what is the integral of (-1)^x from 0 to 1?
what is the integral of (-1)^x from 0 to 1?

I once saw a single blade from a turbofan jet engine in the New York City Museum of Modern Art (MOMA). It was an object of stunning beauty, an engineering and fabrication marvel, and I was compelled to stare at it for a long time. It was smooth and seemingly simple, yet at the same time clearly immensely complicated.

The jet engine fan blades on the left (below) and the single windmill blade on the right are three dimensional compound curves, designed for maximum efficiency and the immense strength needed to do their jobs.

These complicated shapes have to be modeled mathematically, input into simulators, tweaked when new information or new materials become available, and they have to be formed using computer controlled robots because they’re just too difficult for a free-hand machinist to make and reproduce.

Integral calculus and particularly the definite integral is the mathematics of such curves. It allows us to determine cross-sectional areas, volumes and surface areas of odd shapes, and much much more. Here we go …

Take your time with this section, there’s a lot to digest, but it’s worth your time. With a good baseline knowledge of the fundamentals of the integral, you’ll be able to apply it to solve some amazing problems, and even invent new kinds of functions.

We’re going to go at the definite integral first from a geometric perspective. That part is pretty easy; the integral represents the area between a curve and the x-axis (we’ll have to think about that more carefully later), and between domain points a and b, that is, in the interval [a, b].

We’ll calculate that area by dividing it up into small rectangles, and in the fashion of calculus, increasing our accuracy by decreasing the size and adding rectangles, even “to infinity” to get at the exact area. We’ll look at the geometric interpretation for some time before shifting to the analytic one for most of our future work.

In the analytic representation, we just have to introduce the limits of integration, a and b, attached to the integral sign, below.

We’ll develop the two-part Fundamental Theorem of Calculus in this section. It will allow us to take a huge shortcut in the calculation of the areas under and between complicated curves.

Our approach to finding the area under a curve will be to divide the area into rectangles of equal width, call that Δx. On the left (below), you can see that four rectangles, each of which is “pinned” to the function at its upper left corner, provide an OK estimate of the area, but it’s clearly an underestimate.

As we increase the number of rectangles by decreasing their width, our estimate becomes better and better. In the limit that each rectangle is infinitely narrow and there are infinitely many rectangles, it’s clear that we’ll arrive at the exact area under the curve.

We need to develop a mathematical language for the sum of all of these rectangular areas. It turns out we can do it in a number of subtly-different ways. Three are shown below. On the left, the height of each rectangle is specified by setting the height of its upper left corner to the value of the function there. In the other two panels, we set that height at the right corner and the middle of the top of the rectangle, respectively.

While it might be tempting (and probably correct) to suggest that the true area under the curve for a given number of rectangles might be something like an average of these three, we don’t really need to do that. It turns out that any of these approaches will work, so long as we let the width of the rectangles shrink to zero and their number grow to infinity.

When we add these areas, the sums will be called the left-, right- and middle Riemann Sums, respectively, after the 19th century mathematician who developed this way of thinking.

Using the Left-Riemann picture, we set the width of each rectangle to Δx, where Δx = (b-a)/N, and N is the number of rectangles. The heights of the rectangles are f(a), f(a + Δx), f(a + 2Δx), and so on. The sum is:

$$L_N = \sum_{i = 0}^{N – 1} \, f(x + i\Delta x) \, \Delta x$$

where L stands for “left.”

Similarly, we can define the Right- and Middle-Riemann sums:

$$R_R = \sum_{i = 1}^N \, f(x + i\Delta x) \, \Delta x$$


$$M_N = \Delta x \sum_{i = 1}^N \, f\left( x + \left( i – \frac{1}{2} \right) \right) \Delta x$$

Now we formally suggest that as n → ∞, all of these sums approach the area under the curve so divided into rectangles:

$$\lim_{n\to\infty} L_N = \lim_{n\to\infty} R_N = \lim_{n\to\infty} M_N$$

and that this limit is the exact area under the curve. We use a new notation to represent that limit at infinity:

$$\lim_{N\to \infty} \sum_{i=1}^N \, f(x + i\Delta x) \, \Delta x = \int_a^b f(x) \, dx,$$

where $dx$ is an infinitessimally-small bit of the x-coordinate, a and b are the lower and upper bounds of our sum, and the symbol $\int$ represents this new kind of infinite sum.

