Integration

Lessons

Senior 6 · Mathematics

Integration

5 lessons

Lesson 4 of 5

Definite integrals: the limit of a sum

35 min

Area as a sum of thin rectangles

How much area lies under the curve $y = x^{2}$ between $x = 0$ and $x = 2$ ? No formula from geometry answers this, because the top edge is curved. Archimedes, around 250 BC, already had the method we still use: cut the region into thin strips, treat each strip as a rectangle, add, and let the strips become thinner and thinner. Eight strips give the picture below.

Right-endpoint Riemann sum for ∫₀² x² dx with n = 8: the rectangles overshoot, giving 3.1875. As n grows the overshoot shrinks to 0 and the sum tends to 8 ÷ 3 ≈ 2.667.

Definition

Riemann sum: divide $[a, b]$ into $n$ subintervals of width $\Delta x = \frac{b - a}{n}$ , choose a point $x_{i}^{*}$ in the $i$ th subinterval, and form

S_n = \sum_{i=1}^{n} f(x_i^*)\,\Delta x .

Definite integral: the limit of the Riemann sums, when it exists:

\int_a^b f(x)\,dx = \lim_{n \to \infty}\sum_{i=1}^{n} f(x_i^*)\,\Delta x .

$a$ is the lower limit and $b$ the upper limit. If $f$ is continuous on $[a, b]$ , the limit exists and does not depend on the choice of the points $x_{i}^{*}$ ; $f$ is then integrable on $[a, b]$ . The integral sign is a stretched S, for "sum", and $d x$ is what $\Delta x$ becomes in the limit.

Worked example 1: an integral as the limit of a sum

Use right endpoints to evaluate $\int_0^2 x^2\,dx$ as a limit.

Solution $\Delta x = \frac{2}{n}$ and $x_i = \frac{2i}{n}$ . With $\sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6}$ from Unit 3,

S_n = \sum_{i=1}^{n}\left(\frac{2i}{n}\right)^2\frac{2}{n} = \frac{8}{n^3}\sum_{i=1}^{n} i^2 = \frac{8}{n^3}\cdot\frac{n(n + 1)(2n + 1)}{6} = \frac{4}{3}\left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right).

For $n = 8$ this is $3.1875$ , the area of the rectangles in the picture. As $n \to \infty$ , $S_n \to \frac{4}{3} \times 1 \times 2 = \frac{8}{3}$ . So $\int_0^2 x^2\,dx = \frac{8}{3} \approx 2.667$ .

The fundamental theorem of calculus

Limits of sums are slow. The great discovery of Newton and Leibniz in the 1660s and 1670s is that the area problem and the antiderivative problem are the same problem. Fix $a$ and let $A(x) = \int_a^x f(t)\,dt$ be the area from $a$ up to a moving edge $x$ .

Moving the right edge from x to x + h adds a strip of area A(x + h) − A(x) ≈ f(x) · h. Dividing by h and letting h → 0 gives A′(x) = f(x).

Important

Fundamental theorem of calculus. Let $f$ be continuous on $[a, b]$ .

Part 1. $A(x) = \int_a^x f(t)\,dt$ is differentiable, and $A^{'} (x) = f (x)$ : the area function is an antiderivative of $f$ .
Part 2. If $F$ is any antiderivative of $f$ , then $\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a) .$

Reason: for Part 1, $\frac{A(x + h) - A(x)}{h}$ is the area of the thin strip divided by its width; it lies between the smallest and largest values of $f$ on $[x, x + h]$ , and both tend to $f (x)$ as $h \to 0$ because $f$ is continuous. For Part 2, $A$ and $F$ are antiderivatives of the same function, so $A (x) = F (x) + C$ . Since $A (a) = 0$ , $C = - F (a)$ , and then $A (b) = F (b) - F (a)$ .

Worked example 2: evaluating with antiderivatives

Evaluate (a) $\int_1^3 (3x^2 - 2x)\,dx$ (b) $\int_0^{\pi}\sin x\,dx$ (c) $\int_0^1 \frac{x}{x^2 + 1}\,dx$ .

Solution

$\Big[x^3 - x^2\Big]_1^3 = (27 - 9) - (1 - 1) = 18$ .
$\Big[-\cos x\Big]_0^{\pi} = -\cos\pi + \cos 0 = 1 + 1 = 2$ : one arch of the sine curve has area exactly 2.
$\Big[\frac{1}{2}\ln(x^2 + 1)\Big]_0^1 = \frac{1}{2}\ln 2 - 0 \approx 0.3466$ .

Note that the constant $C$ is not needed: it would cancel in $F (b) - F (a)$ .

Properties of definite integrals

property	meaning
$\int_a^a f(x)\,dx = 0$	a strip of zero width has no area
$\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$	reversing the limits changes the sign
$\int_a^b [kf(x) \pm g(x)]\,dx = k\int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$	linearity
$\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx$	areas over touching intervals add
if $f(x) \le g(x)$ on $[a, b]$ , then $\int_a^b f(x)\,dx \le \int_a^b g(x)\,dx$	comparison
$f$ odd: $\int_{-a}^{a} f(x)\,dx = 0$ ; $f$ even: $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$	symmetry

The definite integral is a signed area: parts of the graph below the $x$ -axis count as negative. That is why $\int_0^{2\pi}\sin x\,dx = 0$ , although the two arches have a total area of 4.

Substitution and parts with limits

With a substitution $u = g (x)$ , change the limits too, and never return to $x$ :

\int_a^b f\big(g(x)\big)g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du .

