The Area Between Two Curves

A definite integral measures the area trapped between one curve and the $x$ -axis. Replacing the height $f (x)$ by a difference of two heights extends the idea to any region caught between two graphs. Every problem in this section reduces to the same three steps: draw a typical rectangle, write down its area $d A$ , and integrate.

Item	Statement
Element of area (vertical)	$d A = [f (x) - g (x)] d x$
Vertical rectangles	$A = \int_{a}^{b} [f (x) - g (x)] d x$ , valid when $f (x) \geq g (x)$ on $[a, b]$
Element of area (horizontal)	$d A = [h (y) - k (y)] d y$
Horizontal rectangles	$A = \int_{c}^{d} [h (y) - k (y)] d y$ , valid when $h (y) \geq k (y)$ on $[c, d]$
Height of a vertical rectangle	top curve minus bottom curve
Length of a horizontal rectangle	right curve minus left curve
Finding $a$ and $b$	solve $f (x) = g (x)$
Finding $c$ and $d$	solve $h (y) = k (y)$
Curves that cross inside the interval	split at every crossing and add the pieces
Region that needs two $d x$ integrals	often needs only one $d y$ integral, and the reverse
Symmetry	if the region is symmetric about the $y$ -axis, integrate over half and double
Sign check	an area is never negative, so always integrate top minus bottom

Vertical Rectangles: Integrating with Respect to x

Suppose two given curves $y = f (x)$ and $y = g (x)$ intersect at $x = a$ and $x = b$ , and

f (x) \geq g (x) (a < x < b) .

To find the area of the region bounded between these two curves, consider a vertical rectangle as shown in the figure below. The height of this rectangle is $f (x) - g (x)$ and its width is $d x$ . The area of this infinitesimal rectangle is

d A = [f (x) - g (x)] d x,

and the total area is

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{A=\int_{a}^{b}dA=\int_{a}^{b}[f(x)-g(x)]\,dx.}

The region between an upper curve y = f(x) and a lower curve y = g(x) from x = a to x = b, with a typical vertical rectangle of height f(x) minus g(x) and width dx. — A typical vertical rectangle in the region between $y = f (x)$ and $y = g (x)$ has height $f (x) - g (x)$ and width $d x$ .

Notice that we integrate from the smaller limit $a$ to the larger one $b$ , and $f (x)$ is greater than $g (x)$ in the entire interval of integration; otherwise the integral (and the area) becomes negative, which is meaningless.
Because the curves $y = f (x)$ and $y = g (x)$ intersect at $x = a$ and $x = b$ , they have the same $y$ -values at these points. If in a problem $a$ and $b$ are not given, we can find them by solving the equation $f (x) = g (x)$ .

Find the area of the region bounded by the curves $y = \frac{1}{2} x^{2}$ and $y = 2$ .

The parabola y = x squared over 2 and the horizontal line y = 2 meeting at the points (-2, 2) and (2, 2), with the region they enclose shaded. — The region bounded by $y = \frac{1}{2} x^{2}$ and $y = 2$ .

Solution

Using vertical rectangles, we realize that the height of the typical rectangle is $2 - \frac{1}{2} x^{2}$ and its width is $d x$ . So the element of area is

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

The region between the parabola y = x squared over 2 and the line y = 2, with a typical vertical rectangle of height 2 minus x squared over 2 and width dx. — The typical vertical rectangle has height $2 - \frac{1}{2} x^{2}$ and width $d x$ .

These two curves intersect where

\frac{1}{2} x^{2} = 2 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2,

so the total area bounded by $y = x^{2} / 2$ and $y = 2$ is

\begin{aligned} A=\int_{-2}^{2}dA &= \int_{-2}^{2}\left(2-\frac{1}{2}x^{2}\right)dx\\ &= \left[2x-\frac{1}{6}x^{3}\right]_{x=-2}^{x=2}\\ &= 2\left(4-\frac{1}{6}\cdot8\right)\\ &= \frac{16}{3}. \end{aligned}