$$\color{#E90F89}{\bf{1.}} \int_0^b x^2 \, dx \phantom{00000000} \color{#E90F89}{\bf{2.}} \int_0^b x^1 \, dx \phantom{00000000} \color{#E90F89}{\bf{3.}} \int_0^b x^0 \, dx$$

We need to work through one example of Riemann-sum integration of a “complicated” function in order to see how it works. Doing two more carefully-chosen examples will help make the transition between Rieman integrals and calculating integrals using the fundamental theorem of calculus more clear. We’re working here toward making the transition between the geometric and analytic interpretations of the definite integral.

The first of these examples will take just a bit, so have a beverage and get ready. The second two are very easy and will take about a minute each to understand. The pattern that emerges will tell us a valuable story. We’re letting the lower limit of integration be zero for convenience, but we can show later that it can be any number. Here we go …

The width of each rectangle is b/N, which will approach zero as N→ ∞. The height of the first three rectangles are shown. The sum of these areas for a fixed N is:

$$R_R = \Delta x \sum_{i = 1}^N \, f(x + i\Delta x)$$

Now, in a couple of steps, we’ll plug in what we know from the diagram (remembering that f(x) = x2):

$$ \begin{align} A &= \frac{b}{N} f\left( \frac{b}{N} \right) + \frac{b}{N} f\left( \frac{2b}{N} \right) + \frac{b}{N} f\left( \frac{3b}{N} \right) + \dots + \frac{b}{N} f\left( \frac{Nb}{N} \right) \\ \\ &= \frac{b}{N}\left( \frac{b}{N}\right)^2 + \frac{b}{N}\left( \frac{2b}{N}\right)^2 + \frac{b}{N}\left( \frac{3b}{N}\right)^2 + \dots + \frac{b}{N}\left( \frac{Nb}{N}\right)^2 \end{align}$$

Recognizing that (b/N)3 is in every term, we can simplify this to:

$$A = \left( \frac{b}{N} \right)^3 \left( 1^2 + 2^2 + 3^2 + \dots + N^2 \right)$$

Now what to do about the infinite sum, 12 + 22 + 32 + . . . + N2 ? First, it’s not going to be infinite if it’s divided by N3 (which is hidden in that (b/N)3 term), so there’s some hope that our formula for the area A will be bounded (won’t be infinite). In order to figure this out, we can take a well known digression … to Egypt, if you will.

Consider a pyramid built of 1 x 1 blocks. A top and side view are shown below. The volume of the base, a square layer of 1 x 1 blocks, is N blocks on a side, with volume N2. The second layer up has volume (N-1)2, the third (N-2)2, and so on.

The volume of a general square pyramid, of the pyramid formed by tracing the lower corners (green lines) of each layer and the upper corners (red lines) of each layer are shown. Notice that the actual volume of the pyramid (which is the sum 12 + 22 + 32 + . . . + N2, must be less than the volume of the red pyramid and greater than that of the green one.

That notion gives us the inequality

$$\frac{1}{3} n^3 \lt 1^2 + 2^2 + \dots + N^2 \lt \frac{1}{3}(N + 1)^3$$

Dividing by N3 and making a slight rearrangement of the right side gives:

$$\frac{1}{3} \lt \frac{1^2 + 2^2 + \dots + N^2}{N^3} \lt \frac{1}{3} \left( \frac{N + 1}{N} \right)^3 \color{#E90F89}{\rightarrow} \frac{1}{3} \left(1 + \frac{1}{N} \right)^3$$

Now if we take the limit as N→ ∞, the limit on the right is just 1/3:

$$\frac{1}{3} \lt \lim_{N \to\infty} \frac{1^2 + 2^2 \dots + N^2}{N^3} \lt \frac{1}{3}$$

The limit is “squeezed between 1/3 and 1/3, so it has to equal 1/3. Well, all of that pyramid stuff led us right to what we needed. The value of that sum divided by N3 is just 1/3.

Now we’ve shown that

$$\int_0^b \, x^2 \, dx = \frac{b^3}{3}$$

It took a while, but we won’t have to do any (many?) more examples of full-on Riemann integration like this, but first … our other two examples.

