Math · Calculus I · Worked example

Split the area where two curves cross

Find the total area enclosed by y = x³ and y = x.

y = x^{3}, y = x

Find the crossings

x³ = x gives x³ − x = x(x − 1)(x + 1) = 0, so x = −1, 0 or 1.

x^{3} - x = x (x - 1) (x + 1) = 0

Check the order on each piece

At x = −1/2, x³ = −1/8 lies above x = −1/2. At x = 1/2, x = 1/2 lies above x³ = 1/8. So x³ is on top on [−1, 0] and x is on top on [0, 1].

Integrate the left piece

An antiderivative of x³ − x is x⁴/4 − x²/2, which is 0 at x = 0 and 1/4 − 1/2 = −1/4 at x = −1.

\int_{- 1}^{0} (x^{3} - x) d x = 0 - (\frac{1}{4} - \frac{1}{2}) = \frac{1}{4}

\begin{array}{c} \int_{- 1}^{0} (x^{3} - x) d x \\ = 0 - (\frac{1}{4} - \frac{1}{2}) = \frac{1}{4} \end{array}

Integrate the right piece and add

On [0, 1], x − x³ gives 1/2 − 1/4 = 1/4, so the total is 1/4 + 1/4 = 1/2. Integrating x − x³ straight across [−1, 1] gives 0 instead, because the two pieces cancel.

\int_{0}^{1} (x - x^{3}) d x = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

Result

The total area is 1/2. The signed integral of x − x³ over [−1, 1] is 0.

Your turn

Find the area between y = sin x and y = cos x on [0, π/2].

Show the answer and explanation

2√2 − 2 ≈ 0.828.

The curves cross at x = π/4. Cosine is on top on [0, π/4] and sine on [π/4, π/2]. The first piece is √2 − 1, and by symmetry about x = π/4 so is the second, so the area is 2(√2 − 1) ≈ 0.828.

\begin{array}{l} sin x = cos x \\ x = \frac{π}{4} \\ \int_{0}^{π / 4} (cos x - sin x) d x = (sin \frac{π}{4} + cos \frac{π}{4}) - (0 + 1) = \sqrt{2} - 1 \\ 2 (\sqrt{2} - 1) \approx 0.828 \end{array}

\begin{array}{c} sin x = cos x ⟹ x = \frac{π}{4} \\ \int_{0}^{π / 4} (cos x - sin x) d x = \sqrt{2} - 1 \\ A = 2 (\sqrt{2} - 1) \approx 0.828 \end{array}

Keep exploring

In Graph, the curves enclose two matching loops, one on each side of the origin. Turning the picture half a turn about the origin swaps them, which is why both have area 1/4.

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