The Definite Integral: Learn It 2

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The Definite Integral: Learn It 2

Area and the Definite Integral

When we defined the definite integral, we lifted the requirement that $f(x)$ be nonnegative. But how do we interpret “the area under the curve” when $f(x)$ is negative?

Net Signed Area

Let us return to the Riemann sum.

Consider, for example, the function $f(x)=2-2x^2$ (shown in Figure 2) on the interval $[0,2]$ . Use $n=8$ and choose $x_i^*$ as the left endpoint of each interval. Construct a rectangle on each subinterval of height $f(x_i^*)$ and width $\Delta x$ .

A graph of a downward opening parabola over [-1, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). Eight rectangles are drawn evenly over [0,2] with heights determined by the value of the function at the left endpoints of each. — Figure 2. For a function that is partly negative, the Riemann sum is the area of the rectangles above the $x$ -axis minus the area of the rectangles below the $x$ -axis.

When $f(x_i^*)$ is positive, the product $f(x_i^*) \Delta x$ represents the area of the rectangle, as before. When $f(x_i^*)$ is negative, however, the product $f(x_i^*) \Delta x$ represents the negative of the area of the rectangle.

The Riemann sum then becomes

\sum_{i = 1}^{8} f (x_{i}^{*}) Δ x =

$\displaystyle\sum_{i=1}^{8}f(x_i^*) \Delta x =$ (Area of rectangles above the

x

$x$ -axis)

-

$-$ (Area of rectangles below the

x

$x$ -axis)

Taking the limit as $n\to \infty$ , the Riemann sum approaches the area between the curve above the $x$ -axis and the $x$ -axis, minus the area between the curve below the $x$ -axis and the $x$ -axis, as shown in Figure 3.

A graph of a downward opening parabola over [-2, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). The area in quadrant one under the curve is shaded blue and labeled A1. The area in quadrant four above the curve and to the left of x=2 is shaded blue and labeled A2. — Figure 3. In the limit, the definite integral equals area $A_1$ minus area $A_2$ , or the net signed area.

Then,

\begin{array}{ll} \int_{0}^{2} f (x) d x & = lim_{n \to \infty} \sum_{i = 1}^{n} f (c_{i}) Δ x \\ = A_{1} - A_{2} \end{array}

$\begin{array}{ll} \displaystyle\int_0^2 f(x) dx & =\underset{n\to \infty}{\lim} \displaystyle\sum_{i=1}^{n} f(c_i) \Delta x \\ & =A_1-A_2 \end{array}$

The quantity $A_1-A_2$ is called the net signed area.

Notice that net signed area can be positive, negative, or zero. If the area above the $x$ -axis is larger, the net signed area is positive. If the area below the $x$ -axis is larger, the net signed area is negative. If the areas above and below the $x$ -axis are equal, the net signed area is zero.

Find the net signed area between the curve of the function $f(x)=2x$ and the $x$ -axis over the interval $[-3,3]$ .

The function produces a straight line that forms two triangles: one from $x=-3$ to $x=0$ and the other from $x=0$ to $x=3$ (Figure 4). Using the geometric formula for the area of a triangle, $A=\frac{1}{2}bh$ , the area of triangle $A_1$ , above the axis, is

$A_1=\frac{1}{2}3(6)=9$ ,

where $3$ is the base and $2(3)=6$ is the height. The area of triangle $A_2$ , below the axis, is

A_{2} = \frac{1}{2} (3) (6) = 9

$A_2=\frac{1}{2}(3)(6)=9$ ,

where $3$ is the base and $6$ is the height. Thus, the net area is

$\displaystyle\int_{-3}^3 2x dx=A_1-A_2=9-9=0$ .

A graph of an increasing line over [-6, 6] going through the origin and (-3, -6) and (3,6). The area under the line in quadrant one over [0,3] is shaded blue and labeled A1, and the area above the line in quadrant three over [-3,0] is shaded blue and labeled A2. — Figure 4. The area above the curve and below the $x$ -axis equals the area below the curve and above the $x$ -axis.

