Area between curves

Worked solutions

Problem 1

1. $x = x^2$ gives $x(x - 1) = 0$, so $x = 0$ or $x = 1$.The region is enclosed by the two curves, so its limits are where they meet.
2. On $(0, 1)$, $x > x^2$.Test a point: at $x = \tfrac12$, $\tfrac12 > \tfrac14$. The line is on top.
3. Area $= \displaystyle\int_0^1 (x - x^2)\,dx = \Big[\tfrac{x^2}2 - \tfrac{x^3}3\Big]_0^1$Top minus bottom.
4. $\tfrac16$$\tfrac12 - \tfrac13 = \tfrac16$.

Problem 2

1. $4 - x^2 = x + 2$ gives $x^2 + x - 2 = (x + 2)(x - 1) = 0$, so $x = -2$ or $x = 1$.The intersections are the limits.
2. On $(-2, 1)$, $4 - x^2 > x + 2$.At $x = 0$: $4 > 2$. The parabola is on top.
3. Area $= \displaystyle\int_{-2}^1 \big((4 - x^2) - (x + 2)\big)\,dx = \int_{-2}^1 (2 - x - x^2)\,dx$Top minus bottom; the parentheses around $x + 2$ make both of its terms subtracted.
4. $\Big[2x - \tfrac{x^2}2 - \tfrac{x^3}3\Big]_{-2}^1 = \Big(2 - \tfrac12 - \tfrac13\Big) - \Big(-4 - 2 + \tfrac83\Big) = \tfrac76 + \tfrac{10}3$The antiderivative at the upper limit minus at the lower one.
5. $\tfrac92$$\tfrac76 + \tfrac{20}6 = \tfrac{27}6 = \tfrac92$.

Problem 3

1. $x^3 = x$ gives $x(x - 1)(x + 1) = 0$, so $x = -1, 0, 1$.Three intersections: two enclosed regions, $[-1, 0]$ and $[0, 1]$.
2. On $(-1, 0)$, $x^3 > x$; on $(0, 1)$, $x > x^3$.At $x = -\tfrac12$: $-\tfrac18 > -\tfrac12$. At $x = \tfrac12$: $\tfrac12 > \tfrac18$. The top curve changes at 0.
3. $\displaystyle\int_{-1}^0 (x^3 - x)\,dx = \Big[\tfrac{x^4}4 - \tfrac{x^2}2\Big]_{-1}^0 = 0 - \Big(\tfrac14 - \tfrac12\Big) = \tfrac14$Top minus bottom on the left piece.
4. $\displaystyle\int_0^1 (x - x^3)\,dx = \tfrac12 - \tfrac14 = \tfrac14$Top minus bottom on the right piece. Both curves are odd functions, so the two pieces are mirror images and the areas match.
5. $\tfrac12$$\tfrac14 + \tfrac14$. A single integral $\int_{-1}^1 (x - x^3)\,dx$ gives 0, because its two halves cancel.

Problem 4

1. $\sin x = \cos x$ on $[0, \tfrac\pi2]$ at $x = \tfrac\pi4$ only.$\tan x = 1$; the curves cross inside the interval, so it must be split there.
2. On $(0, \tfrac\pi4)$, $\cos x > \sin x$; on $(\tfrac\pi4, \tfrac\pi2)$, $\sin x > \cos x$.At 0, $\cos 0 = 1 > 0 = \sin 0$; at $\tfrac\pi2$ the order is reversed.
3. $\displaystyle\int_0^{\pi/4}(\cos x - \sin x)\,dx = \Big[\sin x + \cos x\Big]_0^{\pi/4} = \sqrt2 - 1$$\sin\tfrac\pi4 + \cos\tfrac\pi4 = \sqrt2$ and $\sin 0 + \cos 0 = 1$.
4. $\displaystyle\int_{\pi/4}^{\pi/2}(\sin x - \cos x)\,dx = \Big[-\cos x - \sin x\Big]_{\pi/4}^{\pi/2} = -1 - (-\sqrt2) = \sqrt2 - 1$Top minus bottom on the second piece. The picture is symmetric about $x = \tfrac\pi4$, so the pieces are equal.
5. $2\sqrt2 - 2 \approx 0.83$The sum of the two pieces. Integrating $\cos x - \sin x$ over the whole interval gives $(1 + 0) - (0 + 1) = 0$ instead (mistake 2).

