Calculus 1

Area between curves

A region is fenced in by two curves, and neither of them is the x-axis. How much area does it hold, and what changes when the curves cross?

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The fundamental theorem of calculusGrade 11 Systems with a curveAlgebra 1

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Arc lengthCalculus 2 Average value of a functionCalculus 1 The disk methodCalculus 1

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Area between curves

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When you reach for it

A region in the plane is fenced in by two or more curves, and you want its area: a flower bed between two parabolic edges, a leaf-shaped logo between two waves, the gap between a cubic and a line. None of these is a rectangle, triangle or circle, but each is filled by thin strips whose length can be read off the curves. Signed area and integrals from geometry showed that one integral measures area above the axis minus area below it. Here the axis is replaced by a second curve, and every piece of area is counted as positive.

The procedure

Day 1: vertical strips

Strips and a Riemann sum. Let f and g be continuous on [a, b] with f(x) \ge g(x) there. Cut [a, b] into n pieces of width \Delta x and pick a point x_i^* in each. Over the ith piece stands a thin strip from the lower curve to the upper one, with height f(x_i^*) - g(x_i^*) and width \Delta x (Figure 1). The strips nearly fill the region, so

A \approx \sum_{i=1}^{n} \big(f(x_i^*) - g(x_i^*)\big)\,\Delta x.

This is a Riemann sum for f - g (The definite integral as a limit of Riemann sums). As the strips get thinner the sum tends to the definite integral, which exists because f - g is continuous:

A = \int_a^b \big(f(x) - g(x)\big)\,dx.

This holds when f(x) \ge g(x) on all of [a, b]. Read it as top minus bottom, added across the region. With g = 0 it is the familiar area under a curve. Neither curve has to stay above the x-axis: moving both up by the same constant c leaves every height (f + c) - (g + c) = f - g unchanged, so only the gap between the curves matters. The integral is evaluated with an antiderivative (The second fundamental theorem of calculus, part of the The fundamental theorem of calculus).

Figure 1The region of Example 1. A vertical strip at x runs from the bottom curve to the top curve, so its height is (x + 6) - x^2 and its width is \Delta x.

Finding the limits. When a region is “bounded by” two curves and no interval is given, its left and right ends are where the curves meet. Setting f(x) = g(x) is a system with a curve (Systems with a curve), and its solutions are the limits. To see which curve is on top, test one point between consecutive meeting points. Two continuous curves cannot swap order between meeting points, because f - g would have to pass through 0 to change sign (the intermediate value theorem).

Curves that cross. If the curves cross at c inside [a, b], the top curve becomes the bottom one and f - g changes sign. Then \int_a^b (f - g)\,dx counts the strips on one side of c as negative, exactly as Signed area and integrals from geometry counts area below the axis: it is a net area, not an area. Split at each crossing so that the top stays on top in every piece, and add the pieces (Properties of definite integrals). If f \ge g on [a, c] and g \ge f on [c, b],

\begin{aligned} A &= \int_a^c \big(f(x) - g(x)\big)\,dx + \int_c^b \big(g(x) - f(x)\big)\,dx \\ &= \int_a^b \lvert f(x) - g(x) \rvert\,dx. \end{aligned}

The absolute value form holds in every case, since each strip has length \lvert f - g \rvert whichever curve is on top. Evaluating it still means splitting at the zeros of f - g, because an antiderivative of \lvert f - g \rvert is found one piece at a time.

Three or more curves. When a region is fenced by three curves, the top (or bottom) of a vertical strip may be one curve on part of the region and another curve on the rest. Split at every x where the boundary changes, write top minus bottom on each piece, and add.

Day 2: horizontal strips

Some regions are simpler from the side. A horizontal strip at height y runs from the left boundary x = L(y) to the right boundary x = R(y) and has thickness \Delta y. The same Riemann sum argument, now in y, gives

A = \int_c^d \big(R(y) - L(y)\big)\,dy.

This holds when R(y) \ge L(y) on all of [c, d]. Read it as right minus left, added from bottom to top. Each boundary must be written as x in terms of y, and the limits c and d are heights: the y-values where the boundaries meet.

