Corrected exercises: area between two curves, 201-SN3-RE (201-NYB)

Bounds: $x^{2}-2x=x+4\iff x^{2}-3x-4=0\iff(x+1)(x-4)=0$, so $x=-1$ and $x=4$. Test point $x=0$: the line gives $4$, the parabola $0$, so the line is on top. $A=\int_{-1}^{4}\left(-x^{2}+3x+4\right)dx=\left[-\frac{x^{3}}{3}+\frac{3x^{2}}{2}+4x\right]_{-1}^{4}=\frac{56}{3}-\left(-\frac{13}{6}\right)=\frac{125}{6}\approx 20.833$.

Bounds: $3-x^{2}=x^{2}-5\iff x^{2}=4$, so $x=\pm 2$. At $x=0$: $3>-5$, so the parabola $y=3-x^{2}$ is on top. $A=\int_{-2}^{2}\left(8-2x^{2}\right)dx=\left[8x-\frac{2x^{3}}{3}\right]_{-2}^{2}=2\left(16-\frac{16}{3}\right)=\frac{64}{3}\approx 21.333$. Pitfall: $(3-x^{2})-(x^{2}-5)=8-2x^{2}$, the two $x^{2}$ terms add up instead of cancelling.

Bounds: $2\sqrt{x}=x\iff 4x=x^{2}\iff x(x-4)=0$, so $x=0$ and $x=4$. At $x=1$: $2>1$, so the square root is on top. $A=\int_{0}^{4}\left(2\sqrt{x}-x\right)dx=\left[\frac{4}{3}x^{3/2}-\frac{x^{2}}{2}\right]_{0}^{4}=\frac{32}{3}-8=\frac{8}{3}\approx 2.667$. Pitfall: dividing $2\sqrt{x}=x$ by $\sqrt{x}$ and losing the bound $x=0$.

$x^{3}-x^{2}-2x=x(x+1)(x-2)$ vanishes at $-1$, $0$ and $2$. At $x=-\frac{1}{2}$ the function equals $\frac{5}{8}>0$, at $x=1$ it equals $-2<0$: you must split. With $F(x)=\frac{x^{4}}{4}-\frac{x^{3}}{3}-x^{2}$, we have $F(-1)=-\frac{5}{12}$, $F(0)=0$ and $F(2)=-\frac{8}{3}$. $A=\int_{-1}^{0}f-\int_{0}^{2}f=\frac{5}{12}+\frac{8}{3}=\frac{37}{12}\approx 3.083$. Pitfall: $\int_{-1}^{2}f=-\frac{9}{4}$ is a signed integral, not an area.

The bounds are given. The curves touch at $x=0$, where $y=x+1$ is the tangent line to $y=e^{x}$, and the exponential stays above its tangent line: at $x=1$, $e>2$. $A=\int_{0}^{2}\left(e^{x}-x-1\right)dx=\left[e^{x}-\frac{x^{2}}{2}-x\right]_{0}^{2}=(e^{2}-4)-1=e^{2}-5\approx 2.389$. Pitfall: forgetting the value of the antiderivative at $0$, which is $e^{0}=1$ and not $0$.

Bounds: $\frac{4}{x}=5-x\iff x^{2}-5x+4=0$, so $x=1$ and $x=4$. At $x=2$: the line gives $3$, the hyperbola $2$, so the line is on top. $A=\int_{1}^{4}\left(5-x-\frac{4}{x}\right)dx=\left[5x-\frac{x^{2}}{2}-4\ln x\right]_{1}^{4}=\left(12-4\ln 4\right)-\frac{9}{2}=\frac{15}{2}-8\ln 2\approx 1.955$.

The curve is a sideways parabola, $x$ as a function of $y$: integrate with respect to $y$. It meets the $y$-axis, $x=0$, when $y^{2}-2y=0$, that is $y=0$ and $y=2$. At $y=1$, $x=-1<0$: the curve is to the LEFT of the axis. $A=\int_{0}^{2}\left(0-(y^{2}-2y)\right)dy=\left[y^{2}-\frac{y^{3}}{3}\right]_{0}^{2}=4-\frac{8}{3}=\frac{4}{3}\approx 1.333$. Pitfall: integrating $y^{2}-2y$ as it stands, which gives $-\frac{4}{3}$; it is “right minus left”, and here the right-hand boundary is the axis.