Analysis

If $A_1$ is the area above the $x$ -axis and $A_2$ is the area below the $x$ -axis, then the net area is $A_1-A_2$ . Since the areas of the two triangles are equal, the net area is zero.

Starting at timestamp 8:40, watch the following video to see the worked solution to this example.

LECTURER: Section 5.2, the definite integral.

In 5.1, we determined that the area underneath our curve is exactly equal to the limit as n goes to infinity of the sum as i goes from 1 to n f of f of xi star times delta x.

The restriction that we placed here was that f of x must be continuous and non-negative.

We want to move on just a bit.

Largely the functions we're going to deal with are going to be continuous until calculus two, potentially.

But in general, we're going to be dealing with continuous functions that are not necessarily non-negative.

So the definite integral is defined over an interval a to b as the integral from a to b of the limit as n goes to infinity of i to equals 1 to n of f of xi star times delta x, provided that that limit exists.

If the limit exists, the function is said to be integrable.

Now, our theorem here states that continuous functions are integrable.

This does not necessarily mean that all integrable functions are continuous, not by a long shot.

But continuous functions certainly are.

Now we're going to apply that idea of a Riemann integral here in just a second.

And so we need these two summation forms.

Specifically, we'll need the second one.

All right, so let's dive into this.

Find the area between the curve of the function f of x equals x squared and the x-axis over the interval 0 to 2.

Use the graph and area formulas, as well as the definition of a Riemann sum.

All right, so first we want to know the area between 0 to 2 from the points x equals 0 to x equals 2 of x squared d x, and this is going to be equal to the limit as n goes to infinity sum i equals 1 to n of f of x i delta x.

Now, our delta x in this case is 2 minus 0 over n.

So delta x is 2 over n.

Now we're going to use that delta x, it's not a constant there.

It changes.

It is going to be a fixed width.

However, as n increases that x will get- the distance between all of our points will actually get smaller and smaller.

Our xi is a plus i delta x So for us, our a value is zero.

So this is 2i over n.

That is our xi.

That is a generic i value based on what we did before, we went from 0 to 1/2 2/2, that sort of thing.

Well we're going from 0 to 2 over to 4 over n to 3 over n, et cetera.

Or to 1 over n, et cetera.

Now, f of xi, we need to find out what this is.

Our function is x squared.

So this would be our xi squared.

But xi is 2i over n squared.

Making our f of xi equal to 4 over n squared i squared is how I'm going to write that.

Actually, let's put that in the numerator.

So this would be the limit as n goes to infinity.

Sum i equals 1 n of 4i squared over n squared times our delta x.

Delta x was 2 over n.

Now, being that i is what we want to sum, we're going to go ahead and factor out a few things.

Take out a, 8 over n cubed.

The sum of i squared, which we know from our formula earlier.

That is equal to n times n plus 1 times 2n plus 1 over 6.

Yes.

Now we want to know the limit of this.

Now I want to go ahead and make the observation that we are going to have a cubic polynomial, a cubic polynomial in our numerator, a cubic in our denominators.

All that's really important is the coefficients here.

All right, well our coefficients are going to be let me just go and add this in here.

We were going to do a 2 in the numerator so times 8 so that is 16 n cubed plus the rest of our polynomial.

And a 6n cubed in our denominator.

So that the limit is 16 over 6, or 8/3.

So the area from 0 to 2 underneath the curve x squared is 8/3.

I want to make a note about this statement here.

First off, the notation here actually goes back to the 1700s.

And it's attributed to Leibniz .

Gottfried Wilhelm Leibniz.

And he's considered to be the co discoverer of calculus, along with Isaac Newton.

And this elongated s, that's exactly what it is.

The integral symbol, the elongated s stands for summation.

Just like the Greek sigma stands for summation.

There's one thing.

