Area between two curves

When two curves $y = f(x)$ and $y = g(x)$ intersect, they enclose a bounded region. The area of that region is the integral of the vertical gap between them, $f - g$, over the interval where they bound the region.

Setup

If $f(x) \ge g(x)$ on $[a, b]$, the area between them is $A = \int_a^b [f(x) - g(x)]\,dx.$

This formula works regardless of whether $f$ and $g$ are positive or negative , the key quantity is the difference, which is non-negative.

Two cases to keep separate:

• Curves intersect at $x = a$ and $x = b$. The limits are determined by the curves themselves; you solve $f(x) = g(x)$ to find $a$ and $b$.
• Curves are bounded by extra vertical lines $x = a$ and $x = b$. The limits come from those lines.

When the top curve switches

If $f \ge g$ on part of the interval and $g \ge f$ on another part, you must split at the crossover. On each piece, the integrand is the appropriate non-negative difference. $A = \int_a^c [f - g]\,dx + \int_c^b [g - f]\,dx \quad \text{if } f = g \text{ at } x = c.$

In one stroke this is $A = \int_a^b |f(x) - g(x)|\,dx$.

Using horizontal slices

When curves are easier to describe as $x = u(y)$ and $x = v(y)$ (e.g. parabolas opening sideways), use horizontal slices: $A = \int_c^d [u(y) - v(y)]\,dy \quad \text{with } u(y) \ge v(y) \text{ (right of left)}.$

Strategy

1. Sketch both curves.
2. Find intersections by solving $f(x) = g(x)$ , these often become the limits.
3. Identify top and bottom on each sub-interval.
4. Set up the integral of (top $-$ bottom).
5. Evaluate.

Worked examples

Example 1. Area enclosed by $y = x^2$ and $y = x$.

Intersections: $x^2 = x \Rightarrow x(x - 1) = 0$, so $x = 0, 1$. On $[0, 1]$, $x \ge x^2$ (try $x = 1/2$: $1/2 > 1/4$). Area $= \int_0^1 (x - x^2)\,dx = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$.

Example 2. Area enclosed by $y = x^2$ and $y = 2x$.

Intersections: $x^2 = 2x \Rightarrow x = 0, 2$. On $[0, 2]$, $2x \ge x^2$. Area $= \int_0^2 (2x - x^2)\,dx = \left[x^2 - \dfrac{x^3}{3}\right]_0^2 = 4 - \dfrac{8}{3} = \dfrac{4}{3}$.

Example 3. Area enclosed by $y^2 = x$ and $y = x - 2$.

The parabola opens to the right; the line is straight. Use horizontal slices. Intersections in $y$-coords: $y^2 = y + 2 \Rightarrow y^2 - y - 2 = 0 \Rightarrow (y-2)(y+1) = 0 \Rightarrow y = -1, 2$. At height $y \in [-1, 2]$, the line is at $x = y + 2$ (right) and the parabola at $x = y^2$ (left). Area $= \int_{-1}^2 [(y + 2) - y^2]\,dy = \left[\dfrac{y^2}{2} + 2y - \dfrac{y^3}{3}\right]_{-1}^2 = (2 + 4 - 8/3) - (1/2 - 2 + 1/3) = \dfrac{10}{3} - (-7/6) = \dfrac{27}{6} = \dfrac{9}{2}$.

Example 4. Area enclosed by $y = \sin x$ and $y = \cos x$ on $[0, \pi/2]$.

These cross at $x = \pi/4$. On $[0, \pi/4]$, $\cos x \ge \sin x$; on $[\pi/4, \pi/2]$, $\sin x \ge \cos x$. $A = \int_0^{\pi/4}(\cos x - \sin x)\,dx + \int_{\pi/4}^{\pi/2}(\sin x - \cos x)\,dx.$ First: $[\sin x + \cos x]_0^{\pi/4} = \sqrt{2} - 1$. Second: $[-\cos x - \sin x]_{\pi/4}^{\pi/2} = (-0 - 1) - (-\dfrac{\sqrt 2}{2} - \dfrac{\sqrt 2}{2}) = -1 + \sqrt 2 = \sqrt 2 - 1$. Total: $2(\sqrt 2 - 1)$.

Example 5. Area enclosed by $y = x$ and $y = x^3$.

Intersections: $x^3 = x \Rightarrow x(x^2 - 1) = 0 \Rightarrow x = 0, \pm 1$. The region splits into two: one in $[-1, 0]$ where $x^3 \ge x$ (check $x = -0.5$: $-0.125 > -0.5$), one in $[0, 1]$ where $x \ge x^3$. By symmetry of $|x - x^3|$ (odd-odd inside absolute value), the two pieces are equal: each $= \int_0^1 (x - x^3)\,dx = \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{1}{4}$. Total: $\dfrac{1}{2}$.

Example 6. Area enclosed by $y = x^2$ and $y = 4 - x^2$.

Intersections: $x^2 = 4 - x^2 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt 2$. On $[-\sqrt 2, \sqrt 2]$, $4 - x^2 \ge x^2$. By symmetry: $A = 2\int_0^{\sqrt 2} (4 - 2x^2)\,dx = 2[4\sqrt 2 - \dfrac{2 \cdot 2\sqrt 2}{3}] = 2[4\sqrt 2 - \dfrac{4\sqrt 2}{3}] = 2 \cdot \dfrac{8\sqrt 2}{3} = \dfrac{16\sqrt 2}{3}$.

Try it yourself

1. Area enclosed by $y = x^2$ and $y = 4$.
2. Area enclosed by $y = x^2$ and $y = x + 2$.
3. Area enclosed by $y^2 = 4x$ and $x^2 = 4y$.
4. Area enclosed by $y = x^2 - 4$ and $y = 4 - x^2$.
5. Area enclosed by $y = \sqrt{x}$ and $y = x^2$.
6. Area enclosed by $x = y^2 - 2$ and $x = y$.
7. Area enclosed by $y = x$, $y = x^2$, and $x = 2$.
8. Area enclosed by $y = e^x$, $y = e^{-x}$, and $x = 1$.
9. Area enclosed by $y = 2x - x^2$ and $y = x^2$.
10. Area enclosed by $y = x^2$ and $y = 2 - x^2$.
11. Area enclosed by $y = \ln x$ and $y = x - 1$ on $[1, e]$.
12. Area between $y = \sin x$ and $y = \sin 2x$ on $[0, \pi]$.
13. Area enclosed by $y^2 = x$ and $y = x^2$.
14. Area enclosed by $|y| = x$ and $x^2 + y^2 = 8$ in the right half plane.

Pitfalls and tricks

• Always identify which curve is on top on each sub-interval , a sign error here ruins the answer.
• Solve for intersection points first; the limits often come from those.
• Don't mix slice directions within one integral. Pick vertical or horizontal slices and stick with it.
• Symmetry simplifies dramatically. If the region is symmetric about an axis, compute half.
• For parabolas $y^2 = 4ax$ and similar, horizontal slices ($dy$) usually give cleaner limits.