NARRATOR: Exercise 12.
Part A.
Compute the integral from –2 to 2 of (x cubed – x) dx.
Part B.
What is the total area enclosed between the graph of y = x cubed – x
and the x-axis, from x = –2 to x = 2?
Part A. We wish to compute this integral.
So the antiderivative of x cubed
is x to the power of 4 divided by 4.
Add one to the power and divide by the new power.
And the antiderivative of x is x squared divided by 2.
We wish to evaluate that integral between –2 and 2.
So we're first substituting 2.
We're looking at 2 to the power of 4, which is 16,
divided by 4,
minus 2 squared, which is 4,
divided by 2.
Substituting –2.
-2 to the power of 4 is also 16.
And –2 to the power of 2 is also 4.
And so we have (4 – 2) – (4 – 2).
That is 2 – 2, and hence, 0.
So the integral from –2 to 2
of (x cubed – x) dx is 0.
Part B asks us to consider the area
between the graph of y = x cubed – x and the x-axis, from x = –2 to x = 2.
So first, we look at the graph of y = x cubed – x,
and we establish the regions that we need to calculate
in order to find the total area.
We can see from this diagram why indeed our answer to Part A was 0
in that each of the regions here will cancel out with another region
and you would get a total of 0
when integrating just straight from –2 to 2.
In this case, we wish to find the area,
so we need to consider the sined regions.
So we note that we can divide this into four regions.
The first region is that between – 2 and –1,
and that would be found by evaluating the negative integral from – 2 to –1
of x cubed – x dx.
It's a negative integral since the region is below the x-axis
and, hence, it's negatively sined.
The second region would be that from –1 to 0,
and that would be found by evaluating the integral from –1 to 0
of x cubed – x dx.
The third region is the region from 0 to 1,
and again, it's negatively sined because it's underneath the x-axis
so would be evaluated by finding the negative integral from 0 to 1
of x cubed – x dx.
And the final region
is that from 1 to 2,
so we can evaluate that
by finding the integral from 1 to 2
of x cubed – x dx.
Alternatively, we can use the symmetry of the total area
so that we only need to calculate two integrals.
We know that area one is equivalent to area four
and also area two is equivalent to area three.
So rather than adding up
four integrals,
we can simply double the area
from 0 to 1
and also double the area from 1 to 2.
So we now need to calculate this total area.
So we know we'll have –2.
And we've already established this integral,
the integral of x cubed – x
is x to the power of 4 on 4 minus x squared on 2.
And we're calculating that between 0 and 1.
Plus the same integral ...
... between 1 and 2.
So making our substitutions,
we'll have one-quarter – one-half – 0.
Here we'll have 16 on 4 – 4 on 2 - one-quarter – one-half.
And so we have -2 times (one-quarter – one half),
which is negative a quarter.
Plus 2 times 16 on 4 – 4 on 2,
if we recall from earlier, is equivalent to 2.
And then we've got minus (minus one-quarter).
So we have –2 times negative a quarter,
which is positive two-quarters, which is positive one-half.
And we have 2 times 2 minus minus a quarter,
so that's 2 + one-quarter.
2 is equivalent to eight-quarters
so we have nine-quarters.
So we've got one-half + 2 times nine-quarters
which is equivalent to nine-halves.
So that is ten-halves,
which is equal to 5 square units.
So the total area enclosed between
the graph of y = (x cubed – x) and the x-axis from x = –2 to x = 2
is 5 square units.