This definite integral is considerably simpler because we know that the area of a triangle (this is a right triangle) is

$$A = \frac{1}{2}bh$$

Because f(b) = b, we have

$$A = \frac{b^2}{2}$$

… and that’s it. That’s the integral. Now for the third and final example.

The area under this simple curve is just 1 x b = b.

Also notice that if we made the left hand limit of this integration a number, a, instead of zero, that area would be A = b – a.

This table lists our results. In the second column, the endpoint b is removed and the constant of integration is added so you can easily see that what we’ve come up with are just the indefinite integrals of the functions f(x). In the rightmost column is the exact area of each region, the area between the function and the x-axis between x=0 and x=b.

Because of our convenient choice of zero for our starting point (a) we’ve overlooked an important part of the pattern, but we’ll recover that below.

The main result is that doing a Riemann sum is the same as finding an indefinite integral, then evaluating it at the limits of integration, a & b.

This is a remarkable result, and worth repeating. The value of a definite integral, which represents the area under a curve between two points in the function domain, is calculated by performing a Riemann sum. However, that process can be cut short by finding the indefinite integral or antiderivative of the function, F(x) (dropping the constant of integration) and evaluating the result between the two limits of integration: F(b) – F(a).

This result is summed up by one part of the fundamental theorem of calculus, which is covered in another section.

The relevant part of the FTOC is given in the box below.

If f(x) is a continuous function on the interval [a, b], and F(x) is an antiderivative of f(x), then the definite integral of f(x) between x=a and x=b is

$$\int_a^b f(x) \, dx = F(b) – F(a)$$

You can learn more about the fundamental theorem of calculus (there are two parts; this is just one) here.

When writing the result of a definite integral, in which we need to evaluate an antiderivative at x = b and subtract the corresponding value at x = a, it’s customary to use the “bar” notation shown here. The bar means “evaluate at the top number and the bottom number, then subtract the lower result from the upper.

$$ \begin{align} \int_a^b \, f(x) \, dx &= F(b) = F(a) = F(x) \bigg|_a^b \\[5pt] &= F(x) \, \bigg|_{x = a}^{x = b} \end{align}$$

Calculate the area under $f(x) = x^2$ between x = 3 and x = 5.

$$ \begin{align} \int_3^5 x^2 \, dx &= \frac{x^3}{3} \bigg|_3^5 \\[5pt] &\color{#E90F89}{\frac{x^3}{3} \: \text{ is the antiderivative of} \: x^2} \\[5pt] &= \frac{5^3}{3} – \frac{3^3}{3} = \frac{98}{3} \approx 33 \: \text{units}^2 \end{align}$$

Calculate the area under $f(x) = x^3$ between x = 3 and x = 5.

Now compare that last integral with the definite integral of f(x) = x3 between x=3 and x=5. The procedure is the same, just find the antiderivative of x3, F(x), then evaluate between the limits by subtracting F(3) from F(5).

Also notice in this example that x3 > x2 for all positive x, and the value of the integral is larger, too.

$$ \begin{align} \int_3^5 x^3 \, dx &= \frac{x^4}{4} \bigg|_3^5 = \frac{5^4}{4} – \frac{3^4}{4}\\[5pt] &\color{#E90F89}{\frac{x^4}{4} \: \text{ is the antiderivative of} \: x^3} \\[5pt] &= \frac{544}{4} \approx 136 \: \text{units}^2 \end{align}$$

Calculate the area under $f(x) = -2x^2 + 2x + 4$

$$ \begin{align} \int_{-1}^2 &(-2x^2 + 2x + 4) \,dx \\ &= \frac{-2x^3}{3} + x^2 + 4x \bigg|_{-1}^2 \\ &= \frac{-16}{3} + 4 + 8 – \left(\frac{2}{3} + 1 – 4 \right) \\ &= \bf 9 \end{align}$$

All of the properties of the indefinite integral, including that the integral of a sum is the sum of integrals of the parts (graph), and that a multiplicative constant can be removed from the integral, still hold (see the table below).

The definite integral, however, has limits, and those bring in new properties. An important one is illustrated on the right. If c lies in [a, b], then the integral of a function between a and b can be broken into the sum of integrals from a to c and from c to b.