Bounds: $x^{3}-2x^{2}=3x\iff x(x^{2}-2x-3)=0\iff x(x+1)(x-3)=0$, so $x=-1$, $0$ and $3$. At $x=-\frac{1}{2}$, the cubic equals $-\frac{5}{8}$ and the line $-\frac{3}{2}$: the cubic is on top; at $x=1$, the cubic equals $-1$ and the line $3$: the line is on top. With $G(x)=\frac{x^{4}}{4}-\frac{2x^{3}}{3}-\frac{3x^{2}}{2}$, an antiderivative of the cubic minus the line, $G(-1)=-\frac{7}{12}$, $G(0)=0$ and $G(3)=-\frac{45}{4}$. $A=\left(G(0)-G(-1)\right)-\left(G(3)-G(0)\right)=\frac{7}{12}+\frac{135}{12}=\frac{71}{6}\approx 11.833$. Pitfall: a single integral from $-1$ to $3$, where the two pieces partly cancel each other.

$\int_{-2}^{2}\left(x^{3}-4x\right)dx=\left[\frac{x^{4}}{4}-2x^{2}\right]_{-2}^{2}=(4-8)-(4-8)=0$. Without computing: $f$ is ODD, since $f(-x)=-x^{3}+4x=-f(x)$, and the interval is symmetric about the origin. The integral of an odd function over a symmetric interval is always zero, which saves a minute and earns the justification mark.

The area cannot be zero: the curve clearly moves away from the axis. You must split at the ZEROS of $f$, that is where $x(x-2)(x+2)=0$, so at $-2$, $0$ and $2$. On $\left(-2, 0\right)$, $f>0$ (for example $f(-1)=3$); on $\left(0, 2\right)$, $f<0$ (for example $f(1)=-3$). So $\mathcal{A}=\int_{-2}^{0}f(x)\,dx-\int_{0}^{2}f(x)\,dx$. The first piece equals $\left[\frac{x^{4}}{4}-2x^{2}\right]_{-2}^{0}=0-(4-8)=4$ and the second $\left[\frac{x^{4}}{4}-2x^{2}\right]_{0}^{2}=-4$. Hence $\mathcal{A}=4-(-4)=8$.

The formula is $\mathcal{A}=\int_{a}^{b}\left|f(x)\right|dx$, and it is ALWAYS computed by splitting at the zeros of $f$, never by putting absolute values inside an antiderivative. “Compute the integral”, “compute the displacement”, “compute the net change” call for the signed integral; “compute the area”, “compute the distance travelled”, “compute the total amount” call for the integral of the absolute value. The most frequent error on exams is the question that says “area” and the paper that answers with the integral: the whole question is lost, even though the computation is correct.

On $\left[0, 3\right]$, $f$ changes sign once more, at $x=2$. On $\left(0, 2\right)$, $f<0$ and on $\left(2, 3\right)$, $f>0$ (for example $f(3)=15$). So $\mathcal{A}=-\int_{0}^{2}f+\int_{2}^{3}f=4+\left[\frac{x^{4}}{4}-2x^{2}\right]_{2}^{3}=4+\left(\frac{81}{4}-18\right)-(4-8)=4+2.25+4=10.25$. The reflex to build: before any antiderivative, look for the zeros in the interval. A single forgotten zero throws off everything else, and the omission does not show in the answer.

Set them equal: $x^{2}=x+2$, so $x^{2}-x-2=0$, so $(x-2)(x+1)=0$. The $x$-coordinates of the intersections are $x=-1$ and $x=2$. These two numbers are the BOUNDS of the integral, which is why you look for them first: without them, there is nothing to integrate. No bound is given in the problem, and that is the most frequent case on exams.

Test point inside $\left(-1, 2\right)$, the simplest being $x=0$: the line equals $2$ and the parabola equals $0$. The LINE is therefore on top over the whole interval, and it stays there since the two curves do not meet again between $-1$ and $2$. A single test point is enough, provided it is strictly inside: that is exactly the argument the grader expects, and it is worth half a mark.