Now we have our boundaries on our interval, a to b, where this is our lower limit of integration and this is our upper limit of integration.

Now we actually have two kinds of limits here.

We have our limits of integration here, 0 and 2, and we have our limit as n goes to infinity.

Those are two separate things.

Two separate things, so keep that straight.

Now we have what we call the integrand.

And that is this section of stuff here, generally our function, our f of x is our integrand.

x is our variable of integration we have here.

And this dx actually is the change in x.

Very similar to this.

These two.

This is a d for delta.

And I believe that's all the terminology we can throw out there.

All right, so let's move on to number two.

Use the formula for area of a circle to evaluate the integral from 3 to 6 of the square root of 9 minus x minus 3 squared.

All right, so this equation actually is an upper half circle going from 0 to 6 is this endpoint.

And the middle point of that is 3.

So here is the area I am seeking to find.

The area from 3 to 6 of that function right there.

Now if you notice, this is a semicircle with a radius of 3.

Actually what we want is the quarter circle.

So the integral of square root of 9 minus x minus 3 squared equals 1/4 pi r squared.

In this case our r is 3.

So this is 9/4 pi, which is roughly 7.069.

So that would be the area.

So it pays, certainly, to be able to recognize functions and what the graphs are going to look like from them.

For example three, find the net signed area in the total area between the curve of the function f of x equals 2x over the x-axis over the interval negative 3 to 3.

Use the graph and area formulas.

Now for the next few of these, we are going to have very good shapes that are going to fit.

These are actually going to form some triangles.

So we'll go ahead and do this.

At point 3 and negative 3.

Our function is at negative 6 and 6, respectively.

And the area we want to find is this area.

Now, the area for this the, net signed area is going to be- noticed we actually have some underneath and some above the x-axis.

So the area of the one on the right, we're going to call that a1.

And that a2.

That is a base of 3, a height of 6 divided by 2 minus the area of a2, which is 3 times 6 over 2 equals zero.

So we have a zero net area.

I think of this as if I begin walking across the room to my right.

Let's just say I start here, here is my 0 position, my equilibrium there.

Say I move to my right 3 feet, but I turn back and I walk 6 feet going in the opposite direction.

My net gain is actually negative 3 feet.

I've gone 3 feet to the left as a net.

OK.

How far have I actually walked?

My total distance?

Well that's 9 if I add the two together.

So you could think of this as negative 6 feet.

And that is 3 minus 6 feet, which is negative 3 feet gain.

But if I want to know my net, or my total, then I actually need to deal with only positive values and then add those together.

So my total area is the next thing.

The total area.

That is going to be 3 times 6 over 2 plus 3 times 6 over 2.

Which would be 18.

So my total area is the 18 but my net is 0 because those two actually have the same area.

Let's see if we can hammer that out a little bit more.

Find the total area of f of x equals x minus 2 to underneath this curve over the interval from 0 to 6.

At a value of 0, this function's at negative 2, and at a value of 6, this is at 4.

It's a linear function, so we have something that looks like that.

Now the area underneath this is this are plus this area.

Now, the place that intersects there if we set this equal to zero, we find that x equals 2.

x equals 2.

So what this is going to come down to, our total area is the integral from 0 to 2 of x minus 2 dx plus the integral from 2 to 6 of x minus 2 dx.

Now to find the total area, what I'm going to do is make both of these positive, so I'm going to take the absolute value of whatever I get out of those area functions.

So the first, that is a triangle with a base of 2 and a height of 2.

Now, because it's below the x-axis, it's going to end up being negative, so 2 times 2 over 2 plus the absolute value of that triangle.

It has a base of 4, a height of 4, and dividing that by 2.

So that would be an area of positive 2 plus 8.

And so my area is 10.

We have some properties of the definite integral.

If the limits of integration are the same, the integral is just a line and has no area.

So imagine a function here and I have a value a and I want to find the area of that line.

Well that line has- is 0.

