]> IB HL2 - Chapter 04

Integration

4.1: Antiderivatives and Indefinite Integration

The Idea

You now can take a function and find its derivative—can you take a derivative and find the function? Except for a minor detail, I should hope so—just reverse the formulas that you know! You know that the derivative of $y = x^{2}$ is $y^{'} = 2 x$ ; thus, if I ask which function has $y = 2 x$ as its derivative, you should be able to come up with $y = x^{2}$ .

Now for the detail: what's the derivative of $y = x^{2} + 1$ ? How about $y = x^{2} - 5$ ?

Aha! There are a lot of functions whose derivative is $y = 2 x$ …what do these functions have in common? What's the only thing that "disappears" when a derivative is taken?

Constants! So, when we take an antiderivative, the only thing we don't know is the value of any constant on the end. Let's define some notation and make all of this more rigorous.

Basic Rules

Let the derivative of $F (x)$ be $f (x)$ . Then the antiderivative of $f (x)$ is $F (x) + C$ . Notation: $\int f (x) d x = F (x) + C$ . In this, $f (x)$ is the integrand, $d x$ is the variable of integration (read "dee ex," or "with respect to x."), and C is the constant of integration. The symbol ∫ is the integral symbol—antidifferentiation is also called integration.

There is a nice list of basic antiderivatives/integrals on page 243.

Initial Conditions

If more information is given, then an integral can be made more specific—in particular, the constant of integration can be determined. The required information is an initial condition—for example, a point on the original function ( $F (x)$ ). Knowing a bit more information will allow you to set up another equation, which you can solve to find the value of C.

Examples

[1.] Find the antiderivative of $y = 6 x^{2} + 2 x - 1$ .

Work it backwards—instead of lowering the exponent, raise it…and instead of multiplying by the old exponent, divide by the new exponent. Thus, $\int (6 x^{2} + 2 x - 1) d x = \frac{1}{3} 6 x^{3} + \frac{1}{2} 2 x^{2} - x + C = 2 x^{3} + x^{2} - x + C$ .

[2.] $\int (cos x) d x = ?$

This must be memorized…there is no pattern (like there is for polynomials). $\int (cos x) d x = sin x + C$ .

[3.] Find $f (x)$ if $f^{'} (x)$ and $f (e) = - 1$ .

Well, $\int \frac{1}{x} d x = ln x + C$ . Since $f (e) = - 1$ , $ln (e) + C = - 1 ⇒ C = - 1 - ln (e) = - 1 - 1 = - 2$ . Thus, $f (x) = ln (x) - 2$ .

4.2: Area

The Idea

Finding the area of regular geometric shapes isn't terribly difficult…finding the area under non-regular shapes can be daunting! Calculus to the rescue!

Well, let's do what people used to do—estimate with regular shapes! For example, let's try to find the area under the curve $y = x^{2} + 1$ between x = 0 and x = 2. And, let's use rectangles to estimate the area. Let the first rectangle have a base from x = 0 to x = 1, and the second from x = 1 to x = 2. For both rectangles, use the function value on the right edge for the height.

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The area is 7—that's too much, though…maybe if we use more rectangles. How about 4?

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Now the area is 5.75…better, but still too much. How about 10 rectangles?

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That gives an area of 5.08. Hmmm…I wonder what I'd get with 20 rectangles?

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Now I've got an area of 4.87. Getting closer! Those rectangle tops are getting closer and closer to the curve…okay, one more: 50 rectangles!

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That's an area of 4.7472—and there's hardly any bits of rectangles above the curve.

Upper/Lower Sums

We did this with rectangles as high as the right edge—for this function, that gave us the maximum function value on that subinterval. That produces the upper sum—an estimate of the area that is too large. It is also possible to use lower sums—to set the rectangles to the minimum function value on the interval. Here's a picture:

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This estimate (4.28) is too low. But, as with upper sums, the more rectangles, the more accurate the estimate…

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…in this case, 4.5872.

It turns out that the upper and lower sums approach each other as the number of intervals increases…

Take it to the Limit!

