Properties of Definite Integrals, calculus lesson

Computing every integral straight from Riemann sums would be slow. Most of the work is done instead by a handful of rules that come directly from the definition, much as the limit laws spared you from epsilon arguments. They also let you estimate an integral you cannot compute at all.

Key terms

Reversed bounds

Definition For $f$ integrable on $[a, b]$ with $a < b$ , define

\int_b^a f(x) \, dx = -\int_a^b f(x) \, dx, \qquad \int_a^a f(x) \, dx = 0.

In plain words Running the interval backward flips the sign, because every width $\Delta x$ becomes negative. An interval of length zero contributes nothing.

Example Since $\int_0^1 x^2 \, dx = \frac{1}{3}$ , we get

\int_1^0 x^2 \, dx = -\frac{1}{3} \qquad \text{and} \qquad \int_5^5 x^2 \, dx = 0.

Linearity

Definition If $f$ and $g$ are integrable on $[a, b]$ and $c$ is a constant, then $cf + g$ is integrable on $[a, b]$ and

\int_a^b \big( c f(x) + g(x) \big) \, dx = c \int_a^b f(x) \, dx + \int_a^b g(x) \, dx.

In particular the integral of a constant $c$ is

\int_a^b c \, dx = c(b - a).

In plain words You can pull constants out and split an integral of a sum into a sum of integrals. Both moves are true for each Riemann sum, so they survive the limit.

Example Using $\int_0^1 x^2 \, dx = \frac{1}{3}$ ,

\int_0^1 (3x^2 + 2) \, dx = 3 \cdot \frac{1}{3} + 2 \cdot (1 - 0) = 3.

Interval additivity

Definition If $f$ is integrable on a closed interval containing the numbers $a$ , $b$ and $c$ , then

\int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx.

With the reversed bounds convention this holds for any order of $a$ , $b$ and $c$ .

In plain words The signed area from $a$ to $c$ is the signed area from $a$ to $b$ plus the signed area from $b$ to $c$ .

Example If $\int_0^5 f(x) \, dx = 10$ and $\int_0^2 f(x) \, dx = 4$ , then

\int_2^5 f(x) \, dx = \int_0^5 f(x) \, dx - \int_0^2 f(x) \, dx = 10 - 4 = 6.

Comparison

Definition Let $f$ and $g$ be integrable on $[a, b]$ with $a \le b$ .

If $f(x) \ge 0$ for all $x$ in $[a, b]$ , then $\int_a^b f(x) \, dx \ge 0$ .
If $f(x) \le g(x)$ for all $x$ in $[a, b]$ , then $\int_a^b f(x) \, dx \le \int_a^b g(x) \, dx$ .
If $m \le f(x) \le M$ for all $x$ in $[a, b]$ , then

m(b - a) \le \int_a^b f(x) \, dx \le M(b - a).

In plain words A bigger function has a bigger integral, as long as the bounds are in the usual order. Boxing $f$ between two heights boxes its integral between two rectangle areas.

Example On $[0, 1]$ we have $e^{-1} \le e^{-x^2} \le 1$ , because $0 \le x^2 \le 1$ . So

0.367 < e^{-1} \le \int_0^1 e^{-x^2} \, dx \le 1.

The true value is about $0.7468$ . No antiderivative from the usual functions exists here, yet the bound came for free.

Absolute-value inequality

Definition If $f$ is integrable on $[a, b]$ with $a \le b$ , then $\lvert f \rvert$ is integrable on $[a, b]$ and

\left\lvert \int_a^b f(x) \, dx \right\rvert \le \int_a^b \lvert f(x) \rvert \, dx.

In plain words Cancellation between positive and negative parts can only make the integral smaller in size. The integral of $\lvert f \rvert$ has no cancellation, so it is at least as large.

Example Take $f(x) = x$ on $[-1, 2]$ . The two integrals are

\int_{-1}^{2} x \, dx = 2 - \frac{1}{2} = \frac{3}{2}, \qquad \int_{-1}^{2} \lvert x \rvert \, dx = \frac{1}{2} + 2 = \frac{5}{2},

and indeed $\frac{3}{2} \le \frac{5}{2}$ .

Symmetry

Definition Let $a > 0$ and let $f$ be integrable on $[-a, a]$ .

If $f$ is even, meaning $f(-x) = f(x)$ for all $x$ in $[-a, a]$ , then $\int_{-a}^{a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ .
If $f$ is odd, meaning $f(-x) = -f(x)$ for all $x$ in $[-a, a]$ , then $\int_{-a}^{a} f(x) \, dx = 0$ .