You could also think of delta x being a minus a over n, which is 0.

So the change in x is 0, therefore the area of my rectangles is going to be 0.

All right, the next is if we change the limits of integration.

If we go from b to a versus a to b, we are going to have a negative sign in front of our interval.

The next- actually these next several have a whole lot to do with the limit laws.

Because the definite integral is a limit.

A limit of a Riemann sum.

So, the limit of the sums is the sum of the limits.

The difference of the limits is the difference of those two functions, but then we apply the limit to it.

Same idea here with a constant, we can factor a constant out.

That doesn't change anything.

And this last one probably needs a picture.

That if we have- I'm going to draw one case of this.

Just take a function, a, b, and c.

The value of the integral from a to b.

Let's switch these two.

Switch our two limits.

I really want b on the end.

All right, so the value of the integral from a to b is equal to the value of the area from a to c plus the area from c to b.

And this will work regardless of what our limits are.

Now, this actually works when c is between, that's when it just makes sense, but if it's at some other place it actually works as well.

It holds.

Use the properties of definite integral to express the definite integral of f of x equals negative 3x cubed plus 2x plus 2 over the interval negative 2 to 1 as the sum of three definite integrals.

Sum of f of x dx from negative 2 to 1 is equal to negative 3.

There's our constant.

Integral negative 2 to 1 of x cubed dx plus 2 integral from negative 2 to 1 of x dx plus the integral negative 2 to 1 of 2 dx.

And that's all there is for that.

So if it's known that the integral from 0 to 8 of f of x dx equals 10, and we know that the integral from 0 to 5 is 5, then what is the area from 5 to 8.

So actually apply that last part of that theorem, our rules.

The integral from 0 to 8 f of x dx equals- because 5 is between those two, or even if it wasn't, we could make this work.

The integral from 0 to 5 f of x dx plus the integral from 5 to 8 f of x dx.

We know that 0 to 8, that is 10.

We know that the value from 0 to 5 is 5.

And we don't know what that value is.

Well, some solving there, some simple solving algebra there, would tell us that the integral from 5 to 8 f of x dx is 5.

Imagine you don't know what the function is, what it looks like, but 0, 5, 8.

Just like in my example.

We know this area is 10, we know that this area equals 5, so then that area must also be 5.

Next we have the comparison theorem.

One specific case.

If f of x is greater than or equal to 0 on the interval a to b, then the integral from a to b of f of x is greater than or equal to 0.

If there's a positive function, , then we're going to have a positive area.

That's the way to think about that.

Positive f of x, positive area.

Next.

If f of x is greater than or equal to g of x on the interval a to b, then the integral from a to b of f of x is going to be greater than or equal to the integral from a to b of g of x.

Again, think of that as a larger function.

A function with larger values is going to have area that is larger.

OK.

This is actually called the domination theorem as well.

That if one function dominates, if it's larger than the other, then the area, the integral, is going to have to also be larger.

And then we have our max-min inequality.

If f of x is between little, lower case m and capital M over our interval, then lowercase m times b minus a is less than that integral, which is less than capital M times b minus a.

Here is my example.

If I have points a and b.

I'm going to box this in.

My lowest function value is m.

My largest function value is capital M. All right.

So let's suppose I box this in.

The area of the lower box would be m times b minus a.

The area of the bigger box would be capital M times b minus a, so certainly the area is between those two.

So let's actually apply that one real quick.

Compare these two functions.

Square root of 1 plus x squared and square root of 1 plus x over the interval 0 to 1.

I would really like to say that one of them is smaller or larger than the other.

That would be great.

I could look at the graphs, and I'm going to sketch that in just a second.

But how I want to start with this is that we know x is less than or equal to 1, and it's greater than or equal to 0, which means if I multiply by x on all sides, and since x is positive, and my signs say the same, x squared is less than x.

Plus than or equal to x.

Let's add 1 to all sides.

So 1 is less than or equal to 1 plus x squared, which is less than or equal to 1 plus x.

