### Video Transcript

In this video, weβll look at an
important application of integration to finding the area underneath the curve, by
which we mean the area bounded by a curve, the π₯-axis, and two vertical lines π₯
equals π and π₯ equals π. Finding areas below curves is a
useful skill because it often has a practical application. For example, in the case of a
velocityβtime graph, the area below the curve gives the total distance traveled.

Weβll also see how this method can
be adapted to find the area between the curve of a function in the form π₯ equals π
of π¦ and the π¦-axis. And weβll then extend these
techniques to finding the areas of regions enclosed by a curve and a straight line
which is parallel to either the π₯- or π¦-axes.

It is also possible to use
integration to find volumes of solids. And in fact, many of the formulae
we already know for finding the volumes of three-dimensional solids, such as cones
or spheres, can be proved using integration, although thatβs beyond the scope of
what weβll consider here.

Letβs consider a simple example
first of all, where the area weβre looking to find is actually bounded by a straight
line rather than a curve. And the straight line has equation
π¦ equals π of π₯, with π of π₯ equal to four π₯ plus seven. The area weβre looking to find is
actually a trapezoid. So we can do this without using
integration. The lengths of the two parallel
sides of this trapezoid will be the values of the function π evaluated at π and
π. Those are the two values of π₯
which bound this region. π of π is equal to four π plus
seven, and π of π is equal to four π plus seven. The perpendicular height of this
trapezoid will be the difference between the two π₯-values which bound this
region. So itβs equal to π minus π.

To find the area of a trapezoid, we
take half the sum of its parallel sides β so thatβs four π plus seven plus four π
plus seven all over two β and then multiply by the perpendicular height of the
trapezoid β thatβs π minus π. Simplifying gives two π plus two
π plus seven multiplied by π minus π. And then we can distribute the
parentheses, giving two ππ minus two π squared plus two π squared minus two ππ
plus seven π minus seven π.

And now we see that the two ππ
and the negative two ππ will cancel one another out, leaving just negative two π
squared plus two π squared plus seven π minus seven π, which we can rewrite as
two π squared plus seven π minus two π squared plus seven π. So weβve been able to find the
area. But whatβs the link with
integration?

Well, we notice that the
antiderivative of our function π of π₯ is two π₯ squared plus seven π₯ plus a
constant of integration if necessary. So what we have is the
antiderivative of our function π of π₯ evaluated at π minus the antiderivative
evaluated at π, which we can express as a definite integral. Itβs equal to the integral from π
to π of our function π of π₯ with respect to π₯. This suggests then that, in order
to find the area between a straight line and the π₯-axis bounded by the two vertical
lines π₯ equals π and π₯ equals π, we can just integrate the equation of the
function between these limits and evaluate. Weβve seen this work for one
example of a straight line. But how do we know itβll also work
when the function π of π₯ is a curve?

Well, letβs now suppose weβre
looking to find this new area which is bounded by the curve π¦ equals π of π₯, the
π₯-axis, and the two vertical lines π₯ equals π and π₯ equals π. Rather than attempting to find the
whole area to begin with, letβs consider instead taking just a very small slice of
this area, with a width of Ξπ₯, Ξπ₯ meaning a very small change in π₯. This portion of the area can be
approximated by a rectangle with a width of Ξπ₯ and a height of π of π₯.

To find the area of a rectangle, we
multiply its two dimensions together. So this area is approximately equal
to π of π₯ multiplied by Ξπ₯. If we then took a large number of
such slices, each of width Ξπ₯, the total area could be approximated by the sum from
π₯ equals π to π₯ equals π of π of π₯ multiplied by Ξπ₯. In order to improve our
approximation, we have to take a larger number of slices so that their width becomes
smaller and smaller.

In order to make our approximation
into an exact answer, we have to make these rectangles infinitely thin. And so the exact area is equal to
the limit as Ξπ₯ tends to zero of this sum. Now if youβre familiar with using
Ξπ₯ notation when you first learned about differentiation, youβll know that as Ξπ₯
approaches zero, we use the notation dπ₯ to represent its limit. We use an integral sign to
represent this infinite sum of the areas of infinitely thin rectangles.

And so we have that the area is
equal to the integral from π₯ equals π to π₯ equals π of π of π₯ dπ₯. The symbol we use to represent an
integral is in fact an elongated S shape. And in fact, was originally just an
S standing for βsumβ to represent the fact that we are taking the sum of an infinite
number of infinitely thin slices. We see then that, just as in our
example with a straight line, we can find the area bounded by a curve, the π₯-axis,
and the two vertical lines π₯ equals π and π₯ equals π by evaluating the definite
integral of the function π of π₯ between these two π₯-values. Letβs now have a look at some
examples.