Because the region is symmetric with respect to the $y$ -axis, we could simply integrate from $0$ to $2$ and multiply the result by $2$ . That is,

\begin{aligned} A &= 2\int_{0}^{2}\left(2-\frac{1}{2}x^{2}\right)dx\\ &= 2\left[2x-\frac{1}{6}x^{3}\right]_{x=0}^{x=2}\\ &= 2\left(4-\frac{8}{6}\right)=\frac{16}{3}. \end{aligned}

Horizontal Rectangles: Integrating with Respect to y

Sometimes, instead of vertical rectangles, we could (or should) consider horizontal rectangles. To find the area of the region bounded by the curves $x = h (y)$ and $x = k (y)$ (with $k (y) \leq h (y)$ ) and the horizontal lines $y = c$ and $y = d$ (with $c \leq d$ ), shown in the figure below, we consider horizontal rectangles of length $h (y) - k (y)$ and width $d y$ . The element of area is then

d A = [h (y) - k (y)] d y

and the total area is

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{A=\int_{c}^{d}dA=\int_{c}^{d}[h(y)-k(y)]\,dy.}

A region bounded on the right by the curve x = h(y) and on the left by the curve x = k(y), between the horizontal lines y = c and y = d, with a typical horizontal rectangle of length h(y) minus k(y) and width dy. — A typical horizontal rectangle between $x = k (y)$ and $x = h (y)$ has length $h (y) - k (y)$ and width $d y$ .

For instance, in the above example, if we use horizontal rectangles, the length of the rectangle is the $x$ -value (in terms of $y$ ) of the right curve, $x = \sqrt{2 y}$ , minus the $x$ -value of the left curve, $x = - \sqrt{2 y}$ . Because the width of the rectangle is $d y$ , the element of area is

\begin{aligned} dA &= \left[\sqrt{2y}-\left(-\sqrt{2y}\right)\right]dy\\ &= 2\sqrt{2}\sqrt{y}\,dy. \end{aligned}

As the horizontal rectangle sweeps across the region, $y$ varies between $0$ and $2$ . Therefore, the total area is

\begin{aligned} A &= \int_{0}^{2}2\sqrt{2}\sqrt{y}\,dy\\ &= 2\sqrt{2}\int_{0}^{2}y^{1/2}\,dy\\ &= \frac{2\sqrt{2}}{3/2}\left.y^{3/2}\right|_{y=0}^{y=2}\\ &= \frac{4\sqrt{2}}{3}\sqrt{8}\\ &= \frac{16}{3}, \end{aligned}

in agreement with the answer found using vertical rectangles.

Both methods always give the same number. Choose the one that makes the algebra easier: use $d x$ when each vertical line meets the region in a single segment whose top and bottom are given by $y = f (x)$ and $y = g (x)$ , and use $d y$ when each horizontal line meets the region in a single segment whose right and left ends are given by $x = h (y)$ and $x = k (y)$ .

Find the area of the region bounded by the curves $y^{2} = x$ and $y = x - 2$ .

The sideways parabola x = y squared and the line y = x minus 2 crossing at the points (1, -1) and (4, 2), with the region they enclose shaded. — The region bounded by $y^{2} = x$ and $y = x - 2$ .

Solution

First we need to find where these curves intersect. Writing both equations with $x$ alone on the left,

x = y^{2}, x = y + 2 \Rightarrow y^{2} = y + 2

y^{2} - y - 2 = 0

\Rightarrow y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{1 \pm \sqrt{1 + 4 (2)}}{2} = \frac{1 \pm 3}{2}

\Rightarrow y = 2 or y = - 1.

The corresponding $x$ -values come from $x = y^{2}$ :

x = {y^{2} |}_{y = 2} = 4, x = {y^{2} |}_{y = - 1} = 1.

Therefore, these curves intersect at $(4, 2)$ and $(1, - 1)$ . Note that the $x$ -coordinate is listed first, so the upper intersection is $(4, 2)$ and not $(2, 4)$ .

Method (a): horizontal rectangles. To compute the area of the region, it is easier to consider horizontal thin rectangles, as in the figure below. Because $x = y + 2$ is on the right, the length of the typical rectangle is $y + 2 - y^{2}$ and its width is $d y$ . Therefore, the element of area is

d A = (y + 2 - y^{2}) d y .