Choosing the direction. The area of a region does not depend on how it is sliced, so both directions give the same number; the choice is about effort. Vertical strips suit a region whose top and bottom are each one formula y = \ldots all the way across. Horizontal strips suit a region whose left and right sides are each one formula x = \ldots all the way up. A sideways parabola such as x = y^2 - 2y gives two y-values for most x, so vertical strips need both square root branches, while in y it is a single polynomial (Figure 4). A direction is costly when its boundary changes partway (more integrals), when a boundary must be solved with a \pm, or when the solved boundary is hard to integrate. When both directions are workable, doing both is a strong check.

Worked examples

Example 1

Day 1: a parabola and a line

Find the area of the region bounded by y = x + 6 and y = x^2.

Solution

The curves meet where x^2 = x + 6, that is x^2 - x - 6 = 0, or (x - 3)(x + 2) = 0. So x = -2 and x = 3, at the points (-2, 4) and (3, 9). At x = 0 the line gives 6 and the parabola 0, so the line is on top across (-2, 3). A strip at x has height (x + 6) - x^2 (Figure 1), and

\begin{aligned} A &= \int_{-2}^{3} \left(x + 6 - x^2\right)dx \\ &= \left(\frac{x^2}{2} + 6x - \frac{x^3}{3}\right)\Big|_{-2}^{3} \\ &= \frac{27}{2} - \left(-\frac{22}{3}\right) = \frac{125}{6}. \end{aligned}

As a check, the region fits inside the box [-2, 3] \times [0, 9] of area 45, and \displaystyle \frac{125}{6} \approx 20.8 is a plausible share of it.

Example 2

Day 1: Sakshi’s flower bed

Sakshi lays out a flower bed whose edges, measured in meters, are the parabolas y = x^2 - 2x and y = 4x - x^2. Find its area.

Solution

The edges meet where x^2 - 2x = 4x - x^2, so 2x^2 - 6x = 0, 2x(x - 3) = 0, and x = 0 or x = 3. At x = 1 the second parabola gives 3 and the first gives -1, so y = 4x - x^2 is on top. A strip has height (4x - x^2) - (x^2 - 2x) = 6x - 2x^2, and

\begin{aligned} A &= \int_0^3 \left(6x - 2x^2\right)dx \\ &= \left(3x^2 - \frac{2x^3}{3}\right)\Big|_0^3 = 27 - 18 = 9. \end{aligned}

The bed covers 9 square meters. Part of it lies below the x-axis, since x^2 - 2x < 0 on (0, 2), but that needs no special care: only the difference of the two curves enters.

Example 3

Day 1: net area and area for curves that cross

The curves y = x^3 - x^2 and y = 2x enclose two regions. Find \int_{-1}^{2} \big((x^3 - x^2) - 2x\big)\,dx and the total area of the two regions.

Solution

The curves meet where x^3 - x^2 - 2x = 0, and x^3 - x^2 - 2x = x(x - 2)(x + 1), so at x = -1, 0 and 2 (Figure 2). Let h(x) = x^3 - x^2 - 2x, with antiderivative \displaystyle H(x) = \frac{x^4}{4} - \frac{x^3}{3} - x^2. Then \displaystyle H(-1) = -\frac{5}{12}, H(0) = 0 and \displaystyle H(2) = -\frac{8}{3}, and the integral is

H(2) - H(-1) = -\frac{8}{3} + \frac{5}{12} = -\frac{9}{4}.

A negative number cannot be an area. Test each piece. At \displaystyle x = -\frac{1}{2}, \displaystyle h = \frac{5}{8} > 0, so the cubic is on top on (-1, 0); at x = 1, h = -2 < 0, so the line is on top on (0, 2). Hence

\begin{aligned} A &= \int_{-1}^{0} h(x)\,dx + \int_0^2 \big(-h(x)\big)\,dx \\ &= \big(H(0) - H(-1)\big) - \big(H(2) - H(0)\big) \\ &= \frac{5}{12} + \frac{8}{3} = \frac{37}{12}. \end{aligned}

The integral \displaystyle -\frac{9}{4} is \displaystyle \frac{5}{12} - \frac{8}{3}: the left piece minus the right piece. This is the net area of Signed area and integrals from geometry, with the line y = 2x playing the part of the axis.

