EPPS Math and Coding Camp

Integral

Instructor: Prajyna Barua and Azharul Islam

List of Rules of Integration

Fundamental theorem of calculus	\(\int_a^b f(x)\,dx = F(b) - F(a)\)
Rules for bounds	\(\int_a^b f(x)\,dx = - \int_b^a f(x)\,dx\)
	\(\int_a^a f(x)\,dx = 0\)
	\(\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx\) for \(c \in [a, b]\)
Linear rule	\(\int (a f(x) + b g(x))\,dx = a \int f(x)\,dx + b \int g(x)\,dx\)
Integration by substitution	\(\int_a^b f(g(u))g'(u)\ du = \int_{g(a)}^{g(b)} f(x)\,dx\)
Integration by parts	\(\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx\)

Power rule 1	\(\int x^n \,dx = \frac{x^{n+1}}{n+1} + C\) if \(n \neq -1\)
Power rule 2	\(\int x^{-1}\,dx = \ln \|x\| + C\)
Exponential rule 1	\(\int e^x\,dx = e^x + C\)
Exponential rule 2	\(\int a^x\,dx = \frac{a^x}{\ln(a)} + C\)
Logarithm rule 1	\(\int \ln(x)\,dx = x \ln(x) - x + C\)
Logarithm rule 2	\(\int \log_a(x)\,dx = \frac{x \ln(x) - x}{\ln(a)} + C\)
Trigonometric rules	\(\int \sin(x)\,dx = -\cos(x) + C\)
	\(\int \cos(x)\,dx = \sin(x) + C\)
	\(\int \tan(x)\,dx = -\ln \|\cos(x)\| + C\)
Piecewise rules	Split definite integral into corresponding pieces

7.1 THE DEFINITE INTEGRAL AS A LIMIT OF SUMS

Use of integration

What if instead we were concerned with the area under the curve?

That is, we want to know what the area is between the curve and the x-axis, from some point \(x_1\) to another point \(x_2\).

As we alluded to earlier in this chapter, calculating such areas is central to statistical inference, as we will see in Part III. If we let \(x_1 = 1\) and \(x_2 = 2\) and use \(f(x) = x^2\), then we can visualize this area by dropping lines down from \(f(x)\) at the two endpoints of the secant in the figure in Chapter 5. We show this in this figure, in which the shaded area is the area of interest.

Area under \(y = x_2\) from \(x = 1\) to \(x = 2\)

Definition

The integral symbol \(\int\) looks like an \(S\) and is meant to, to remind you that it is in essence a sum.
The \(dx\) in the integral tells you the variable of integration.
It is exactly analogous to the \(dx\) in the derivative. The f(x), or more generally, the expression multiplying the \(dx\), is known as the integrand. The \(a\) and \(b\) are the bounds (or limits) of integration.
They tell you the value of \(x\) at which to start and the value at which to stop.
Unlike the derivative, which can be represented using a variety of notation, this is the single, common, agreed-upon notation for the integral.

For our case, \(x^2\) is the integrand, \(a = 1\) and \(b = 2\) are the bounds of integration, and the definite integral is \(\int_1^2 x^2 dx\)

7.2 INDEFINITE INTEGRALS AND THE FUNDAMENTAL THEOREM OF CALCULUS

Well see that the derivative and the antiderivative (aka indefinite integral) are similarly related.

7.2.1 Antiderivatives and the Indefinite Integral

Take \(F(x) = x + 10\). The derivative of that is also one, as is the derivative of \(F(x) = x + 1,000\).

In general, because the derivative of a constant is zero, there are lots (an infinite number, actually) of antiderivatives that all produce the same \(f(x)\).

Luckily, they’re all of the same form: \(F(x) = x + C\). We call \(C\) the constant of integration, and it can be any constant value.

So, if \(f(x) = x\), then we try to figure out what function \(F(x)\), when differentiated, yields \(x\).

Well, we know that the derivative of \(x^2\) is \(2x\), so that’s pretty close. We also know that the derivative is linear, so we can divide \(x^2\) by \(2\) to get a derivative of \(x\).

Thus, the antiderivative of \(x\) is \(\frac{1}{2}x^2 + C\). (Don’t forget the \(C\)!)

For a slightly more complicated example, consider \(\frac{1}{x}\).

There’s no polynomial that when differentiated produces this; check for yourself.

What about other functions? While \(e^x\) certainly doesn’t work, its inverse function, \(\ln(x)\), does.

Since \((\ln(x))' = \frac{1}{x}\), the antiderivative of \(\frac{1}{x}\) is \(\ln|x| + C\).