Because $y$ varies between $- 1$ and $2$ , the area is

\begin{aligned} A &= \int_{-1}^{2}dA=\int_{-1}^{2}(y+2-y^{2})\,dy\\ &= \left[\frac{1}{2}y^{2}+2y-\frac{y^{3}}{3}\right]_{-1}^{2}\\ &= \frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{9}{2}. \end{aligned}

The region between the parabola x = y squared and the line x = y + 2, with a typical horizontal rectangle running from the parabola on the left to the line on the right. — With horizontal rectangles, the typical rectangle has length $(y + 2) - y^{2}$ and width $d y$ .

Method (b): vertical rectangles. To compute the area, we can also consider thin vertical rectangles. However, we need to divide the region into two subregions, because the bottom boundary changes at $x = 1$ . If $x > 1$ , then the height of the rectangle is $\sqrt{x} - (x - 2)$ , and if $0 < x < 1$ , the height of the rectangle is $\sqrt{x} - (- \sqrt{x}) = 2 \sqrt{x}$ :

The same region divided by the vertical line x = 1 into two parts, each with its own typical vertical rectangle, because the lower boundary changes from the parabola to the line at x = 1. — With vertical rectangles, the region must be split at $x = 1$ , where the lower boundary changes.

Therefore

dA=\begin{cases} 2\sqrt{x}\,dx & (\text{if }0\leq x\leq1)\\ (\sqrt{x}-x+2)\,dx & (\text{if }1\leq x\leq4) \end{cases}

and the total area is

\begin{aligned} A &= \int_{0}^{4}dA\\ &= \int_{0}^{1}2\sqrt{x}\,dx+\int_{1}^{4}(\sqrt{x}-x+2)\,dx\\ &= \left[\frac{4}{3}x^{3/2}\right]_{0}^{1}+\left[\frac{2}{3}x^{3/2}-\frac{1}{2}x^{2}+2x\right]_{1}^{4}\\ &= \frac{4}{3}+\left(\frac{16}{3}-8+8\right)-\left(\frac{2}{3}-\frac{1}{2}+2\right)\\ &= \frac{4}{3}+\frac{16}{3}-\frac{13}{6}=\frac{9}{2}. \end{aligned}

The two methods agree, but the horizontal one needed a single integral instead of two.

Regions That Must Be Split

If the curves cross somewhere strictly between the endpoints, then neither curve is on top throughout, and a single integral of $f (x) - g (x)$ would count part of the area as negative. The cure is to find every crossing point, and integrate the positive difference separately on each subinterval.

If $f$ and $g$ cross at $x = p$ inside $[a, b]$ , then

A = \int_{a}^{p} | f (x) - g (x) | d x + \int_{p}^{b} | f (x) - g (x) | d x,

where on each piece the absolute value is removed by putting the larger function first.

Find the area between the graphs of $y = \cos 2 x$ and $y = \sin x$ on the interval $[0, \frac{π}{2}]$ .

Solution

It is a good idea to sketch these two curves. To graph $y = \cos 2 x$ we start with the graph of $y = \cos x$ and compress it horizontally by a factor of $2$ , as in the figure below.

The graphs of y = sin x, y = cos 2x, and y = cos x drawn together, showing that cos 2x is the graph of cos x compressed horizontally by a factor of 2. — Graphs of $\sin x$ , $\cos 2 x$ , and $\cos x$ .

As we can see from the figure, the region between the curves consists of two parts. In one part, the curve $y = \cos 2 x$ is the upper curve, and in the other part, the curve $y = \sin x$ is the upper one. To find exactly where these two curves intersect, we must solve the equation $\cos 2 x = \sin x$ . We do this by writing

\begin{aligned} \cos2x &= \sin x\\ \cos^{2}x-\sin^{2}x &= \sin x &&{\small(\cos2x=\cos^{2}x-\sin^{2}x)}\\ 1-2\sin^{2}x &= \sin x &&{\small(\cos^{2}x+\sin^{2}x=1)} \end{aligned}

2 \sin^{2} x + \sin x - 1 = 0.