And then the square root function is a continuous- it's actually monotonic.

Can take the square root of all sides.

So this is the square root of 1 plus x squared, which is less than or equal to the square root of 1 plus x.

I'm going to add one more restriction back here.

x is less than or equal to 1, which means this is less than or equal to 2, which means this is less than or equal to the square root of 2.

OK.

So what I can say, based on this statement of fact, is that the integral from 0 to 1 of f of x dx is less than the integral from 0 to 1 of g of x dx.

I can also say that the area of both of these is bounded between 1 times 1.

So my lower case m is 1, and my capital M is the square root of 2.

Now, my a and b are 0 and 1, respectively.

So I can say that both of these are between.

That would be 1 times 1, and the square root of 2.

Because our width was 1.

So I can actually use the max-min inequality.

But what I'm really after here is that the integral- OK, I'm going to box that in again.

What I'm really after here is to say this.

But I can actually apply the max-min inequality to give myself sort of a bound.

I have an upper bound and a lower bound, and that's what these are.

A lower bound on my area and an upper bound on my area.

One last thing we need to touch on, and that is the average value of a function.

The average value of a function is given by f average equals 1 over b minus a times the integral from a to b of f of x dx.

So let's find the average value here.

What we're doing is we're finding the area and then dividing by our interval.

So that should make some sense.

So our f average equals 1 over b minus a.

In this case, were from 5 to 0, or 0 to 5 times the integral of x plus 1 dx, going from 0 to 5.

So at a function value of 0, this is at 1.

x value of 5, that means we are at six.

So we're actually looking for this area here.

Now, based on the fact this is made up of a triangle and a- we have a triangle here.

Blue triangle and a rectangle.

The area of that is going to be 17.5.

So the average value is 3.5.

What that means, I'll do that in a different color, is if I go to a y value of 3.5 and I make a rectangle here of width 5, going from 0 to 5, the area of that rectangle would be equal to the integral from 0 to 5.

So another way to write that formula we had.

f average equals 1 over b minus a integral a to b f of x dx.

Another way to write that is that the integral from a to b of f of x dx equals f average times b minus a.

So rather than being the lower bound or the upper bound, f average is the exact bound to tell us what the integral is.

All right, well that is the last problem from this section.

Section 5.2 on definite integrals.

—

You can view the transcript for this segmented clip of “5.2 The Definite Integral” here (opens in new window).

Total Area

One application of the definite integral is finding when given a velocity function. If $v(t)$ represents the velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original position.

This is a very important application of the definite integral, and we examine it in more detail later in the chapter. For now, we’re just going to look at some basics to get a feel for how this works by studying constant velocities.

When velocity is a constant, the area under the curve is just velocity times time. This idea is already very familiar.

If a car travels away from its starting position in a straight line at a speed of $70$ mph for $2$ hours, then it is $140$ mi away from its original position (Figure 6). Using integral notation, we have

\int_{0}^{2} 70 d t = 140

$\displaystyle\int_0^2 70 dt=140$

A graph in quadrant 1 with the x-axis labeled as t (hours) and y-axis labeled as v (mi/hr). The area under the line v(t) = 75 is shaded blue over [0,2]. — Figure 6. The area under the curve $v(t)=75$ tells us how far the car is from its starting point at a given time.

In the context of displacement, net signed area allows us to take direction into account.

If a car travels straight north at a speed of $60$ mph for $2$ hours, it is $120$ mi north of its starting position. If the car then turns around and travels south at a speed of $40$ mph for $3$ hours, it will be back at it starting position (Figure 7).

A graph in quadrants one and four with the x-axis labeled as t (hours) and the y axis labeled as v (mi/hr). The first part of the graph is the line v(t) = 60 over [0,2], and the area under the line in quadrant one is shaded. The second part of the graph is the line v(t) = -40 over [2,5], and the area above the line in quadrant four is shaded. — Figure 7. The area above the axis and the area below the axis are equal, so the net signed area is zero.

