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NARRATOR: Exercise 12.

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Part A.

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Compute the integral from –2 to 2
of (x cubed – x) dx.

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Part B.

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What is the total area enclosed
between the graph of y = x cubed – x

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and the x-axis, from x = –2 to x = 2?

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Part A. We wish to compute
this integral.

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So the antiderivative of x cubed

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is x to the power of 4
divided by 4.

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Add one to the power
and divide by the new power.

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And the antiderivative of x
is x squared divided by 2.

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We wish to evaluate that integral
between –2 and 2.

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So we're first substituting 2.

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We're looking at 2 to the power of 4,
which is 16,

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divided by 4,

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minus 2 squared, which is 4,

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divided by 2.

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Substituting –2.

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-2 to the power of 4 is also 16.

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And –2 to the power of 2 is also 4.

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And so we have (4 – 2) – (4 – 2).

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That is 2 – 2, and hence, 0.

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So the integral from –2 to 2

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of (x cubed – x) dx is 0.

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Part B asks us to consider the area

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between the graph of y = x cubed – x
and the x-axis, from x = –2 to x = 2.

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So first, we look at
the graph of y = x cubed – x,

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and we establish the regions
that we need to calculate

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in order to find the total area.

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We can see from this diagram why
indeed our answer to Part A was 0

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in that each of the regions here
will cancel out with another region

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and you would get a total of 0

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when integrating
just straight from –2 to 2.

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In this case,
we wish to find the area,

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so we need to consider
the sined regions.

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So we note that we can divide this
into four regions.

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The first region is that
between – 2 and –1,

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and that would be found by evaluating
the negative integral from – 2 to –1

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of x cubed – x dx.

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It's a negative integral since
the region is below the x-axis

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and, hence, it's negatively sined.

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The second region would be that
from –1 to 0,

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and that would be found by evaluating
the integral from –1 to 0

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of x cubed – x dx.

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The third region
is the region from 0 to 1,

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and again, it's negatively sined
because it's underneath the x-axis

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so would be evaluated by finding
the negative integral from 0 to 1

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of x cubed – x dx.

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And the final region

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is that from 1 to 2,

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so we can evaluate that

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by finding the integral from 1 to 2

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of x cubed – x dx.

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Alternatively, we can use
the symmetry of the total area

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so that we only need
to calculate two integrals.

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We know that area one
is equivalent to area four

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and also area two
is equivalent to area three.

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So rather than adding up

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four integrals,

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we can simply double the area

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from 0 to 1

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and also double the area from 1 to 2.

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So we now need to calculate
this total area.

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So we know we'll have –2.

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And we've already
established this integral,

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the integral of x cubed – x

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is x to the power of 4 on 4
minus x squared on 2.

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And we're calculating that
between 0 and 1.

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Plus the same integral ...

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... between 1 and 2.

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So making our substitutions,

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we'll have
one-quarter – one-half – 0.

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Here we'll have 16 on 4 – 4 on 2 -
one-quarter – one-half.

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And so we have
-2 times (one-quarter – one half),

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which is negative a quarter.

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Plus 2 times 16 on 4 – 4 on 2,

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if we recall from earlier,
is equivalent to 2.

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And then we've got
minus (minus one-quarter).

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So we have –2
times negative a quarter,

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which is positive two-quarters,
which is positive one-half.

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And we have 2 times
2 minus minus a quarter,

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so that's 2 + one-quarter.

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2 is equivalent to eight-quarters

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so we have nine-quarters.

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So we've got
one-half + 2 times nine-quarters

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which is equivalent to nine-halves.

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So that is ten-halves,

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which is equal to 5 square units.

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So the total area enclosed between

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the graph of y = (x cubed – x)
and the x-axis from x = –2 to x = 2

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is 5 square units.