The above equation is a quadratic equation in terms of $\sin x$ . That is, if $u = \sin x$ , then

2 u^{2} + u - 1 = 0

\Rightarrow u = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 1 \pm \sqrt{1 + 8}}{4} = \frac{- 1 \pm 3}{4},

\sin x = - 1 or \sin x = \frac{1}{2} .

\sin x = \frac{1}{2} \Rightarrow x = \frac{π}{6}

The equation $\sin x = - 1$ has no solution in the interval $[0, \frac{π}{2}]$ , so the only crossing is at $x = π / 6$ .

The region between y = cos 2x and y = sin x on the interval from 0 to pi over 2, split at x = pi over 6 where the two curves cross and swap which one is on top. — Finding the area between the graphs.

Therefore

dA=\begin{cases} (\cos2x-\sin x)\,dx & 0\leq x\leq\frac{\pi}{6}\\ (\sin x-\cos2x)\,dx & \frac{\pi}{6}\leq x\leq\frac{\pi}{2} \end{cases}

and the desired area is

\begin{aligned} A &= \int_{0}^{\frac{\pi}{6}}dA+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}dA\\ &= \int_{0}^{\frac{\pi}{6}}(\cos2x-\sin x)\,dx+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(\sin x-\cos2x)\,dx\\ &= \left[\frac{1}{2}\sin2x+\cos x\right]_{0}^{\frac{\pi}{6}}+\left[-\cos x-\frac{1}{2}\sin2x\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\\ &= \left(\frac{1}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)-(0+1)+\left(-0-\frac{1}{2}\cdot0\right)-\left(-\frac{\sqrt{3}}{2}-\frac{1}{2}\cdot\frac{\sqrt{3}}{2}\right)\\ &= \frac{3\sqrt{3}}{4}-1+\frac{3\sqrt{3}}{4}\\ &= \frac{3\sqrt{3}}{2}-1. \end{aligned}

Numerically $A \approx 1.598$ , which is a plausible size for a region roughly $1.5$ units wide and about one unit tall.

Exercises

Find the area of the region bounded by $y = x$ and $y = x^{2}$ .

Solution

The curves meet where $x = x^{2}$ , that is $x (x - 1) = 0$ , so $x = 0$ and $x = 1$ . On $(0, 1)$ we have $x > x^{2}$ , so the line is on top and the height of the typical rectangle is $x - x^{2}$ . The element of area is

d A = (x - x^{2}) d x,

and therefore

\begin{aligned} A &= \int_{0}^{1}\left(x-x^{2}\right)dx\\ &= \left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}\\ &= \frac{1}{2}-\frac{1}{3}=\frac{1}{6}. \end{aligned}

Find the area of the region bounded by $y = 4 - x^{2}$ and the $x$ -axis.

Solution

The parabola meets $y = 0$ where $4 - x^{2} = 0$ , so $x = \pm 2$ . The height of the typical rectangle is $(4 - x^{2}) - 0$ , so the element of area is

d A = (4 - x^{2}) d x .

The region is symmetric about the $y$ -axis, so

\begin{aligned} A &= \int_{-2}^{2}\left(4-x^{2}\right)dx=2\int_{0}^{2}\left(4-x^{2}\right)dx\\ &= 2\left[4x-\frac{x^{3}}{3}\right]_{0}^{2}\\ &= 2\left(8-\frac{8}{3}\right)=\frac{32}{3}. \end{aligned}

Find the area of the region enclosed by $y = \sqrt{x}$ and $y = x^{2}$ .

Solution

The curves meet where $\sqrt{x} = x^{2}$ . Squaring, $x = x^{4}$ , so $x (x^{3} - 1) = 0$ and $x = 0$ or $x = 1$ . For $0 < x < 1$ we have $\sqrt{x} > x^{2}$ , so the square root is the upper curve and the element of area is

d A = (\sqrt{x} - x^{2}) d x .

Therefore

\begin{aligned} A &= \int_{0}^{1}\left(\sqrt{x}-x^{2}\right)dx\\ &= \left[\frac{2}{3}x^{3/2}-\frac{x^{3}}{3}\right]_{0}^{1}\\ &= \frac{2}{3}-\frac{1}{3}=\frac{1}{3}. \end{aligned}

Find the area of the region bounded by $x = 4 - y^{2}$ and the $y$ -axis, using horizontal rectangles.