Figure 2The curves of Example 3 cross at x = 0, so the top curve on [-1, 0] is the bottom curve on [0, 2].

Example 4

Day 1: a region bounded by three curves

Find the area of the region in the first quadrant bounded by \displaystyle y = \sqrt{x}, y = 6 - x and the x-axis.

Solution

The curve and the line meet where \displaystyle \sqrt{x} = 6 - x. Squaring gives x = 36 - 12x + x^2, so x^2 - 13x + 36 = 0 and (x - 4)(x - 9) = 0. At x = 9 the left side is 3 and the right side is -3, so x = 9 was introduced by squaring; the corner is (4, 2). The curve leaves the axis at x = 0 and the line reaches it at x = 6. A vertical strip runs from the axis up to \displaystyle \sqrt{x} when 0 < x < 4 and up to 6 - x when 4 < x < 6 (Figure 3). The top changes at x = 4, so

\begin{aligned} A &= \int_0^4 \sqrt{x}\,dx + \int_4^6 (6 - x)\,dx \\ &= \frac{2}{3}x^{3/2}\Big|_0^4 + \left(6x - \frac{x^2}{2}\right)\Big|_4^6 \\ &= \frac{16}{3} + 2 = \frac{22}{3}. \end{aligned}

The second piece is a triangle with base 2 and height 2, whose area is indeed 2.

Figure 3The region of Example 4. The top of a vertical strip is \displaystyle \sqrt{x} left of x = 4 and 6 - x right of it, so the integral splits at x = 4.

Example 5

Day 2: a sideways parabola

Find the area of the region bounded by x = y^2 - 2y and x = 2y.

Solution

Both boundaries give x in terms of y, so slice horizontally. They meet where y^2 - 2y = 2y, so y(y - 4) = 0, at (0, 0) and (8, 4). At y = 1 the parabola gives x = -1 and the line x = 2, so the line is on the right. A strip has length 2y - (y^2 - 2y) = 4y - y^2 (Figure 4), and

\begin{aligned} A &= \int_0^4 \left(4y - y^2\right)dy = \left(2y^2 - \frac{y^3}{3}\right)\Big|_0^4 \\ &= 32 - \frac{64}{3} = \frac{32}{3}. \end{aligned}

Vertical strips give the same number with more work. Completing the square, (y - 1)^2 = x + 1, so \displaystyle y = 1 \pm \sqrt{x + 1}, and the leftmost point is (-1, 1). For -1 \le x \le 0 a strip runs between the two branches; for 0 \le x \le 8 it runs from the line \displaystyle y = \frac{x}{2} up to the upper branch:

\begin{aligned} A &= \int_{-1}^{0} 2\sqrt{x + 1}\,dx \\ &\quad + \int_0^8 \left(1 + \sqrt{x + 1} - \frac{x}{2}\right)dx \\ &= \frac{4}{3} + \left(8 + \frac{52}{3} - 16\right) \\ &= \frac{4}{3} + \frac{28}{3} = \frac{32}{3}. \end{aligned}

Figure 4The region of Example 5. A horizontal strip at height y runs from the left boundary to the right boundary, so its length is 2y - (y^2 - 2y).

Example 6

Day 2: Hina checks Example 4 sideways

Find the area of the region of Example 4 again, this time with horizontal strips.

Solution

The region runs from height y = 0 up to the corner (4, 2). On its left is the curve \displaystyle y = \sqrt{x}, which for y \ge 0 is x = y^2; on its right is the line, x = 6 - y. At every height in (0, 2) the left side is the parabola and the right side is the line, so the boundary never changes and one integral does it:

\begin{aligned} A &= \int_0^2 \left(6 - y - y^2\right)dy = \left(6y - \frac{y^2}{2} - \frac{y^3}{3}\right)\Big|_0^2 \\ &= 12 - 2 - \frac{8}{3} = \frac{22}{3}. \end{aligned}

This agrees with Example 4. The x-axis, which was the bottom of every vertical strip, is now only the lower limit y = 0.