Let π of π₯ equal two π₯ squared
plus three. Determine the area bounded by the
curve π¦ equals π of π₯, the π₯-axis, and the two lines π₯ equals negative one and
π₯ equals five.

Letβs begin with a sketch of the
region whose area weβre looking to calculate. Itβs bounded by a quadratic curve
with a positive leading coefficient and a π¦-intercept of three. Itβs also bounded by the two
vertical lines π₯ equals negative one and π₯ equals five and the π₯-axis. So itβs this area here that weβre
looking to find.

We recall that the area bounded by
the curve π¦ equals π of π₯, the π₯-axis, and the two vertical lines π₯ equals π
and π₯ equals π can be found by evaluating the definite integral from π to π of
π of π₯ with respect to π₯. Our function π of π₯ is two π₯
squared plus three. The lower limit for our integral,
the value of π, is the lower value of π₯. Thatβs negative one. And the upper limit, the value of
π, is the upper limit of π₯. Thatβs five. So the area weβre looking for is
equal to the integral from negative one to five of two π₯ squared plus three with
respect to π₯.

We recall that, in order to
integrate powers of π₯ not equal to negative one, we increase the power by one and
then divide by the new power. So the integral of two π₯ squared
is two π₯ cubed over three, and the integral of three is three π₯. We have then that the area is equal
to two π₯ cubed over three plus three π₯ evaluated between negative one and
five.

Remember, thereβs no need for a
constant of integration here, as this is a definite integral. We then substitute the limits,
giving two multiplied by five cubed over three plus three multiplied by five minus
two multiplied by negative one cubed over three plus three multiplied by negative
one. Thatβs 250 over three plus 15 minus
negative two-thirds minus three. That simplifies to 102. And so we can say that the area of
the region bounded by the curve π¦ equals π of π₯, the π₯-axis, and the two lines
π₯ equals negative one and π₯ equals five found by evaluating the definite integral
of our function π of π₯ between the limits of negative one and five is 102 square
units.

In our next example, weβll see what
happens when the area weβre trying to find lies below the π₯-axis.

The curve shown is π¦ equals one
over π₯. What is the area of the shaded
region? Give an exact answer.

Now we recall that the area of the
region bounded by the curve π¦ equals π of π₯, the lines π₯ equals π and π₯ equals
π, and the π₯-axis is given by the integral from π to π of π of π₯ with respect
to π₯. In this case, weβre told that the
function π of π₯ is one over π₯. And from the graph, we can see that
the values of the limits for this integral are negative one for the lower limit and
negative one-third for the upper limit. So we have the definite integral
from negative one to negative one-third of one over π₯ with respect to π₯.

We then recall that the integral of
one over π₯ with respect to π₯ is equal to the natural logarithm of the absolute
value of π₯ plus the constant of integration. And that absolute value is really
clear here because the two values for our limits are both negative. And we recall that the natural
logarithm of a negative value is undefined. So we must make sure we include
those absolute value signs.

So weβre taking the natural
logarithm of a positive value. We donβt need the constant of
integration π here as weβre performing a definite integral. Substituting the limits gives the
natural logarithm of the absolute value of negative one-third minus the natural
logarithm of the absolute value of negative one. Thatβs the natural logarithm of
one-third minus the natural logarithm of one. And at this point, we can recall
that the natural logarithm of one is just zero. So our answer has simplified to the
natural logarithm of one-third.

Now this may not be immediately
obvious to you. But in fact, the natural logarithm
of one-third is a negative value. We can see this if we recall one of
our laws of logarithms, which is that the logarithm of π over π is equal to the
logarithm of π minus the logarithm of π. And so the natural logarithm of
one-third is the natural logarithm of one minus the natural logarithm of three. And again, we recall that the
natural logarithm of one is equal to zero. So our answer appears to be that
this area is equal to negative the natural logarithm of three.

This doesnβt really make sense
though as areas should be positive. What we see then is that when we
use integration to evaluate an area below the π₯-axis, we will get a negative
result. This doesnβt mean though that the
area itself is negative. The negative sign is just
signifying to us that the area is below the π₯-axis.

Really then, what we shouldβve done
is include absolute value signs around our integral sign at the beginning. And what this means is that
although the value of the integral is negative, the natural logarithm of three, the
value of the area is the absolute value of this. So thatβs just the natural
logarithm of three. The integral is negative to signify
that the area is below the π₯-axis. But the area itself is
positive. So our answer to the question, and
it is an exact answer, is that this area is equal to the natural logarithm of three.

Now this does have important
implications in cases where the area weβre looking to find is split into regions
which are above and regions which are below the π₯-axis. In such a case, we need to split
our integral up into different parts using the linearity of integration and evaluate
each integral separately and then add their absolute values together. In our next example, weβll look at
how we can find the area between a curve given in the form π₯ equals π of π¦ and
the π¦-axis.