Solution

The curve meets the $y$ -axis where $4 - y^{2} = 0$ , so $y = \pm 2$ . A horizontal rectangle runs from the left boundary $x = 0$ to the right boundary $x = 4 - y^{2}$ , so its length is $4 - y^{2}$ and its width is $d y$ . The element of area is

d A = (4 - y^{2}) d y,

and

\begin{aligned} A &= \int_{-2}^{2}\left(4-y^{2}\right)dy=2\int_{0}^{2}\left(4-y^{2}\right)dy\\ &= 2\left[4y-\frac{y^{3}}{3}\right]_{0}^{2}\\ &= 2\left(8-\frac{8}{3}\right)=\frac{32}{3}. \end{aligned}

This is the same region as in Exercise 2, reflected across the line $y = x$ , so the equality of the two answers is no accident.

Find the area of the region bounded by $y = x^{2}$ and $y = 2 x + 3$ .

Solution

Set the two expressions equal:

x^{2} = 2 x + 3 \Rightarrow x^{2} - 2 x - 3 = 0 \Rightarrow (x - 3) (x + 1) = 0,

so $x = - 1$ and $x = 3$ , giving the points $(- 1, 1)$ and $(3, 9)$ . Between these values the line is above the parabola, so the height of the typical rectangle is $(2 x + 3) - x^{2}$ and the element of area is

d A = (2 x + 3 - x^{2}) d x .

Hence

\begin{aligned} A &= \int_{-1}^{3}\left(2x+3-x^{2}\right)dx\\ &= \left[x^{2}+3x-\frac{x^{3}}{3}\right]_{-1}^{3}\\ &= (9+9-9)-\left(1-3+\frac{1}{3}\right)\\ &= 9-\left(-\frac{5}{3}\right)=\frac{32}{3}. \end{aligned}

Find the area of the region bounded by $y = \sqrt{x}$ , the $x$ -axis, and the line $x = 4$ , first with vertical rectangles and then with horizontal rectangles.

Solution

Method (a): vertical rectangles. A vertical rectangle at position $x$ runs from $y = 0$ up to $y = \sqrt{x}$ , so the element of area is

d A = \sqrt{x} d x .

Since $x$ runs from $0$ to $4$ ,

A = \int_{0}^{4} \sqrt{x} d x = {[\frac{2}{3} x^{3 / 2}]}_{0}^{4} = \frac{2}{3} \cdot 8 = \frac{16}{3} .

Method (b): horizontal rectangles. Solving $y = \sqrt{x}$ for $x$ gives $x = y^{2}$ , which is the left boundary; the right boundary is the line $x = 4$ . The length of the rectangle is $4 - y^{2}$ , so now

d A = (4 - y^{2}) d y .

The corner of the region is at $(4, 2)$ , so $y$ runs from $0$ to $2$ and

\begin{aligned} A &= \int_{0}^{2}\left(4-y^{2}\right)dy\\ &= \left[4y-\frac{y^{3}}{3}\right]_{0}^{2}\\ &= 8-\frac{8}{3}=\frac{16}{3}. \end{aligned}

Both methods give $16 / 3$ , as they must.

Find the area of the region bounded by $y = x - 1$ and $y^{2} = 2 x + 6$ .

Solution

Write both boundaries as $x$ in terms of $y$ :

x = \frac{y^{2} - 6}{2} and x = y + 1.

They meet where

\frac{y^{2} - 6}{2} = y + 1 \Rightarrow y^{2} - 6 = 2 y + 2 \Rightarrow y^{2} - 2 y - 8 = 0,

that is $(y - 4) (y + 2) = 0$ , so $y = - 2$ and $y = 4$ , giving the points $(- 1, - 2)$ and $(5, 4)$ . The line is on the right, so the length of the horizontal rectangle is

(y + 1) - \frac{y^{2} - 6}{2} = - \frac{y^{2}}{2} + y + 4,

so the element of area is

d A = (- \frac{y^{2}}{2} + y + 4) d y,

and

\begin{aligned} A &= \int_{-2}^{4}\left(-\frac{y^{2}}{2}+y+4\right)dy\\ &= \left[-\frac{y^{3}}{6}+\frac{y^{2}}{2}+4y\right]_{-2}^{4}\\ &= \left(-\frac{64}{6}+8+16\right)-\left(\frac{8}{6}+2-8\right)\\ &= \frac{40}{3}+\frac{14}{3}=18. \end{aligned}

Vertical rectangles would require splitting the region at $x = - 1$ , where the bottom boundary changes from the lower half of the parabola to the line, so the horizontal setup is much shorter here.