Example 7

Day 2: a leaf between two waves

Rafael’s logo is the region between y = \sin x and y = \cos x from \displaystyle x = \frac{\pi}{4} to \displaystyle x = \frac{5\pi}{4}, in centimeters. Find its area.

Solution

The curves meet where \tan x = 1, at \displaystyle x = \frac{\pi}{4} + k\pi for integers k; the given ends are consecutive meeting points, so the curves do not cross in between. At \displaystyle x = \frac{\pi}{2} the sine is 1 and the cosine 0, so the sine is on top:

\begin{aligned} A &= \int_{\pi/4}^{5\pi/4} (\sin x - \cos x)\,dx \\ &= \big(-\cos x - \sin x\big)\Big|_{\pi/4}^{5\pi/4} \\ &= \sqrt{2} - \left(-\sqrt{2}\right) = 2\sqrt{2}. \end{aligned}

The leaf covers \displaystyle 2\sqrt{2} \approx 2.83 square centimeters. Horizontal strips would need \arcsin y and \arccos y on several branches with splits at more than one height, so vertical strips are the clear choice here.

Boundaries

\int_a^b (f - g)\,dx is the area only when f \ge g on all of [a, b]. If the curves cross inside, it is a net area; split at the crossings or integrate \lvert f - g \rvert.
Both curves must be continuous on a closed, bounded interval. With a vertical asymptote inside or an endless interval, as for \displaystyle y = \frac{1}{x^2} near x = 0, the integral is improper and needs a separate limit argument.
Each boundary must be a function of the variable of integration on each piece: one y for each x with vertical strips, one x for each y with horizontal strips. A sideways parabola or a circle has to be cut into branches first.
“The region bounded by” two curves makes sense only when they enclose one. The curves y = x^2 + 1 and y = x never meet, so they enclose nothing, and an area between them needs a stated interval.
The result is in square units of the axes, square meters if both axes are in meters.

Where this goes next

The next lesson, Average value of a function, divides an area by the width of its interval. A strip between two curves becomes the base of a slice in Volumes with known cross sections, and spun around an axis it sweeps out the disks of the The disk method and the rings of the The washer method. Adding up small pieces along a curve, rather than strips under it, gives arc length.

Exercises

Day 1: one curve on top

Find the area between y = x^2 + 2 and y = x from x = 0 to x = 3.
Find the area of the region bounded by y = 2 - x^2 and y = -x.
Find the area of the region bounded by y = x^2 and y = 2x + 8.
Find the area of the region bounded by y = x^2 - 4 and y = 2 - x^2.
Find the area of the region bounded by \displaystyle y = 2\sqrt{x} and y = x.
Find the area of the region bounded by y = x, \displaystyle y = \frac{1}{x} and x = 3.

Day 1: curves that cross

Find \int_0^3 \left(x^2 - 2x\right)dx and the area between y = x^2 and y = 2x from x = 0 to x = 3. Why do they differ?
Find the total area of the regions enclosed by y = x^3 and y = 4x.
Find the total area of the regions enclosed by y = x^3 - 6x and y = -x^2.
Find the area between y = 4 - x^2 and y = x + 2 from x = 0 to x = 2.

Day 1: three boundaries

Find the area of the region in the first quadrant bounded by y = x^2, y = 2 - x and the x-axis.
Find the area of the region in the first quadrant bounded by y = x, \displaystyle y = \frac{x}{4} and \displaystyle y = \frac{1}{x}.

Day 2: horizontal strips

Find the area of the region bounded by x = 2y^2 and x = y + 1.
Find the area of the region bounded by x = 3 - y^2 and x = 2y^2.
Find the total area of the regions enclosed by x = y^3 and x = y.
Find the area of the region bounded by y = \ln x, the y-axis, the x-axis and the line y = 2.
Find the area of the region bounded by y = x^3, the line y = 8 and the y-axis, once with vertical strips and once with horizontal strips.