Problem 5

1. In $y$: $x = y^2$ is the left boundary and $x = y + 2$ the right one, for $0 \le y \le 2$.Solve each curve for $x$. They meet where $y^2 = y + 2$, at $y = 2$ ($y = -1$ is below the $x$-axis, outside the region), and the $x$-axis is $y = 0$.
2. Area $= \displaystyle\int_0^2 \big((y + 2) - y^2\big)\,dy = \Big[\tfrac{y^2}2 + 2y - \tfrac{y^3}3\Big]_0^2 = 2 + 4 - \tfrac83$Right minus left, one integral.
3. In $x$: the bottom is $y = 0$ for $0 \le x \le 2$ and $y = x - 2$ for $2 \le x \le 4$; the top is $\sqrt x$ throughout.The line crosses the $x$-axis at $x = 2$ and meets $\sqrt x$ at $x = 4$ (where $\sqrt4 = 2 = 4 - 2$), so the bottom changes formula at $x = 2$.
4. $\displaystyle\int_0^2 \sqrt x\,dx + \int_2^4 \big(\sqrt x - x + 2\big)\,dx = \tfrac23 2^{3/2} + \Big[\tfrac23x^{3/2} - \tfrac{x^2}2 + 2x\Big]_2^4$Two integrals, one for each bottom.
5. $= \tfrac{4\sqrt2}3 + \Big(\tfrac{16}3 - 8 + 8\Big) - \Big(\tfrac{4\sqrt2}3 - 2 + 4\Big) = \tfrac{16}3 - 2$The $\tfrac{4\sqrt2}3$ terms cancel.
6. $\tfrac{10}3$Both routes: $6 - \tfrac83 = \tfrac{10}3$ and $\tfrac{16}3 - 2 = \tfrac{10}3$. In $y$ it is one integral of a polynomial.

Problem 6

1. $\tfrac1x = \tfrac52 - x$ gives $2 = 5x - 2x^2$, so $2x^2 - 5x + 2 = (2x - 1)(x - 2) = 0$.Multiply both sides by $2x$, which is not 0 at an intersection.
2. $x = \tfrac12$ or $x = 2$, and on $(\tfrac12, 2)$ the line is on top.At $x = 1$: $\tfrac52 - 1 = \tfrac32 > 1$.
3. Area $= \displaystyle\int_{1/2}^2 \Big(\tfrac52 - x - \tfrac1x\Big)\,dx = \Big[\tfrac52x - \tfrac{x^2}2 - \ln x\Big]_{1/2}^2$$x > 0$ on the interval, so $\ln\lvert x\rvert = \ln x$.
4. $= (5 - 2 - \ln 2) - \Big(\tfrac54 - \tfrac18 - \ln\tfrac12\Big) = (3 - \ln 2) - \Big(\tfrac98 + \ln 2\Big)$$\ln\tfrac12 = -\ln 2$, so $-\ln\tfrac12 = +\ln 2$.
5. $\tfrac{15}8 - 2\ln 2 \approx 0.489$$3 - \tfrac98 = \tfrac{15}8$, and the two $\ln 2$ terms add. Sanity check: the gap is at most $\tfrac12$ (at $x = 1$) over an interval of length $\tfrac32$, so the area is below $\tfrac34$.

Problem 7

1. Total area $= \displaystyle\int_{-2}^2 (4 - x^2)\,dx = 16 - \tfrac{16}3 = \tfrac{32}3$.The curves meet at $x = \pm2$; top minus bottom.
2. At height $y$ the region runs from $x = -\sqrt y$ to $x = \sqrt y$, a width of $2\sqrt y$.The cut is horizontal, so slice horizontally: integrate in $y$.
3. Area below $y = c$: $\displaystyle\int_0^c 2\sqrt y\,dy = \tfrac43c^{3/2}$.Right minus left is $2\sqrt y$, from the bottom of the region, $y = 0$, up to the cut.
4. $\tfrac43c^{3/2} = \tfrac{16}3$, so $c^{3/2} = 4$.Half of $\tfrac{32}3$. As a check, $c = 4$ gives $\tfrac43\cdot8 = \tfrac{32}3$, the whole region.
5. $c = 4^{2/3} = 2^{4/3} \approx 2.52$Raise both sides to the power $\tfrac23$. The cut is above the halfway height $y = 2$, because the region is narrow at the bottom (mistake 4).