$\mathcal{A}=\int_{-1}^{2}\left[(x+2)-x^{2}\right]dx=\left[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}\right]_{-1}^{2}$. At $2$: $2+4-\frac{8}{3}=\frac{10}{3}$. At $-1$: $\frac{1}{2}-2+\frac{1}{3}=-\frac{7}{6}$. So $\mathcal{A}=\frac{10}{3}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}=4.5$. Order-of-magnitude check: the region fits in a rectangle $3$ wide and $4$ high, so less than $12$; and it visibly fills a little less than half of it ✓.

It is the same integral with the two functions SWAPPED, so it equals $-\frac{9}{2}$. An area cannot be negative: the negative sign is the signal, and the only one, that the order “top minus bottom” was reversed. You fix it by changing the sign, not by redoing the computation. Remembering this reading saves time on the exam: a negative area is repaired in three seconds, provided the integral was set up with its bounds in increasing order.

$x^{3}=x$ gives $x\left(x^{2}-1\right)=0$, so $x=-1$, $x=0$ and $x=1$. THREE intersections, not two: that is what the average paper misses, because it divides by $x$ and loses the solution $x=0$. Never divide by an expression that can be zero; factor instead.

Test point in $\left(-1, 0\right)$, for example $x=-\frac{1}{2}$: $x^{3}=-\frac{1}{8}$ and $x=-\frac{1}{2}$, so the CUBIC is on top. Test point in $\left(0, 1\right)$, for example $x=\frac{1}{2}$: $x^{3}=\frac{1}{8}$ and $x=\frac{1}{2}$, so the LINE is on top. The roles of top and bottom switch at $x=0$, which forces you to split the integral there.

$\mathcal{A}=\int_{-1}^{0}\left(x^{3}-x\right)dx+\int_{0}^{1}\left(x-x^{3}\right)dx$. With $H(x)=\frac{x^{4}}{4}-\frac{x^{2}}{2}$: the first equals $H(0)-H(-1)=0-\left(\frac{1}{4}-\frac{1}{2}\right)=\frac{1}{4}$; the second equals $-\left[H(1)-H(0)\right]=-\left(-\frac{1}{4}\right)=\frac{1}{4}$. So $\mathcal{A}=\frac{1}{2}$. You could also use symmetry: both functions are odd, so the region is symmetric about the origin and it is enough to double one of the two pieces.

$\int_{-1}^{1}\left(x-x^{3}\right)dx=0$, because the integrand is odd on a symmetric interval. This number is not the area because the expression $x-x^{3}$ is POSITIVE on $\left(0, 1\right)$ and NEGATIVE on $\left(-1, 0\right)$: the two halves cancel exactly. Writing a single integral over $\left[-1, 1\right]$ amounts to subtracting an area instead of adding it. The check that saves you: a zero area when the drawing shows two well-filled regions always signals a missing split.

Set the two expressions of $x$ equal: $y^{2}=y+2$, so $y^{2}-y-2=0$, so $(y-2)(y+1)=0$. The $y$-coordinates are $y=-1$ and $y=2$, and the points are $(1, -1)$ and $(4, 2)$. Note that you set the $x$ equal because it is $x$ that is expressed as a function of $y$: the variable of integration will be $y$.

A horizontal strip at height $y$ runs from the LEFT curve to the RIGHT curve. Test point $y=0$: the parabola gives $x=0$ and the line $x=2$, so the line is on the right. $\mathcal{A}=\int_{-1}^{2}\left[(y+2)-y^{2}\right]dy=\left[\frac{y^{2}}{2}+2y-\frac{y^{3}}{3}\right]_{-1}^{2}=\frac{10}{3}+\frac{7}{6}=\frac{9}{2}$. With horizontal strips, the formula is “right minus left”, exactly like “top minus bottom” with vertical strips.