The negation property is interesting. If F'(x) = f(x) in the example in the table, then we are comparing F(b) – F(a) to F(a) – F(b). Makes sense.

If we try to find the definite integral of the sine function between 0 and 2π, the answer is zero:

This begs the question: What do we mean by “area under the curve”? Between 0 and π, this integral has a value of 1, the area under that part of the curve. If we calculate the integral between π and 2π using the fundamental theorem of calculus, however, we get -1, and the sum is zero.

Not to worry , there are two different ways to think about this integral and they’re both right, depending on the question you’re asking. First, let’s focus on area between the curve and the x-axis. This suggests that “negative” areas simply need to be subtracted from positive areas (or alternatively, we used absolute values), like this:

$$\int_0^{2\pi} sin(x) \,dx = \int_0^{\pi} sin(x) \, dx – \int_{\pi}^{2\pi} sin(x) \, dx$$

The antiderivative of sin(x) is -cos(x), so our integral is

$$= – cos(x) \, \bigg|_0^{\pi} + cos(x) \, \bigg|_{\pi}^{2\pi}$$

Evaluation at the limits gives

$$= cos(\pi) – cos(0) – cos(2\pi) + cos(\pi)$$

The resulting total area is

$$= 1 + 1 + 1 + (-1) = 2 \: units^2$$

The other way we can think about such an integral is as a pure sum. Perhaps the curve represents money gained and lost over time, and the sum should be zero. In that case,

$$\int_0^{2\pi} sin(x) \, dx = 0$$

In many cases we just do the straightforward integral like this, but be careful if what you’re after is total area; you may have to divide the domain and integrate to obtain the absolute value of the area.

Integrate $f(x) = x^3 – 4x^2 – 4x + 16$ between -2 and 4 to determine the shaded area.

Here’s the calculation. Don’t lose sight of the fact that calculating odd areas like this

amounts to little more than finding an easy antiderivative and doing some arithmetic. The integral is

$$ \begin{align} &\int_{-2}^2 (x^3 – 4x^2 – 4x + 16) \, dx \\ -&\int_2^4 (x^3 – 4x^2 – 4x + 16) \, dx \end{align}$$

Then we evaluate the limits:

$$ \begin{align} = \frac{x^4}{4} – \frac{4x^3}{3} – 2x^2 + 16x \bigg|_{-2}^2 \\ \\ – \left( \frac{x^4}{4} – \frac{4x^3}{3} – 2x^2 + 16x \bigg|_2^4 \right) \\ \\ = \bf 49.33 \: units^2 \end{align}$$

Perform each definite integration and evaluate the result between the limits.

1. $$\int_{-2}^3 \, 8x^3 \, dx$$
2. $$\int_{-1}^1 \, (t^2 – 2) \, dt$$
3. $$\int_1^2 \, \left( \frac{3}{x^2} – 1 \right) \, dx$$
4. $$\int_0^{\pi} \, (1 + sin(t)) \, dt$$
5. $$\int_{\pi/16}^{\pi/8} \, csc^2 (4t) \, dt$$
6. $$\int_2^5 \, (2x + 1)^{-2} \, dx$$
7. $$\int_3^7 \, \sqrt{2x + 3} \, dx$$
8. $$\int_0^4 \, \frac{6}{\sqrt{3x + 4}} \, dx$$
9. $$\int_0^1 \, 2x^2 (x^3 + 5)^4 \, dx$$
10. $$\int_1^3 \, (x + 3)^3 \, dx$$
11. $$\int_0^1 \, \frac{x}{(x^2 + 1)^3} \, dx$$
12. $$\int_0^4 \, x \sqrt{x^2 + 9} \, dx$$
13. $$\int_0^{\pi/2} \, cos^3(x) sin(x) \, dx$$
14. $$\int_0^1 \, \theta tan(\theta^2) \, d\theta$$
15. $$\int_0^2 \, x \sqrt{5 – \sqrt{4 – x^2}} \, dx$$
16. $$\int_{\pi/3}^{\pi/2} cot^2 \left( \frac{x}{2} \right) csc^2 \left( \frac{x}{2} \right) dx$$

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