In plain words An even function’s graph is a mirror image across the $y$ -axis, so the two halves have equal integrals. An odd function’s left half is the right half flipped upside down, so the two halves cancel.

Example The function $x^2$ is even, so

\int_{-1}^{1} x^2 \, dx = 2 \int_0^1 x^2 \, dx = \frac{2}{3}.

The function $x^3 \cos x$ is odd, as the product of an odd and an even function, so $\int_{-\pi}^{\pi} x^3 \cos x \, dx = 0$ without any computing.

Piecewise continuous integrands

Definition A function $f$ on $[a, b]$ is piecewise continuous if it is continuous at all but finitely many points of $[a, b]$ and has finite one-sided limits at each of those points, where at $a$ and $b$ only the limit from inside the interval is required. Such a function is bounded and integrable. If the points $a = c_0 < c_1 < \cdots < c_k = b$ include all its discontinuities, then

\int_a^b f(x) \, dx = \sum_{j=1}^{k} \int_{c_{j-1}}^{c_j} f(x) \, dx,

and changing the value of $f$ at finitely many points does not change any of these integrals.

In plain words A graph with a few jumps still has a well defined integral. Split at the jumps, integrate each piece and add. What happens exactly at a jump point does not matter.

Example Let $f(x) = 1$ for $0 \le x < 1$ and $f(x) = 3$ for $1 \le x \le 2$ . Then

\int_0^2 f(x) \, dx = \int_0^1 1 \, dx + \int_1^2 3 \, dx = 1 + 3 = 4.

Mean value theorem for integrals

Definition The average value of an integrable function $f$ on $[a, b]$ with $a < b$ is

f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x) \, dx.

If $f$ is continuous on $[a, b]$ , then there is a number $c$ in $[a, b]$ with

f(c) = \frac{1}{b - a} \int_a^b f(x) \, dx, \qquad \text{that is} \qquad \int_a^b f(x) \, dx = f(c)(b - a).

Here is why. By the extreme value theorem a continuous $f$ has a minimum $m$ and a maximum $M$ on $[a, b]$ . The comparison property gives

m(b - a) \le \int_a^b f(x) \, dx \le M(b - a),

so the average lies between $m$ and $M$ . Since $f$ takes the values $m$ and $M$ , the intermediate value theorem gives a $c$ between those two points with $f(c)$ equal to the average. Both steps use continuity, and without it the conclusion can fail. With a little more work one can always choose $c$ in the open interval $(a, b)$ .

In plain words A continuous function actually reaches its average value somewhere. There is a rectangle of height $f(c)$ with exactly the same area as the region under the curve.

Example For $f(x) = x^2$ on $[0, 1]$ the average is

\frac{1}{1 - 0} \int_0^1 x^2 \, dx = \frac{1}{3}.

Solving $c^2 = \frac{1}{3}$ in $[0, 1]$ gives $c = \frac{1}{\sqrt{3}} \approx 0.577$ . For a step function this can fail. The function in the previous example has average $2$ on $[0, 2]$ but only takes the values $1$ and $3$ .

Check yourself

Given $\int_1^4 f(x) \, dx = 7$ and $\int_1^4 g(x) \, dx = -2$ , find $\int_4^1 \big( 2f(x) - 3g(x) \big) \, dx$ .

Show the answer

By linearity

\int_1^4 \big( 2f(x) - 3g(x) \big) \, dx = 2 \cdot 7 - 3 \cdot (-2) = 20,

and reversing the bounds gives $-20$ .

Show that $2 \le \int_0^2 \sqrt{1 + x^3} \, dx \le 6$ .

Show the answer

On $[0, 2]$ we have $0 \le x^3 \le 8$ , so $1 \le \sqrt{1 + x^3} \le 3$ . The comparison property with $m = 1$ , $M = 3$ and $b - a = 2$ gives

1 \cdot 2 \le \int_0^2 \sqrt{1 + x^3} \, dx \le 3 \cdot 2.

Find $\int_{-3}^{3} (x^5 - 4x^3 + 2) \, dx$ .

Show the answer

The part $x^5 - 4x^3$ is odd, so its integral over $[-3, 3]$ is $0$ . The constant contributes $2 \cdot 6 = 12$ . The answer is $12$ .

Properties of Definite Integrals

Theory, part 101

Practice, part 102

Key terms

Reversed bounds

Linearity

Interval additivity

Comparison

Absolute-value inequality

Symmetry

Piecewise continuous integrands

Mean value theorem for integrals

Check yourself

How this lesson connects

Builds on