With vertical strips, you need TWO. The reason is that the lower boundary of the region changes nature at $x=1$: for $0\le x\le 1$, the region is bounded above by $y=\sqrt{x}$ and below by $y=-\sqrt{x}$; for $1\le x\le 4$, it is bounded above by $y=\sqrt{x}$ and below by the line $y=x-2$. So $\mathcal{A}=\int_{0}^{1}2\sqrt{x}\,dx+\int_{1}^{4}\left(\sqrt{x}-x+2\right)dx=\frac{4}{3}+\left(\frac{16}{3}-8+8\right)-\left(\frac{2}{3}-\frac{1}{2}+2\right)=\frac{4}{3}+\frac{16}{3}-\frac{13}{6}=\frac{9}{2}$ ✓. Same answer, three times the work: that is the criterion for choosing. Integrate in $y$ as soon as a boundary changes formula with vertical strips, and in $x$ otherwise.

The region here is the image of the one in exercise 2 under the reflection in the line $y=x$, which swaps the roles of $x$ and $y$: $y=x^{2}$ becomes $x=y^{2}$ and $y=x+2$ becomes $x=y+2$. A reflection preserves areas, so both equal $\frac{9}{2}$. This is not a coincidence but a tool: when a region is awkward to compute in one direction, check whether its reflection is easier, which amounts exactly to changing the direction of the strips.

$\sin x=\cos x$ is equivalent to $\tan x=1$ (the cosine is not zero where the equality holds, otherwise the sine would be zero too, which is impossible). So $x=\frac{\pi}{4}$ or $x=\frac{5\pi}{4}$ on $\left[0, 2\pi\right]$. Two solutions per turn, one per period of the tangent.

Test point $x=\frac{\pi}{2}$: $\sin=1$ and $\cos=0$, so the SINE is on top over the whole interval. $\mathcal{A}=\int_{\pi/4}^{5\pi/4}\left(\sin x-\cos x\right)dx=\left[-\cos x-\sin x\right]_{\pi/4}^{5\pi/4}$. At $\frac{5\pi}{4}$: $-\left(-\frac{\sqrt{2}}{2}\right)-\left(-\frac{\sqrt{2}}{2}\right)=\sqrt{2}$. At $\frac{\pi}{4}$: $-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}$. So $\mathcal{A}=2\sqrt{2}\approx 2.83$.

On $\left[0, 2\pi\right]$ the two curves cross at $\frac{\pi}{4}$ and $\frac{5\pi}{4}$, which cuts the interval into three pieces. On $\left[0,\frac{\pi}{4}\right]$ the cosine is on top, on $\left[\frac{\pi}{4},\frac{5\pi}{4}\right]$ the sine, on $\left[\frac{5\pi}{4},2\pi\right]$ the cosine again. The first piece equals $\left[\sin x+\cos x\right]_{0}^{\pi/4}=\sqrt{2}-1$, the second $2\sqrt{2}$, the third $\left[\sin x+\cos x\right]_{5\pi/4}^{2\pi}=1-\left(-\sqrt{2}\right)=1+\sqrt{2}$. Total: $\left(\sqrt{2}-1\right)+2\sqrt{2}+\left(1+\sqrt{2}\right)=4\sqrt{2}\approx 5.66$. The answer comes out round, which is a good sign that the computation is right.

$\int_{0}^{2\pi}\sin x\,dx=\left[-\cos x\right]_{0}^{2\pi}=-1+1=0$: over a full turn, the positive arch and the negative arch cancel. The AREA, on the other hand, is computed by splitting at $x=\pi$: $\mathcal{A}=\int_{0}^{\pi}\sin x\,dx-\int_{\pi}^{2\pi}\sin x\,dx=2+2=4$. It is the same lesson as in exercise 1, on a periodic function: a zero integral never means a zero area, and periodicity makes this pitfall even more frequent, because zeros are numerous.

Intersect the lines two at a time. $y=x$ and $y=2x$ meet at $(0, 0)$. $y=2x$ and $y=6-x$ give $2x=6-x$, so $x=2$ and the point $(2, 4)$. $y=x$ and $y=6-x$ give $x=3$ and the point $(3, 3)$. The region is the triangle with vertices $(0, 0)$, $(2, 4)$ and $(3, 3)$.