Again, using integral notation, we have

\begin{array}{ll} \int_{0}^{2} 60 d t + \int_{2}^{5} - 40 d t & = 120 - 120 \\ = 0 \end{array}

$\begin{array}{ll} \displaystyle\int_0^2 60 dt + \displaystyle\int_2^5 -40 dt & =120-120 \\ & =0 \end{array}$

In this case the displacement is zero.

Suppose we want to know how far the car travels overall, regardless of direction. In this case, we want to know the area between the curve and the $x$ -axis, regardless of whether that area is above or below the axis. This is called the total area.

Graphically, it is easiest to think of calculating total area by adding the areas above the axis and the areas below the axis (rather than subtracting the areas below the axis, as we did with net signed area).

To accomplish this mathematically, we use the absolute value function. Thus, the total distance traveled by the car is

\begin{array}{ll} \int_{0}^{2} | 60 | d t + \int_{2}^{5} | - 40 | d t & = \int_{0}^{2} 60 d t + \int_{2}^{5} 40 d t \\ = 120 + 120 \\ = 240 \end{array}

$\begin{array}{ll} \displaystyle\int_0^2 |60| dt + \displaystyle\int_2^5 |-40| dt & = \displaystyle\int_0^2 60 dt + \displaystyle\int_2^5 40 dt \\ & =120+120 \\ & =240 \end{array}$

Bringing these ideas together formally, we state the following definitions.

net signed area and total area

Let $f(x)$ be an integrable function defined on an interval $[a,b]$ . Let $A_1$ represent the area between $f(x)$ and the $x$ -axis that lies above the axis and let $A_2$ represent the area between $f(x)$ and the $x$ -axis that lies below the axis.

Then, the net signed area between $f(x)$ and the $x$ -axis is given by

\int_{a}^{b} f (x) d x = A_{1} - A_{2}

$\displaystyle\int_a^b f(x) dx = A_1-A_2$

The total area between $f(x)$ and the $x$ -axis is given by

\int_{a}^{b} | f (x) | d x = A_{1} + A_{2}

$\displaystyle\int_a^b |f(x)| dx = A_1+A_2$

Find the total area between $f(x)=x-2$ and the $x$ -axis over the interval $[0,6]$ .

Calculate the $x$ -intercept as $(2,0)$ (set $y=0$ , solve for $x$ ). To find the total area, take the area below the $x$ -axis over the subinterval $[0,2]$ and add it to the area above the $x$ -axis on the subinterval $[2,6]$ (Figure 8).

A graph of a increasing line f(x) = x-2 going through the points (-2,-4), (0,2), (2,0), (4,2), and (6,4). The area under the line in quadrant one and to the left of the line x=6 is shaded and labeled A1. The area above the line in quadrant four is shaded and labeled A2. — Figure 8. The total area between the line and the $x$ -axis over $[0,6]$ is $A_2$ plus $A_1$ .

We have

\int_{0}^{6} | (x - 2) | d x = A_{2} + A_{1}

$\displaystyle\int_0^6 |(x-2)| dx = A_2+A_1$

Then, using the formula for the area of a triangle, we obtain

A_{2} = \frac{1}{2} b h = \frac{1}{2} \cdot 2 \cdot 2 = 2

$A_2=\frac{1}{2}bh=\frac{1}{2} \cdot 2 \cdot 2=2$

A_{1} = \frac{1}{2} b h = \frac{1}{2} \cdot 4 \cdot 4 = 8

$A_1=\frac{1}{2}bh=\frac{1}{2} \cdot 4 \cdot 4=8$

The total area, then, is

A_{1} + A_{2} = 8 + 2 = 10

$A_1+A_2=8+2=10$ .

Starting at timestamp 11:34, watch the following video to see the worked solution to this example.

You can view the transcript for this segmented clip of “5.2 The Definite Integral” here (opens in new window).