Let's formalize what we've done: I've got a function, $f (x)$ , and I'm looking at that function on an interval—let's call it ]a, b[—and I've divided the interval into n subintervals, each of width $\frac{b - a}{n}$ —let's call that Δx. If the left edge of a rectangle is located at x, then the area of that rectangle is $f (x) \cdot Δ x$ (This corresponds with the left edge of the rectangle in our example—a lower sum. It won't matter in a moment). Thus, the total area is $\sum_{x = a}^{b - Δ x} f (x) \cdot Δ x$ . Of course, the accuracy of this interval depends on the number of rectangles—on the magnitude of Δx. Thus, we really want to know what value is approached as Δx approaches zero. So the area under the curve is $lim_{Δ x → 0} [\sum_{x = a}^{b - Δ x} f (x) \cdot Δ x]$ (the -Δx doesn't matter when Δx is close to zero).

Note that this may be positive or negative—thus, it might not actually be the area under a curve…

4.3: Riemann Sums and Definite Integrals

Isn't this familiar?

It turns out that you don't have to divide the area under the function into regular (even; equally spaced) strips…you can divide it into any number of irregular strips! This is called a partition.

Okay; so let Δ be a partition of the interval ]a, b[—in other words, $Δ = {x_{0}, x_{1}, x_{2} \dots x_{n}}$ , where x ₀ = a and x _n = b. For the interval $[x_{i - 1}, x_{i}]$ , let $Δ x_{i}$ be the length, and c _i be any x-value in the interval. The sum $\sum_{i = 1}^{n} f (c_{i}) \cdot Δ x_{i}$ is a Riemann Sum—which is merely a generalization of the upper and lower sums from the previous section.

Definite Integrals

So—let $|| Δ ||$ be the length of the largest interval in the partition Δ. The limit $lim_{|| Δ || → 0} [\sum_{i = 1}^{n} f (c_{i}) \cdot Δ x_{i}]$ will be the same as the upper and lower limits, and we define $lim_{‖Δ‖ → 0} [\sum_{i = 1}^{n} f (c_{i}) \cdot Δ x_{i}] = \int_{a}^{b} f (x) d x$ . This is called a definite integral, with lower limit a and upper limit b.

Properties

Since a definite integral is based on limits, the properties of limits (which we've used with derivatives) still apply…constants can be factored out, and the integral of a sum (or difference) can be separated into two separate integrals. In addition, $\int_{a}^{a} f (x) d x = 0$ , $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ , and $\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$ .

4.4: The Fundamental Theorem of Calculus

The (First) Fundamental Theorem of Calculus

Perhaps you've wondered why we used the same symbol for antidifferentiation and definite integrals…now for the answer!

If $f (x)$ is continuous on the interval [a, b], and $F (x)$ is an antiderivative of $f (x)$ on the interval [a, b], then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .

Whoa! So antidifferentiation not only "undoes" derivatives, it also finds the area under a curve!

The Mean Value Theorem for Integrals

For any interval, there exists some magic spot where the height of the function (at that spot) can be used to create a rectangle with the same area as the area under curve.

If $f (x)$ is continuous on the interval [a, b], then there exists a value c in the interval so that $\int_{a}^{b} f (x) d x = f (c) (b - a)$ .

On any closed interval, a function has a definite range of values. It turns out that calculus can be used to find the average of those values: $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ .

The Second Fundamental Theorem of Calculus

If $f (x)$ is continuous on some open interval I, where a ∈ I, then ∀ x ∈ I, $\frac{d}{d x} [\int_{a}^{x} f (t) d t] = f (x)$ .

This will be particularly useful in the next chapter…

Examples

[4.] Find $\int_{1}^{3} (9 x^{2} - 4 x + 1) d x$ .