Day 2: trigonometric and exponential boundaries

Find the area between y = \sin x and y = \cos x from x = 0 to \displaystyle x = \frac{\pi}{2}.
Find the area of the region bounded by y = e^x, y = e^{-x} and the line x = \ln 2.
Find the area between y = x and y = \sin x from x = 0 to x = \pi.

Reasoning

Explain why the area between y = f(x) + c and y = g(x) + c on [a, b] is the same as the area between y = f(x) and y = g(x), and why this means a region below the x-axis needs no special treatment.
Explain why \int_a^b \lvert f(x) - g(x) \rvert\,dx is the area between the curves whether or not they cross, and use Exercise 7 to show that \left\lvert \int_a^b \big(f(x) - g(x)\big)\,dx \right\rvert can be smaller.
Suppose f and g are continuous on [a, b] and meet only at x = a and x = b. Prove that the area between them is \left\lvert \int_a^b \big(f(x) - g(x)\big)\,dx \right\rvert.
Let p < q. Prove that the region bounded by y = x^2 and the line through (p, p^2) and (q, q^2) has area \displaystyle \frac{(q - p)^3}{6}, and check it against Example 1 and Exercise 3.

Answers

\displaystyle \frac{21}{2}. The gap x^2 - x + 2 is always positive (discriminant 1 - 8 < 0), so \displaystyle \int_0^3 \left(x^2 - x + 2\right)dx = 9 - \frac{9}{2} + 6.
\displaystyle \frac{9}{2}. They meet where x^2 - x - 2 = 0, at x = -1 and x = 2; the parabola is on top, and \displaystyle \int_{-1}^{2} \left(2 + x - x^2\right)dx = \frac{9}{2}.
36. They meet at x = -2 and x = 4; the line is on top, and \displaystyle \left(x^2 + 8x - \frac{x^3}{3}\right)\Big|_{-2}^{4} = \frac{80}{3} + \frac{28}{3}.
\displaystyle 8\sqrt{3}. They meet where 2x^2 = 6, at \displaystyle x = \pm\sqrt{3}, and \displaystyle \int_{-\sqrt{3}}^{\sqrt{3}} \left(6 - 2x^2\right)dx = 12\sqrt{3} - 4\sqrt{3}.
\displaystyle \frac{8}{3}. Squaring \displaystyle 2\sqrt{x} = x gives x = 0 or x = 4, both genuine. The curve is on top: \displaystyle \int_0^4 \left(2\sqrt{x} - x\right)dx = \frac{32}{3} - 8.
4 - \ln 3 \approx 2.90. The curves meet at x = 1, the line is on top on (1, 3), and \displaystyle \int_1^3 \left(x - \frac{1}{x}\right)dx = 4 - \ln 3.
The integral is 0 and the area is \displaystyle \frac{8}{3}. The curves cross at x = 2: \displaystyle \int_0^2 \left(2x - x^2\right)dx = \frac{4}{3} and \displaystyle \int_2^3 \left(x^2 - 2x\right)dx = \frac{4}{3}. The integral counts the first piece as negative, and the two pieces cancel.
8. They meet at x = 0, \pm 2. By symmetry the area is 2\int_0^2 \left(4x - x^3\right)dx = 2(8 - 4).
\displaystyle \frac{253}{12}. The difference is x^3 + x^2 - 6x = x(x + 3)(x - 2). The cubic is on top on (-3, 0), giving \displaystyle \frac{63}{4}; the parabola is on top on (0, 2), giving \displaystyle \frac{16}{3}; and \displaystyle \frac{63}{4} + \frac{16}{3} = \frac{253}{12}.
3. They cross at x = 1. \displaystyle \int_0^1 \left(2 - x - x^2\right)dx = \frac{7}{6} and \displaystyle \int_1^2 \left(x^2 + x - 2\right)dx = \frac{11}{6}.
\displaystyle \frac{5}{6}. The curves meet at x = 1. \displaystyle \int_0^1 x^2\,dx + \int_1^2 (2 - x)\,dx = \frac{1}{3} + \frac{1}{2}.
