-1^4) = (1/4)(1) - (1/4)(1) = 1/4 - 1/4 = 0.

So, the value of the definite integral ∫x^3 dx from -1 to 1 is 0.