### Video Transcript

Let π be a continuous function on
the closed interval negative four to eight, where π is twice differentiable with π
prime prime of π₯ greater than zero on the open interval negative four to eight. Selected values of π are shown in
the table. i) Using the trapezoidal rule approximation with five trapezoids, find
the approximate value of the integral evaluated between negative four and eight of
π of π₯ with respect to π₯. ii) Using the result in part one, find the average
value of π on the closed interval negative four to eight. iii) Find the average
rate of change of π over the closed interval negative four to eight. And iv) Examine π prime of zero
and π prime of four and determine if π prime of zero is greater than, smaller
than, or equal to π prime of four. Explain why.

Weβll begin with part one. The trapezoidal rule approximation
is the technique that allows us to approximate the value of definite integrals. We split the region between the
curve and the π₯-axis into trapezoids. And we evaluate the area of each of
these. Now, usually, the trapezoids are of
equal width. By looking at the π₯-values in our
table though, we can see that our trapezoids are going to have unequal width.

The first trapezoid is from
negative four to negative two. So it will have a width of two
units. The second trapezoid is from
negative two to zero. So that also has a width of two
units. But the third trapezoid will have a
width of one unit, the fourth a width of three units, and the fifth a width of four
units. So rather than use the trapezoidal
rule approximation that we usually do, we have an unsimplified version. We take the difference between two
consecutive π₯-values. Remember this gives us the width of
the trapezoid. And we multiply this by a half of
the sum of the corresponding π of π₯ values. Weβre going to do this for each of
the trapezoids that weβre interested in and find their sum.

The width of the first trapezoid is
negative two minus negative four. And we multiply that by the
corresponding π of π₯ values divided by two. Thatβs 16 plus 10 divided by two
which is 26. The width of our second trapezoid
is zero minus negative two. And we multiply that by half of the
sum of the corresponding π of π₯ values. Thatβs six plus 10 divided by
two. Thatβs 16. The width of our third trapezoid is
one minus zero. And weβre going to multiply that by
five plus six divided by two, which is 11 over two. We add four minus one times three
plus five over two which is 12. And finally, we add eight minus
four times two plus three over two, which is equal to 10. The sum total is therefore 139 over
two.

So we can say that an approximation
for the total area of our five trapezoids is 139 over two square units. And this means the approximate
value of the integral evaluated between negative four and eight of π of π₯ with
respect to π₯ is 139 over two. For part two, weβre going to use
the result from part one to find the average value of π over our closed
interval. The formula for the average value
of a function over a closed interval π to π is one over π minus π times the
integral evaluated between π and π of π of π₯ with respect to π₯.

We can let π be equal to negative
four and π be equal to eight. And remember we already evaluated
this integral between negative four and eight and we got 139 over two. One over eight minus negative four
is one over 12. So the average value of π is given
as a 12 times 139 over two. Thatβs 139 over 24.

Now part three, weβre looking to
find the average rate of change of π over the closed interval negative four to
eight. Once again, we have a formula we
can use. The average rate of change of a
function over a closed interval π to π is given by π of π minus π of π over π
minus π. Once again, we let π be equal to
negative four and π be equal to eight. So this becomes π of eight minus
π of negative four over eight minus negative four. Eight minus negative four is
12.

And we refer to our table for the
values for π of eight and π of negative four. Theyβre two and 16,
respectively. Two minus 16 is negative 14. And then, we simplify our fraction
by dividing both the numerator and the denominator by two. And we see the average rate of
change of π over the closed interval negative four to eight is negative
seven-sixths.

For part four, we need to determine
whether π prime of zero is greater than, smaller than, or equal to π prime of
four. Now, we donβt actually have values
for π prime of zero and π prime of four. But weβre told that π is twice
differentiable and that π prime prime of π₯ is greater than zero on the open
interval negative four to eight. Well, letβs recall what this
actually means.

π prime prime of π₯ is the second
derivative. Itβs the derivative of π prime of
π₯. And in this case, itβs strictly
greater than zero. This means that π prime of π₯ the
derivative of our function must be strictly increasing. In other words, as our value of π₯
increases, so does our value of π prime of π₯. And this tells us that π prime of
four must be strictly greater than π prime of zero. It cannot be equal to it. And we see then that π prime of
zero must be smaller than π prime of four.