Find the area enclosed by the graph
of π₯ equals nine minus π¦ squared, the π¦-axis, and the lines π¦ equals negative
three and π¦ equals three.

We notice that, in this example,
the area weβve been asked to find is enclosed by a curve with the equation π₯ equals
some function of π¦. The area is also bounded by the
π¦-axis rather than the π₯-axis and two lines with equations of the form π¦ equals
some constant, which are horizontal rather than vertical lines.

Now we could repeat the process
from first principles to see how we can use integration to find this area, rather
than an area bounded by a curve in the form π¦ equals π of π₯ and the π₯-axis. But actually, it is as simple as
swapping π₯ and π¦ around. In order to find the area enclosed
by a curve in the form π₯ equals π of π¦, the π¦-axis, and the two horizontal lines
π¦ equals π and π¦ equals π, we evaluate the definite integral from π to π of π
of π¦ with respect to π¦.

In this case then, weβre evaluating
the definite integral from negative three to three of nine minus π¦ squared dπ¦. Notice that everything in the
integral is in terms of π¦, not π₯. To integrate powers of π¦ not equal
to negative one, we increase the power by one and divide by the new power, giving
nine π¦ minus π¦ cubed over three evaluated between negative three and three. We then substitute our limits,
giving nine multiplied by three minus three cubed over three minus nine multiplied
by negative three minus negative three cubed over three. Thatβs 27 minus nine minus negative
27 plus nine, which is equal to 36.

So we find that the given area is
36 square units. And in this problem, weβve seen
that, in order to find an area enclosed by the graph of π₯ equals some function of
π¦, the π¦-axis, and two horizontal lines, we can just perform a definite integral
with everything in terms of π¦ rather than everything in terms of π₯. Notice also that, in this question,
we werenβt concerned that some of the area lay below the π₯-axis as this time we
were integrating with respect to π¦. And all of the area lay on the same
side of the π¦-axis.

Now it may also be the case that,
in other types of questions, weβre asked to find an area between a curve and the
π¦-axis. But the equation of the curve is
given in the form π¦ equals π of π₯ rather than π₯ equals π of π¦. In this case, weβd need to
rearrange the equation to give an equation thatβs in the form π₯ equals π of
π¦. And we need to make sure that any
limits we were given were converted from limits in π₯ to limits in π¦ if
necessary. It is, however, beyond the scope of
this video to consider this in detail here. In our final example, weβll see how
to find the area of a region where none of the lines enclosing it are the π₯- or
π¦-axis.

Find the area of the shaded
region.

Letβs use the given figure to first
determine the equations of each of the lines that bound this region. First, we have the curve π¦ equals
three π₯ squared plus four π₯ minus two. We also have the vertical lines π₯
equals one and π₯ equals two. But rather than our region being
bounded by the π₯- or π¦-axis, the final line which bounds this region is the line
π¦ equals five.

Now if we were just looking to find
the area bounded by the quadratic graph, the lines π₯ equals one and π₯ equals two
on the π₯-axis, we could evaluate the definite integral from one to two of three π₯
squared plus four π₯ minus two with respect to π₯. Notice though that this would
include the area of the rectangle below the line π¦ equals five. And so to find the area of just the
pink region, weβd need to subtract the area of this rectangle from our integral.

The area of this rectangle isnβt
tricky to find. It has a vertical height of five
units and a width of one unit. Thatβs two minus one. So its area is just five multiplied
by one, or five. This then gives us one method that
we can use. We evaluate the definite integral
to find the area all the way down to the π₯-axis and then subtract the area of the
orange rectangle. Working through the integral would
give an answer of six square units.

But there is actually another way
that we could answer this question. The area below the line π¦ equals
five can also be found using integration. Itβs equal to the integral from one
to two of five with respect to π₯. And using the linear property of
integration, we could then express this as a single integral. If you were to form this integral,
it would give the same result.

This also gives us a clue as to how
we could find definite integrals enclosed by a curve and a diagonal line, or even
two curves, although that is beyond the scope of this video.

In this video, weβve seen that the
area bounded by the curve π¦ equals π of π₯, the lines π₯ equals π, π₯ equals π,
and the π₯-axis can be found by evaluating the definite interval from π to π of π
of π₯ with respect to π₯. We also saw that we need to take
the absolute value of this integral if the area is below the π₯-axis. We also saw that a very similar
result can be applied when the region is bounded by the π¦-axis and the function of
the form π₯ equals π of π¦.

And finally, we saw that if the
region is bounded by a horizontal or vertical line other than the π₯- or π¦-axis, we
can subtract the equation of this line from the equation of the curve prior to
evaluating the definite integral.