Find the area of the region between $y = \sin x$ and $y = \cos x$ on $[0, \frac{π}{2}]$ .

Solution

The curves cross where $\sin x = \cos x$ , that is $\tan x = 1$ , so $x = π / 4$ inside the interval. On $[0, \frac{π}{4}]$ the cosine is larger, and on $[\frac{π}{4}, \frac{π}{2}]$ the sine is larger, so the element of area changes form halfway across:

dA=\begin{cases} (\cos x-\sin x)\,dx & 0\leq x\leq\frac{\pi}{4}\\[4pt] (\sin x-\cos x)\,dx & \frac{\pi}{4}\leq x\leq\frac{\pi}{2} \end{cases}

and the region must be split at $x = π / 4$ :

\begin{aligned} A &= \int_{0}^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx\\ &= \left[\sin x+\cos x\right]_{0}^{\pi/4}+\left[-\cos x-\sin x\right]_{\pi/4}^{\pi/2}\\ &= \left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-1\right)+\left(-0-1\right)-\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\right)\\ &= \left(\sqrt{2}-1\right)+\left(\sqrt{2}-1\right)\\ &= 2\sqrt{2}-2\approx0.828. \end{aligned}

By the symmetry of the two graphs about the line $x = π / 4$ , the two pieces have equal area, which is exactly what the calculation shows.

Find the area of the region bounded by $y = \frac{1}{x}$ , $y = \frac{1}{x^{2}}$ , and the line $x = 2$ .

Solution

The two curves meet where $\frac{1}{x} = \frac{1}{x^{2}}$ , that is $x = 1$ . For $x > 1$ we have $\frac{1}{x} > \frac{1}{x^{2}}$ , so the height of the typical rectangle is $\frac{1}{x} - \frac{1}{x^{2}}$ and the element of area is

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

Since $x$ runs from $1$ to $2$ ,

\begin{aligned} A &= \int_{1}^{2}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)dx\\ &= \left[\ln x+\frac{1}{x}\right]_{1}^{2}\\ &= \left(\ln2+\frac{1}{2}\right)-(0+1)\\ &= \ln2-\frac{1}{2}\approx0.193. \end{aligned}

The answer is positive, since $\ln 2 \approx 0.693 > 0.5$ , as it must be.

The region bounded by $y = x^{2}$ and $y = 4$ is cut into two pieces by the horizontal line $y = c$ , where $0 < c < 4$ . Find the value of $c$ for which the two pieces have equal area.

Solution

First find the whole area. The parabola meets $y = 4$ at $x = \pm 2$ , so the element of area for the whole region is $d A = (4 - x^{2}) d x$ and

A_{total} = \int_{- 2}^{2} (4 - x^{2}) d x = 2 {[4 x - \frac{x^{3}}{3}]}_{0}^{2} = \frac{32}{3} .

Now the lower piece is the region between $y = x^{2}$ and $y = c$ . There the element of area is

d A = (c - x^{2}) d x,

and the parabola meets $y = c$ at $x = \pm \sqrt{c}$ , so the lower area is

\begin{aligned} A_{\text{lower}} &= \int_{-\sqrt{c}}^{\sqrt{c}}\left(c-x^{2}\right)dx=2\left[cx-\frac{x^{3}}{3}\right]_{0}^{\sqrt{c}}\\ &= 2\left(c^{3/2}-\frac{c^{3/2}}{3}\right)=\frac{4}{3}c^{3/2}. \end{aligned}

Setting this equal to half the total area,

\frac{4}{3} c^{3 / 2} = \frac{16}{3} \Rightarrow c^{3 / 2} = 4 \Rightarrow c = 4^{2 / 3} = 2 \sqrt[3]{2} \approx 2.520 .

