Mathematics

Calculus: Integration

Definite integrals, area under a curve, and an applied consumer-surplus example.

The definite integral $\int_{a}^{b} f (x) d x$ gives the signed area between $f$ and the $x$ -axis from $a$ to $b$ . By the Fundamental Theorem of Calculus, if $F^{'} (x) = f (x)$ then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ . The power rule for integration is $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ (for $n \neq = - 1$ ).

1. A definite integral
easy
Evaluate $\int_{0}^{3} (2 x + 1) d x$ .
Show solution
An antiderivative is $F (x) = x^{2} + x$ .

$\int_{0}^{3} (2 x + 1) d x = F (3) - F (0) = (9 + 3) - (0 + 0) = 12$ .
2. Area under a parabola
medium
Find the area bounded by $f (x) = 4 - x^{2}$ and the $x$ -axis between the curve's two roots.
Show solution
Roots: $4 - x^{2} = 0 \Rightarrow x = \pm 2$ . On $[- 2, 2]$ the curve is above the axis, so the area is

$\int_{- 2}^{2} (4 - x^{2}) d x = [4 x - \frac{x ^{3}}{3}]_{- 2}^{2}$ .

At $x = 2$ : $8 - \frac{8}{3} = \frac{16}{3}$ . At $x = - 2$ : $- 8 + \frac{8}{3} = - \frac{16}{3}$ .

Area $= \frac{16}{3} - (- \frac{16}{3}) = \frac{32}{3} \approx 10.67$ .
3. Consumer surplus
hard
Demand is $p (q) = 100 - 2 q$ . The market clears at quantity $q^{*} = 30$ and price $p^{*} = 40$ . Consumer surplus is the area between the demand curve and the price line from $0$ to $q^{*}$ : $C S = \int_{0}^{q^{*}} (p (q) - p^{*}) d q$ . Compute it.
Show solution
$p (q) - p^{*} = (100 - 2 q) - 40 = 60 - 2 q$ .

$CS = \int_0^{30}(60 - 2q)\,dq = \left[60q - q^2\right]_0^{30} = (1800 - 900) - 0 = \$ 900$.

This equals the area of the triangle above the price and below the demand curve: $\frac{1}{2} \times 30 \times 60 = 900$ — the total value buyers received above what they paid.

Calculus: Integration

1. A definite integral

2. Area under a parabola

3. Consumer surplus

Go deeper