Definite integrals

This is about the basics of evaluating definite integrals, and surprisingly how they can be used to define functions.

Evaluation
Properties
Variable of integration
Using definite integral to define a function
- Derivation: z-scores of normal distribution
Notes on trig

Evaluation

Suppose $F(x)$ is an antiderivative of $f(x)$ , in this course the definite integral is evaluated using

\int_{x_1}^{x_2} f(x)\d x = F(x_2) - F(x_1)

It can be equivalently notated as

\int_{x_1}^{x_2} f(x)\d x = F(x)\Big\vert_{x_1}^{x_2}

And in cases to be absolutely clear, this can also be written as

\int_{x_1}^{x_2} f(x)\d x = F(x)\Big\vert_{x = x_1}^{x = x_2}

$x_1$ and $x_2$ are the limits of integration.

Connecting back to indefinite integrals, we customarily set the $+C$ to $0$ , but it does not matter as by taking the difference, it cancels.

Example: Evaluate

\int_{- \e^2}^{-1}\frac 1x \d x

\begin{align*}\int_{- \e^2}^{-1}\frac 1x \d x &= \ln\lvert x\rvert \Big\vert_{- \e^2}^{-1} \\ &= \ln\lvert -1\rvert - \ln\lvert - \e^2\rvert \\ &= \ln 1 - \ln \e^2 \\ &= 0 - 2 = -2 \qed\end{align*}

Where differentiation is about instantaneous change, definite integration is about the accumulation of change.

Definite integrals are defined as the signed area between a curve and the $x$ -axis. In this course, it is sufficient to use antiderivatives to find the area under a continuous function.

Properties

These properties assume the function is defined for all $x$ within the limits of integration, or is discontinuous (HL) at a finite number of points.

\int_{x_1}^{x_2} f(x)\d x = -\int_{x_2}^{x_1} f(x)\d x

\int_{x_1}^{x_2} f(x)\d x + \int_{x_2}^{x_3} f(x)\d x = \int_{x_1}^{x_3} f(x)\d x

The properties from indefinite integrals also carry over.

\int_{x_1}^{x_2} f(x) + g(x)\d x = \int_{x_1}^{x_2} f(x) \d x + \int_{x_1}^{x_2} g(x) \d x

\int_{x_1}^{x_2} k f(x)\d x = k\int_{x_1}^{x_2} f(x)\d x

With $F(x)$ being an antiderivative of $f(x)$ ,

\int_{x_1}^{x_2} f(ax+b)\d x = \frac 1a\int_{ax_1+b}^{ax_2+b} f(x)\d x = \frac 1a F(ax_2+b) - \frac 1a F(ax_1+b)

The last property comes directly from substituting in $x_1$ and $x_2$ into the chosen antiderivative. This as a $u$ -substitution.

Example: Find

\int_0^1 (4x+2)^3 \d x

Here we need to apply both integration of $x^n$ and $f(ax + b)$ .

The antiderivative of $x^3$ is $\frac{1}{4}x^4 + C$

Then the $f(4x + 2)$ means we need to divide by another $4$ .

\begin{align*} \int_0^1 (4x+2)^3 \d x &= \frac {1}{16}(4x+2)^4 \Big\vert_0^1 \\ &= \frac{1}{2^4}(6^4 - 2^4) \\ &= 3^4 - 1 \\ &= 80 \qed \end{align*}

Alternatively we could have factored out a $2^3$ early on

\begin{align*} \int_0^1 (4x+2)^3 \d x &= 8\int_0^1 (2x+1)^3 \d x \\ &= 8\cdot\frac{1}{2\cdot 4}\left[(2x+1)^4\right]_0^1 \\ &= 81 - 1 \\ &= 80 \qed \end{align*}

which in retrospect is easier, but not by a lot.

Variable of integration

A key difference between indefinite and definite integration is that indefinite integrals return families of functions, but definite integration return a value. Just like how in $f(x)$ , $x$ just means the input, the $x$ in definite integrals is very much a placeholder in the same way.

As such all of the following are equivalent

\int_{-3}^{5} f(x)\d x

\int_{-3}^{5} f(w)\d w

\int_{-3}^{5} f(\tau)\d \tau

Here $w$ , $x$ , and $\tau$ are called dummy variables or variables of integration.

Note to HL candidates: for $u$ -substitution, typically the limits of integration change. Here, they do not change because the substitutions are $w = x$ and $\tau = x$

Using definite integral to define a function

There is a way to specify the antiderivative function $F(x)$ by expressing it as a definite integral.

F(x) = \int_{x_1}^{x} f(\tau)\d \tau

On a calculator, the solver should be used; graphing takes too long for definite integrals.

Example: Simplify

f(x) = \int_{0}^{x} \cos^2\theta\d \theta

Use double angle identity for $\cos 2\theta$ involving $\cos^2\theta$ .

\begin{align*}\cos 2\theta &= 2\cos^2\theta - 1 \\ \frac12 + \frac12\cos 2\theta &= \cos^2\theta \end{align*}

We find the antiderivative, using the $+C = 0$ trick

\begin{align*}f(x) &= \int_0^x \frac12 + \frac12\cos 2\theta \d\theta \\ &= \left[\frac12\theta + \frac14\sin2\theta \right]_{\theta=0}^{\theta=x} \\ &= \frac12x + \frac14\sin2x - \left(\frac12(0) + \frac14\sin2(0)\right) \\ &= \frac12x + \frac14\sin2x \qed\end{align*}

In this particular question the second part entirely cancels. This should not assumed to be the case.

See an example on TI-84 Plus.

Derivation: z-scores of normal distribution

Derivation: Let

f(x) = \frac{2}{\sqrt\pi} \int_0^x \e^{-t^2} \d t

and given the normal distribution probability density function

g(x) = \frac{1}{\sigma\sqrt{2\pi}} \e^{-\frac{(x-\mu)^2}{2\sigma^2}}

Show that

\int_{\mu}^{\mu+z\sigma} g(x) \d x = \frac12f\left(\frac{z}{\sqrt2}\right) \text{ for all } z\in\mathbb R

\begin{align*} I &= \frac{1}{\sigma\sqrt{2\pi}} \int_{\mu}^{\mu+z\sigma} \e^{-\frac{(x-\mu)^2}{2\sigma^2}} \d x \\ &= \frac{1}{\sigma\sqrt{2\pi}} \int_{\mu}^{\mu+z\sigma} \e^{-\left(\frac{x-\mu}{\sigma\sqrt2}\right)^2} \d x \\ &= \frac{\sigma\sqrt2}{\sigma\sqrt{2\pi}} \int_{\frac{\mu - \mu}{\sigma\sqrt2}}^{\frac{\mu + z\sigma - \mu}{\sigma\sqrt2}} \e^{-t^2} \d t \\ &= \frac{1}{\sqrt\pi} \int_{0}^{\frac{z}{\sqrt2}} \e^{-t^2} \d t \\ &= \frac12 \left(f\left(\frac{z}{\sqrt2}\right) - f(0) \right) \\ &= \frac12f\left(\frac{z}{\sqrt2}\right) \qed \end{align*}

$f$ is called the “error function”.

We made use of

\int_{x_1}^{x_2} f(ax+b)\d x = \frac 1a\int_{ax_1+b}^{ax_2+b} f(x)\d x

and

\int_0^0 f(x)\d x = 0

With $x = \mu + z\sigma$ , we obtain

z = \frac{x - \mu}{\sigma}

In other words, probability from mean to $z$ standard deviations more than the mean, depends only on $z$ .

Notes on trig

Calculus of trigonometric functions are only available in radians. Ensure that your calculator is in radians mode.