Problem 8

1. $\displaystyle\int_0^1 \big(p(x) - q(x)\big)\,dx = 1 - 1 = 0$.Every density integrates to 1, so the signed integral of the difference of two densities is always 0. It says nothing about how different they are.
2. $2x = 1$ at $x = \tfrac12$; $q$ is on top on $(0, \tfrac12)$ and $p$ on $(\tfrac12, 1)$.The crossing splits the interval.
3. $\displaystyle\int_0^{1/2}(1 - 2x)\,dx = \Big[x - x^2\Big]_0^{1/2} = \tfrac14$ and $\displaystyle\int_{1/2}^1(2x - 1)\,dx = \Big[x^2 - x\Big]_{1/2}^1 = 0 - \big(-\tfrac14\big) = \tfrac14$Top minus bottom on each piece. The two pieces are equal, as step 1 forces: their difference is the signed integral, 0.
4. Area $\tfrac12$; total variation distance $\tfrac14$The area is $\tfrac14 + \tfrac14$, and the distance is half of it. It is also the largest difference in probability the two densities give any event: $P(x > \tfrac12)$ is $\tfrac34$ under $p$ and $\tfrac12$ under $q$.

Problem 9

1. The line is $y = \dfrac{x - 1}{e - 1}$, and it meets $\ln x$ at $x = 1$ and $x = e$.Slope $\tfrac{1 - 0}{e - 1}$ through $(1, 0)$. Both points are on $y = \ln x$, since $\ln 1 = 0$ and $\ln e = 1$.
2. On $(1, e)$, $\ln x$ is above the line.$\ln x$ is concave, so it lies above every chord. At $x = 2$: $\ln 2 \approx 0.693 > \tfrac1{e - 1} \approx 0.582$.
3. $\displaystyle\int_1^e \ln x\,dx = \Big[x\ln x - x\Big]_1^e = (e - e) - (0 - 1) = 1$By parts, with $u = \ln x$ and $dv = dx$.
4. $\displaystyle\int_1^e \frac{x - 1}{e - 1}\,dx = \frac{1}{e - 1}\cdot\frac{(e - 1)^2}{2} = \frac{e - 1}2$$\int_1^e (x - 1)\,dx = \Big[\tfrac{(x - 1)^2}2\Big]_1^e$. Geometrically, a triangle with base $e - 1$ and height 1.
5. $1 - \dfrac{e - 1}2 = \dfrac{3 - e}2 \approx 0.141$Top minus bottom: step 3 minus step 4.

Problem 10

1. $\sin 2x = \sin x$ gives $2\sin x\cos x - \sin x = \sin x\,(2\cos x - 1) = 0$.The double-angle formula, then factor. Dividing by $\sin x$ would lose the crossings at 0 and $\pi$.
2. On $[0, \pi]$: $\sin x = 0$ at $x = 0, \pi$, and $\cos x = \tfrac12$ at $x = \tfrac\pi3$.So the curves cross inside the interval at $\tfrac\pi3$.
3. On $(0, \tfrac\pi3)$, $\sin 2x > \sin x$; on $(\tfrac\pi3, \pi)$, $\sin x > \sin 2x$.$\sin 2x - \sin x = \sin x\,(2\cos x - 1)$, and $\sin x > 0$ on $(0, \pi)$, so the sign is that of $2\cos x - 1$: positive before $\tfrac\pi3$, negative after.
4. $\displaystyle\int_0^{\pi/3}(\sin 2x - \sin x)\,dx = \Big[-\tfrac{\cos 2x}2 + \cos x\Big]_0^{\pi/3} = \Big(\tfrac14 + \tfrac12\Big) - \Big(-\tfrac12 + 1\Big) = \tfrac14$$\cos\tfrac{2\pi}3 = -\tfrac12$ and $\cos\tfrac\pi3 = \tfrac12$.
5. $\displaystyle\int_{\pi/3}^{\pi}(\sin x - \sin 2x)\,dx = \Big[-\cos x + \tfrac{\cos 2x}2\Big]_{\pi/3}^{\pi} = \Big(1 + \tfrac12\Big) - \Big(-\tfrac12 - \tfrac14\Big) = \tfrac94$$\cos\pi = -1$ and $\cos 2\pi = 1$.
6. $\tfrac52$$\tfrac14 + \tfrac94$. The single integral $\int_0^\pi(\sin x - \sin 2x)\,dx$ is $2 - 0 = 2$, short by twice the first piece.