Math · Calculus I · Worked example

Find the area between a parabola and a line

Find the area of the region enclosed by y = 4 − x² and y = x + 2.

y = 4 - x^{2}, y = x + 2

Find where the curves meet

Set 4 − x² = x + 2. Then x² + x − 2 = 0, which factors as (x + 2)(x − 1) = 0, so x = −2 or x = 1. The curves meet at (−2, 0) and (1, 3).

4 - x^{2} = x + 2 ⟹ x = - 2 or x = 1

\begin{array}{c} 4 - x^{2} = x + 2 \\ x = - 2 or x = 1 \end{array}

At x = 0, between the intersections, the parabola gives 4 and the line gives 2. The parabola is on top.

(4 − x²) − (x + 2) = 2 − x − x².

A = \int_{- 2}^{1} (2 - x - x^{2}) d x

An antiderivative is 2x − x²/2 − x³/3. At x = 1 it is 2 − 1/2 − 1/3 = 7/6; at x = −2 it is −4 − 2 + 8/3 = −10/3. Subtracting gives 7/6 + 10/3 = 9/2.

\int_{- 2}^{1} (2 - x - x^{2}) d x = (2 - \frac{1}{2} - \frac{1}{3}) - (- 4 - 2 + \frac{8}{3}) = \frac{9}{2}

\begin{array}{c} \int_{- 2}^{1} (2 - x - x^{2}) d x \\ = (2 - \frac{1}{2} - \frac{1}{3}) - (- 4 - 2 + \frac{8}{3}) \\ = \frac{9}{2} \end{array}

The area is 9/2 = 4.5 square units.

Find the area of the region enclosed by y = x and y = x².

Show the answer and explanation

The curves meet where x = x², at x = 0 and x = 1. Between them the line is on top (at x = 1/2, 1/2 > 1/4), so A = ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.

\begin{array}{l} x = x^{2} \\ x = 0 or x = 1 \\ \int_{0}^{1} (x - x^{2}) d x = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \end{array}

In Slopes, sums & signed area, six midpoint rectangles give 4.5625. Raise the count to 60 and the sum comes within a thousandth of the exact 9/2.

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