Notice that the bisecting line sits well above the midpoint $y = 2$ , because the region is much wider near the top than near the bottom.

Find the total area of the region enclosed by $y = x^{3}$ and $y = x$ .

Solution

The curves meet where $x^{3} = x$ , that is $x (x^{2} - 1) = 0$ , so $x = - 1, 0, 1$ . The two graphs cross at $x = 0$ , so the region consists of two loops and we must integrate separately on each side of the origin. On $[- 1, 0]$ we have $x^{3} \geq x$ , and on $[0, 1]$ we have $x \geq x^{3}$ , so the element of area is

dA=\begin{cases} \left(x^{3}-x\right)dx & -1\leq x\leq0\\[4pt] \left(x-x^{3}\right)dx & 0\leq x\leq1 \end{cases}

Hence

\begin{aligned} A &= \int_{-1}^{0}\left(x^{3}-x\right)dx+\int_{0}^{1}\left(x-x^{3}\right)dx\\ &= \left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0}+\left[\frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{1}\\ &= \left(0-\left(\frac{1}{4}-\frac{1}{2}\right)\right)+\left(\frac{1}{2}-\frac{1}{4}\right)\\ &= \frac{1}{4}+\frac{1}{4}=\frac{1}{2}. \end{aligned}

Had we carelessly written $A = \int_{- 1}^{1} (x - x^{3}) d x$ , the integrand would be an odd function and the answer would have been $0$ : the two loops would have cancelled instead of adding.

Two cars start side by side at time $t = 0$ . Car A has velocity $v_{A} (t) = 6 t$ metres per second and car B has velocity $v_{B} (t) = 3 t^{2}$ metres per second, for $0 \leq t \leq 4$ seconds. (a) At what time after $t = 0$ are the two cars side by side again? (b) How far ahead is car B at $t = 4$ ? (c) What is the total area between the two velocity curves on $[0, 4]$ , and what does it mean?

Solution

The velocity curves cross where $6 t = 3 t^{2}$ , that is $3 t (t - 2) = 0$ , so $t = 0$ and $t = 2$ . Car A is faster on $(0, 2)$ and car B is faster on $(2, 4)$ .

(a) The cars are side by side again when their displacements are equal, that is when the signed area between the curves is zero:

\int_{0}^{T} (6 t - 3 t^{2}) d t = {[3 t^{2} - t^{3}]}_{0}^{T} = T^{2} (3 - T) = 0.

Apart from $T = 0$ , this gives $T = 3$ seconds. Up to $t = 2$ car A pulls ahead by ${[3 t^{2} - t^{3}]}_{0}^{2} = 12 - 8 = 4$ metres, and between $t = 2$ and $t = 3$ car B makes up exactly those $4$ metres.

(b) The lead of car B at $t = 4$ is

\int_{0}^{4} (3 t^{2} - 6 t) d t = {[t^{3} - 3 t^{2}]}_{0}^{4} = 64 - 48 = 16 metres .

(c) Here the element of area is measured between the two velocity curves, with the faster car on top:

dA=\begin{cases} \left(6t-3t^{2}\right)dt & 0\leq t\leq2\\[4pt] \left(3t^{2}-6t\right)dt & 2\leq t\leq4 \end{cases}

The geometric area adds the two pieces with positive signs:

\begin{aligned} A &= \int_{0}^{2}\left(6t-3t^{2}\right)dt+\int_{2}^{4}\left(3t^{2}-6t\right)dt\\ &= \left[3t^{2}-t^{3}\right]_{0}^{2}+\left[t^{3}-3t^{2}\right]_{2}^{4}\\ &= (12-8)+\left[(64-48)-(8-12)\right]\\ &= 4+20=24. \end{aligned}

So the total area is $24$ metres: it is the distance car A gained while it was ahead, $4$ metres, plus the distance car B gained afterwards, $20$ metres. The net lead of $16$ metres from part (b) is the difference $20 - 4$ of the two pieces, not their sum. This is the same distinction as in the previous exercise between a signed integral and a true area.

Vertical Rectangles: Integrating with Respect to x

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