$\int_{1}^{3} (9 x^{2} - 4 x + 1) d x = {(\frac{9 x^{3}}{3} - \frac{4 x^{2}}{2} + x)]}_{1}^{3} = {3 x^{3} - 2 x + x]}_{1}^{3}$ = $(3 {(3)}^{3} - 2 (3) + 3) - (3 {(1)}^{3} - 2 (1) + 1) = (81 - 6 + 3) - (3 - 2 + 1) = 78 - 2 = 76$

[5.] Find the area under the function $f (x) = x^{3} - 4 x$ on the interval [-2, 0].

The area is $\int_{- 2}^{0} (x^{3} - 4 x) d x = {\frac{x^{4}}{4} - 2 x^{2}]}_{- 2}^{0} = (0) - (\frac{16}{4} - 2 \cdot 4) = 4$ .

[6.]Find the average value of the function $g (x) = \frac{1}{x}$ on the interval [1, 2].

The average value of a function is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ ; in this case, $\frac{1}{2 - 1} \int_{1}^{2} \frac{1}{x} d x = \frac{1}{1} {[ln x]}_{1}^{2} = ln (2) - ln (1)$ .

4.5: Integration by Substitution

The Chain Rule allowed us to take the derivative of compositions—there is a similar (reversed) process for integrals. In particular, $\int f (g (x)) g^{'} (x) d x = F (g (x)) + C$ . The key to using this is to recognize $g (x)$ ; usually, we make a substitution— $u = g (x)$ , which makes $\frac{d u}{d x} = g^{'} (x) ⇒ d u = g^{'} (x) d x$ . This changes the integral above to $\int f (u) d u = F (u) + C$ .

If you're working with a definite integral, then be careful—the limits of integration change when you substitute! In particular, $\int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (u) d u$ .

The trick is to see a composition of functions multiplied by the derivative of the inner function. When the inner function is linear, this will be easy, since its derivative is a constant—and constants can be created at will!

Examples

[7.] Find $\int x \cdot \sin (2 x^{2}) d x$ .

The composition should be easy to see— $sin x$ and $2 x^{2}$ . So $u = 2 x^{2}$ and $d u = 4 x \cdot d x$ . Note that we don't have a coefficient of 4 in the integral—since it's a constant, we can throw it in easily; $\frac{d u}{4} = x \cdot d x$ . So, that makes $\int x \cdot \sin (2 x^{2}) d x = \int \sin (2 x^{2}) \cdot x \cdot d x = \int \sin (u) \cdot \frac{d u}{4} = \frac{1}{4} \int \sin (u) d u$ . Time to finish! $\frac{1}{4} \int \sin (u) d u = \frac{1}{4} (- \cos (u)) + C = - \frac{1}{4} \cos (2 x^{2}) + C$ .

[8.] Find $\int \frac{2 x}{x^{2} + 1} dx$ .

So—where's the composition? Not the 2x, I think—how about $\frac{1}{x^{2} + 1}$ ? That's a composition of $\frac{1}{x}$ and $x^{2} + 1$ . So, let $u = x^{2} + 1$ . That makes $d u = 2 x \cdot d x$ . Substitute: $\int \frac{2 x}{x^{2} + 1} d x = \int \frac{1}{u} d u = ln | u | + C = ln (x^{2} + 1) + C$ .

[9.] Find $\int_{1}^{0} x^{3} {(1 + x^{4})}^{3} d x$ .

Another simple composition—but this is a definite integral! If $u = 1 + x^{4}$ and $d u = 4 x^{3} d x ⇒ \frac{d u}{4} = x^{3} d x$ , then the limits of integration change! In particular, when x = 0, u = 1, and when x = 1, u = 2. So $\int_{1}^{0} x^{3} {(1 + x^{4})}^{3} d x = \int_{1}^{0} {(1 + x^{4})}^{3} \cdot x^{3} d x = \int_{1}^{2} u^{3} \cdot \frac{d u}{4}$ . Finish it up! $\frac{1}{4} \int_{1}^{2} u^{3} d u = \frac{1}{4} {[\frac{u^{4}}{4}]}_{1}^{2} = \frac{1}{4} [\frac{16}{4} - \frac{1}{4}] = \frac{15}{16}$ .

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