We write an indefinite integral as \(\int f(x)dx = F(x)\), so it looks the same as a definite integral without the bounds on the integral.

The difference is that the definite integral returns a value, the area under the curve, while the indefinite integral returns a function that, when differentiated, reproduces the integrand. In symbols, \(\frac{d \int f(x)dx}{dx} = f(x)\).

7.2.2 The Fundamental Theorem of Calculus

\[ \int_a^b f(x)dx=F(b)-F(a) \]

7.3.5 The Integral Is Also a Linear Operator

More formally, to say that the integral is also a linear operator is to say that \(\int(af(x) + bg(x))dx = a\int f(x)dx + b\int g(x)dx\).

To show this we need to use the definition of the antiderivative. The proof follows from this definition and the linearity of the derivative discussed in the previous chapter.

Example:

All polynomials can be tackled with this rule. For instance, consider \(f(x) = 4x^5 + 2x^2 + 5\). Linearity implies we can treat each term separately.

The integral \(\int x^5dx = \frac{x^6}{6} + C\),

so \(4\int x^5dx = \frac{2x^6}{3} + C\).

The integral \(\int x^2dx = \frac{x^3}{3} + C\),

so \(2\int x^2dx = \frac{2x^3}{3} + C\).

7.3.6 Integration by Substitution

If \(x = g(u)\), then \(f(x) = f(g(u))\), which is a composite function. Integration by substitution says that

\[\int_{a}^{b} f(g(u))g'(u)du = \int_{g(a)}^{g(b)} f(x)dx\]

First consider the function \(f(x) = \frac{1}{2\pi}xe^{-\frac{x^2}{2}}\). This function is the probability distribution function of a standard normal distribution, multiplied by \(x\).

Its integral is \(\int \left(\frac{1}{2\pi}xe^{-\frac{x^2}{2}}\right)dx\), which happens to be the expected value of \(x\) in a standard normal distribution. We’ll see this more in Chapter 11.

This integral looks complicated, but it turns out that it reduces quite easily via substitution.

The composite function here is \(e^{-\frac{x^2}{2}}\), which we don’t know how to integrate.

But we do know how to integrate \(e^u\), so let’s set \(u = g(x) = -\frac{x^2}{2}\) and see what happens.

We can use the power rule to see that \(g'(x) = -x\), which means we can rewrite the integral as \(\int \left(\frac{-1}{2\pi}g'(x)e^{g(x)}\right)dx\).

Integration by substitution implies that this integral is the same as \(\int \left(\frac{-1}{2\pi}e^u\right)du\), which just equals \(\frac{-1}{2\pi}e^u + C = \frac{-1}{2\pi}e^{-\frac{x^2}{2}} + C\). So we’ve completed our integral.

For example, to compute \(\int x^4e^{x^5}dx\),

we set \(u = g(x) = x^5\),

so \(dx = (g'(x))^{-1}du = \frac{1}{5x^4}dx\),

and thus \(\int x^4e^{x^5}dx = \int x^4e^{x^5} \frac{1}{5x^4}du = \frac{1}{5}\int e^udu\).

This is just \(\frac{1}{5}e^u + C = \frac{1}{5}e^{x^5} + C\).

For example, let’s say we had \(\int_{1}^{3} 3x^2e^{x^3}dx\), i.e., our recent example made into a definite integral.

With \(u = g(x) = x^3\), the integral becomes \(\int_{g(1)}^{g(3)} e^udu = \int_{1}^{27} e^udu = e^{27} - e^1\).

7.3.7 Integration by Parts

Integration by parts states that \[ \int f(x) g'(x) \, dx = f(x) g(x) - \int f'(x) g(x) \, dx. \]

Basically, the rule allows you to change around your functions. If you can’t integrate \(f(x)g'(x)\) but you can integrate \(g'(x)\) and \(f'(x)g(x)\), it’s very useful.

Practically, this occurs when \(g'(x)\) is something like \(e^x\) that does not get more complex when you integrate it, while \(f(x)\) is something like \(x\) that simplifies when you differentiate it. We’ll see this shortly in the examples. First, the proof, which is straightforward.

The product rule states that \((f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\)

Now for why we’d use it. Let’s start with \(\int x e^x dx\).

We set \(f(x) = x\), \(g'(x) = e^x\), and the rule gives us \(\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C\).

We could in theory do this multiple times for more complicated functions.

For example, \(\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx + C\). We can then use our previous example of integration by parts to complete the problem: \(\int x^2 e^x \, dx = x^2 e^x - 2(x e^x - e^x) + C\).

Another example: \(\int x(x+1) \frac{3}{2} \, dx\)