Question

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

Answer 1

We will need two ideas here:

• `intx^ndx=x^(n+1)/(n+1)+C`
• `int_a^bf'(x)dx=f(b)-f(a)`

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`

Answer 2

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`