The Definite Integral

Main idea

The definite integral is what the rectangle sums close in on as the rectangles get infinitely thin. It measures net area: area above the x-axis counts as positive and area below counts as negative.

\int_{\htmlData{sym=0}{a}}^{\htmlData{sym=1}{b}} f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n {\htmlData{sym=2}{f(x_i^*)}}\,{\htmlData{sym=3}{\Delta x}}

What the symbols mean

$a$
Lower limit
Where the region starts.
$b$
Upper limit
Where it ends.
$f(x_i^*)$
Sample height
The function's value in the ith slice. Negative below the axis, so that area subtracts.
$\Delta x$
Slice width
(b − a)/n, shrinking to zero in the limit.

Explanation

A definite integral is simply the exact area under a curve between two points, with any area below the axis subtracted. All we're really doing is taking last lesson's rectangle sums and letting the number of rectangles go to infinity. The notation is built from those sums: $\int$ (a stretched-out S, for "sum"), $a$ and $b$ (where the region starts and stops), $f(x)$ (the height) and $dx$ (the width, now infinitely thin). So $\int_a^b f(x)\,dx$ literally reads "add up height times width from $a$ to $b$ ." It looks fancy, but it's the same sum with a new name.

Unlike an indefinite integral (a family of functions), a definite integral is a number. The two share a symbol for a very good reason, which the next lesson reveals.

Net area. Where $f$ dips below the axis, those rectangles have negative heights, so that area subtracts.

A common mistake. It's tempting to read every definite integral as "the area", because that's how we introduced it. It isn't always: $\int_0^{2\pi}\sin x\,dx = 0$ , even though the curve plainly encloses some area. The part below the axis cancels the part above. If you want total area, split at the crossings and add the absolute values.

Properties:

$\int_a^a f(x)\,dx = 0$ and $\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$
$\int_a^b \big[c f(x) \pm g(x)\big]dx = c\int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$
$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$
If $f(x) \ge g(x)$ on $[a, b]$ , then $\int_a^b f \ge \int_a^b g$ .

Geometry evaluates some integrals directly: lines give triangles and trapezoids; $\sqrt{r^2 - x^2}$ gives part of a circle. An odd function on $[-a, a]$ integrates to 0, because the areas cancel.

See it

This is ∫₀ᵗ sin x dx, the net area under sin x from 0 to t. As you can see, it climbs while the curve is above the axis, reaching 2 at t = π. After that the curve dips below, that area counts as negative, and it pulls the total back down to exactly 0 at t = 2π.

Worked examples

Example 1

Evaluate $\displaystyle\int_0^3(2x - 2)\,dx$ using geometry.

1
The line $y = 2x - 2$ crosses the axis at $x = 1$ . From 0 to 1 it's below the axis, forming a triangle of area $\frac12(1)(2) = 1$ , which counts as $-1$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Evaluate $\displaystyle\int_0^4 3\,dx$ .

Problem 2+5 XP

Evaluate $\displaystyle\int_{-1}^1 x^3\,dx$ .

Problem 3+5 XP

Evaluate $\displaystyle\int_0^2\sqrt{4 - x^2}\,dx$ .

Problem 4+5 XP

If $\displaystyle\int_0^5 f(x)\,dx = 8$ and $\displaystyle\int_0^2 f(x)\,dx = 3$ , find $\displaystyle\int_2^5 f(x)\,dx$ .

Problem 5+5 XP

With the same $f$ , find $\displaystyle\int_5^0 f(x)\,dx$ .

Problem 6+5 XP

A drone's vertical velocity is $v(t) = 3 - t$ m/s for $0 \le t \le 5$ s. Find its net change in altitude, then the total vertical distance it travels.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Evaluate $\displaystyle\int_{-3}^{3}\left(x^3\cos x + \sqrt{9 - x^2}\right)dx$ without finding any antiderivative.

Where it's used

Net area is net change: integrate a velocity that's sometimes negative and you get displacement, not total distance. Integrate a cash flow and you get net profit.

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