The UPPER boundary changes formula at $x=2$: before, it is $y=2x$; after, it is $y=6-x$. The lower boundary stays $y=x$ across the whole width. So $\mathcal{A}=\int_{0}^{2}\left(2x-x\right)dx+\int_{2}^{3}\left[(6-x)-x\right]dx$. The general rule: split at every $x$-value where one of the two boundaries changes expression, and the vertex $(2, 4)$ is precisely such an $x$-value.

$\int_{0}^{2}x\,dx=\left[\frac{x^{2}}{2}\right]_{0}^{2}=2$. $\int_{2}^{3}\left(6-2x\right)dx=\left[6x-x^{2}\right]_{2}^{3}=(18-9)-(12-4)=9-8=1$. So $\mathcal{A}=2+1=3$.

With horizontal strips, the LEFT boundary changes at $y=3$: for $0\le y\le 3$, the strip runs from $x=\frac{y}{2}$ to $x=y$; for $3\le y\le 4$, it runs from $x=\frac{y}{2}$ to $x=6-y$. So $\mathcal{A}=\int_{0}^{3}\frac{y}{2}\,dy+\int_{3}^{4}\left(6-\frac{3y}{2}\right)dy=\frac{9}{4}+\left[6y-\frac{3y^{2}}{4}\right]_{3}^{4}=\frac{9}{4}+\left(24-12\right)-\left(18-\frac{27}{4}\right)=\frac{9}{4}+12-\frac{45}{4}=3$ ✓. Geometric check with the determinant formula on the three vertices: $\mathcal{A}=\frac{1}{2}\left|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right|=\frac{1}{2}\left|0(4-3)+2(3-0)+3(0-4)\right|=\frac{1}{2}\left|6-12\right|=3$ ✓. Three methods, one number: when a region is a polygon, the geometric check costs ten seconds and is worth more than rereading.

Intersections: $x^{2}=4$ gives $x=\pm 2$. On $\left(-2, 2\right)$, the line is on top (test point $x=0$: $4>0$). So $\mathcal{A}=\int_{-2}^{2}\left(4-x^{2}\right)dx$. Since the function is EVEN, double one half: $2\int_{0}^{2}\left(4-x^{2}\right)dx=2\left[4x-\frac{x^{3}}{3}\right]_{0}^{2}=2\left(8-\frac{8}{3}\right)=\frac{32}{3}$.

With horizontal strips, the width of $R$ at height $y$ runs from $x=-\sqrt{y}$ to $x=+\sqrt{y}$, so it equals $2\sqrt{y}$. The upper part is the one with $y$-coordinates between $a$ and $4$: $\mathcal{A}_{\text{upper}}(a)=\int_{a}^{4}2\sqrt{y}\,dy=\left[\frac{4}{3}y^{3/2}\right]_{a}^{4}=\frac{4}{3}\left(8-a^{3/2}\right)$. Check: for $a=0$ we get back $\frac{32}{3}$ ✓, the total area. Horizontal strips are all but compulsory here: with vertical strips, the upper part would need a split at $x=\pm\sqrt{a}$ and would cost three integrals.

We want $\mathcal{A}_{\text{upper}}(a)=\frac{1}{2}\cdot\frac{32}{3}=\frac{16}{3}$, so $\frac{4}{3}\left(8-a^{3/2}\right)=\frac{16}{3}$, so $8-a^{3/2}=4$, so $a^{3/2}=4$. Raising to the power $\frac{2}{3}$: $a=4^{2/3}=\sqrt[3]{16}\approx 2.52$. Check: $\frac{4}{3}\left(8-4\right)=\frac{16}{3}$ ✓, and $2.52$ does lie in $\left(0, 4\right)$ ✓.

No, and that is the whole point of the question. For a rectangle, the width is constant, the area grows linearly with the height, and the horizontal line that halves it is at mid-height, that is $a=2$. Here the width $2\sqrt{y}$ INCREASES with $y$: the upper half of the region is wider, so the cut has to move above $2$ to leave it the same area, and we do find $2.52>2$. The lesson generalizes: the line that cuts an area in half is at mid-height only if the region is as wide at the top as at the bottom.