\ln 2. The top is y = x until it meets \displaystyle y = \frac{1}{x} at x = 1; the bottom \displaystyle y = \frac{x}{4} meets \displaystyle y = \frac{1}{x} at x = 2. \displaystyle \int_0^1 \frac{3x}{4}\,dx + \int_1^2 \left(\frac{1}{x} - \frac{x}{4}\right)dx = \frac{3}{8} + \ln 2 - \frac{3}{8}.
\displaystyle \frac{9}{8}. They meet where 2y^2 - y - 1 = 0, at \displaystyle y = -\frac{1}{2} and y = 1; the line is on the right, and \displaystyle \int_{-1/2}^{1} \left(y + 1 - 2y^2\right)dy = \frac{9}{8}.
4. They meet where 3y^2 = 3, at y = \pm 1, and \int_{-1}^{1} \left(3 - 3y^2\right)dy = 6 - 2.
\displaystyle \frac{1}{2}. They meet at y = 0, \pm 1. On (0, 1), y > y^3, so by symmetry the area is \displaystyle 2\int_0^1 \left(y - y^3\right)dy = 2\cdot\frac{1}{4}.
e^2 - 1 \approx 6.39. With x = e^y on the right and x = 0 on the left, \int_0^2 e^y\,dy = e^2 - 1.
12 both ways. \int_0^2 \left(8 - x^3\right)dx = 16 - 4, and with x = y^{1/3}, \displaystyle \int_0^8 y^{1/3}\,dy = \frac{3}{4}\cdot 16.
\displaystyle 2\sqrt{2} - 2 \approx 0.83. They cross at \displaystyle \frac{\pi}{4}. \displaystyle \int_0^{\pi/4} (\cos x - \sin x)\,dx = \sqrt{2} - 1, and the second piece equals it by symmetry about \displaystyle x = \frac{\pi}{4}.
\displaystyle \frac{1}{2}. On (0, \ln 2), e^x > e^{-x}, and \displaystyle \left(e^x + e^{-x}\right)\Big|_0^{\ln 2} = \frac{5}{2} - 2.
\displaystyle \frac{\pi^2}{2} - 2 \approx 2.93. Since \sin x \le x for x \ge 0, the area is \displaystyle \left(\frac{x^2}{2} + \cos x\right)\Big|_0^{\pi} = \frac{\pi^2}{2} - 1 - 1.
The strip heights are (f + c) - (g + c) = f - g at every x, so the integrand, and with it the area, is unchanged. A region partly below the axis can be lifted by a large c without changing any strip, so the same formula applies to it.
At each x, \lvert f - g \rvert is top minus bottom, whichever curve is on top, so it is the strip height everywhere; splitting at the crossings and adding gives the sum of the pieces’ areas. In Exercise 7 the absolute value of the integral is 0 while the area is \displaystyle \frac{8}{3}: cancellation between pieces lowers \left\lvert \int (f - g) \right\rvert, while \int \lvert f - g \rvert never cancels.
f - g is continuous and nonzero on (a, b). If it were positive at one point and negative at another, the intermediate value theorem would give a zero between them, which is not allowed. So either f \ge g on [a, b] and the area is \int_a^b (f - g)\,dx \ge 0, or g \ge f and the area is \int_a^b (g - f)\,dx = -\int_a^b (f - g)\,dx. Either way it is the absolute value of the integral.
The line has slope p + q and equation y = (p + q)x - pq, so the gap is (p + q)x - pq - x^2 = -(x - p)(x - q) \ge 0 on [p, q]. With u = x - p and h = q - p, the area is \displaystyle \int_0^h u(h - u)\,du = \frac{h^3}{2} - \frac{h^3}{3} = \frac{h^3}{6}. Example 1 has p = -2, q = 3: \displaystyle \frac{125}{6}. Exercise 3 has p = -2, q = 4: \displaystyle \frac{216}{6} = 36.