By parts, the product term is evaluated between the limits: $\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du$ .

Worked example 3: techniques with limits

Evaluate (a) $\int_0^2 x(x^2 + 1)^3\,dx$ (b) $\int_0^{\frac{\pi}{2}} x\cos x\,dx$ (c) $\int_{-2}^{2}(x^3 - 4x + x^2)\,dx$ .

Solution

$u = x^{2} + 1$ , $du = 2x\,dx$ . When $x = 0$ , $u = 1$ ; when $x = 2$ , $u = 5$ . $\frac{1}{2}\int_1^5 u^3\,du = \frac{1}{2}\Big[\frac{u^4}{4}\Big]_1^5 = \frac{625 - 1}{8} = 78 .$
Parts with $u = x$ , $dv = \cos x\,dx$ : $\Big[x\sin x\Big]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}}\sin x\,dx = \frac{\pi}{2} - \Big[-\cos x\Big]_0^{\frac{\pi}{2}} = \frac{\pi}{2} - 1 \approx 0.571 .$
$x^{3} - 4 x$ is odd, so its integral over $[- 2, 2]$ is 0; $x^{2}$ is even. Hence the value is $2\int_0^2 x^2\,dx = \frac{16}{3}$ .

Warning

The integrand must be continuous on the whole interval. Blindly, $\int_{-1}^{1}\frac{dx}{x^2} = \Big[-\frac{1}{x}\Big]_{-1}^{1} = -1 - 1 = -2$ : a negative area for a function that is always positive! The error is that $\frac{1}{x^2}$ is not defined at $x = 0$ , inside the interval, so Part 2 of the theorem does not apply. (The area is in fact infinite.)

Tip

Part 1 lets you differentiate an integral without evaluating it: $\frac{d}{dx}\int_0^x \sqrt{1 + t^3}\,dt = \sqrt{1 + x^3}$ . With a variable upper limit $g (x)$ , the chain rule adds a factor: $\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f\big(g(x)\big)\,g'(x)$ .

3Blue1Brown: Integration and the fundamental theorem of calculus, Chapter 8 of Essence of Calculus (about 20 min). Watch the car example: the distance travelled is an area under the velocity graph, and the thin strip argument shows why that area changes at the rate v(t).

✦

Practice

Questions 7 and 8 need more thought.

Evaluate $\int_0^2 (x^3 - 3x + 1)\,dx$ . Does your answer mean that there is no area between the curve and the axis? Explain.
Evaluate $\int_1^4 \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)dx$ .
Evaluate (a) $\int_0^{\frac{\pi}{4}}\sec^2 x\,dx$ (b) $\int_0^{\frac{\pi}{6}}\sin 3x\,dx$ .
Evaluate $\int_2^3 \frac{2x + 1}{x^2 + x}\,dx$ , giving an exact answer.
Use $u = x + 1$ to evaluate $\int_0^3 \frac{x}{\sqrt{x + 1}}\,dx$ .
Evaluate $\int_1^{e} x\ln x\,dx$ .
Without finding an antiderivative, evaluate (a) $\int_{-\pi}^{\pi} x^2\sin x\,dx$ (b) $\int_{-3}^{3}|x|\,dx$ .
(a) Use right endpoints to show, as a limit of a sum, that $\int_0^3 2x\,dx = 9$ . (b) Find $\frac{d}{dx}\int_1^{x^2}\frac{dt}{1 + t^2}$ .

Show answersHide answers

$\Big[\frac{x^4}{4} - \frac{3x^2}{2} + x\Big]_0^2 = 4 - 6 + 2 = 0$ . No: the curve is above the axis on part of $[0, 2]$ and below it on the rest, and the signed areas cancel.
$\Big[\frac{2}{3}x^{\frac{3}{2}} + 2\sqrt{x}\Big]_1^4 = \frac{28}{3} - \frac{8}{3} = \frac{20}{3}$
(a) $\tan\frac{\pi}{4} - 0 = 1$ (b) $\Big[-\frac{1}{3}\cos 3x\Big]_0^{\frac{\pi}{6}} = 0 + \frac{1}{3} = \frac{1}{3}$
$\Big[\ln(x^2 + x)\Big]_2^3 = \ln 12 - \ln 6 = \ln 2$
$\int_1^4 \left(u^{\frac{1}{2}} - u^{-\frac{1}{2}}\right)du = \Big[\frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}}\Big]_1^4 = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$
$\Big[\frac{x^2}{2}\ln x - \frac{x^2}{4}\Big]_1^{e} = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4} \approx 2.097$
(a) 0, since $x^2\sin x$ is odd (b) $2\int_0^3 x\,dx = 9$ , since $∣ x ∣$ is even
(a) $\Delta x = \frac{3}{n}$ , $x_i = \frac{3i}{n}$ : $S_n = \frac{18}{n^2}\cdot\frac{n(n + 1)}{2} = 9\left(1 + \frac{1}{n}\right) \to 9$ (b) $\frac{2x}{1 + x^4}$

Sources and credits

Written by Ganzaa: the explanations, the proof sketches, the diagrams (drawn from computed values), the worked examples, the practice items and the quiz questions. Every value was checked numerically.
Video: Integration and the fundamental theorem of calculus | Chapter 8, Essence of calculus, 3Blue1Brown, embedded from YouTube.

1Area as a sum of thin rectangles

2The fundamental theorem of calculus

3Properties of definite integrals

4Substitution and parts with limits

Area as a sum of thin rectangles

The fundamental theorem of calculus

Properties of definite integrals

Substitution and parts with limits