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1. 1.Computation · 7 min

   Find each area exactly. Before integrating, state the intersection points, say which curve is on top (or to the right) on each piece and how you decided, and choose the variable of integration deliberately.

   1. (a)The region enclosed by $y=x^3-x$ and $y=3x$.
   2. (b)The region enclosed by $x=y^2$ and $x=3-2y$.
   3. (c)Explain why ${\int }_{-2}^2[3x-(x^3-x)]dx$ is not the answer to (a), and say what number it produces.
   Show the solution to question 1

   (a) Find the crossings, then split. The curves meet where $x^3-x=3x$, that is $x^3-4x=x(x-2)(x+2)=0$, so at $x=-2$, $0$ and $2$. Three crossings mean two pieces, and the curves swap places at $x=0$.

   On $[0,2]$ test $x=1$: the line gives $3$ and the cubic gives $0$, so the line is on top. On $[-2,0]$ test $x=-1$: the line gives $-3$ and the cubic gives $0$, so the cubic is on top. (Two continuous curves that do not cross inside a piece cannot swap, so one test point per piece settles it.)

   Both functions are odd, so the region is symmetric through the origin and the two pieces have equal area. Hence

   $$A=2{\int }_0^2[3x-(x^3-x)]dx=2{\int }_0^2\left(4x-x^3\right)dx=2{\left[2x^2-\frac{x^4}{4}\right]}_0^2=2(8-4)=8.$$

   If you prefer not to trust the symmetry, the left piece is ${\int }_{-2}^0\left(x^3-4x\right)dx={\left[\frac{x^4}{4}-2x^2\right]}_{-2}^0=0-(4-8)=4$, the same as the right piece.

   (b) Integrate in $y$. Both boundaries are already given as $x$ in terms of $y$, and each is a single formula for the whole region, so horizontal strips are the natural choice. The curves meet where $y^2=3-2y$, that is $y^2+2y-3=(y+3)(y-1)=0$, so at $y=-3$ and $y=1$. At $y=0$ the line gives $x=3$ and the parabola $x=0$, so the line is on the right. Right minus left:

   $$A={\int }_{-3}^1[(3-2y)-y^2]dy={\left[3y-y^2-\frac{y^3}{3}\right]}_{-3}^1=\left(3-1-\frac{1}{3}\right)-\left(-9-9+9\right)=\frac{5}{3}+9=\frac{32}{3}.$$

   Why not $x$? The crossings are $(1,1)$ and $(9,-3)$. For $0\le x\le 1$ a vertical strip runs from $y=-\sqrt{x}$ to $y=\sqrt{x}$, but for $1\le x\le 9$ it runs from $y=-\sqrt{x}$ up to the line $y=\frac{3-x}{2}$: the top boundary changes formula at $x=1$, so the $x$-route is two integrals with square roots. It does give the same answer: ${\int }_0^{1}2\sqrt{x}dx+{\int }_1^9\left(\frac{3-x}{2}+\sqrt{x}\right)dx=\frac{4}{3}+\frac{45}{4}-\frac{23}{12}=\frac{32}{3}$.

   (c) It is a signed area. The integrand $3x-(x^3-x)=4x-x^3$ is "top minus bottom" only on $[0,2]$; on $[-2,0]$ it is negative, so the left piece is counted with a minus sign. Since $4x-x^3$ is odd, the two pieces cancel exactly:

   $${\int }_{-2}^2\left(4x-x^3\right)dx={\left[2x^2-\frac{x^4}{4}\right]}_{-2}^2=(8-4)-(8-4)=0.$$

   A region of area $8$ reported as $0$ is the signed-area trap: locate the crossings first, then integrate $|f-g|$ piece by piece.

2. 2.Computation · 7 min

   The base of a solid is the region $\mathcal{B}$ in the $xy$-plane bounded by the parabola $y=4-x^2$ and the $x$-axis. For each solid, say what shape a typical slice is, express its area in terms of the slicing variable, and find the volume exactly.

   1. (a)Cross-sections perpendicular to the $y$-axis are equilateral triangles with one side lying in $\mathcal{B}$.
   2. (b)Cross-sections perpendicular to the $x$-axis are semicircles whose diameter lies in $\mathcal{B}$.
   3. (c)For each solid, give an upper bound for the volume using only the largest slice and the length of the solid, and check your answers against it.
   Show the solution to question 2

   (a) Slice in $y$. The slices are perpendicular to the $y$-axis, so they are indexed by $y$, and $\mathcal{B}$ occupies $0\le y\le 4$. At height $y$ the base is the horizontal chord of the parabola, from $x=-\sqrt{4-y}$ to $x=\sqrt{4-y}$, so the triangle's side is $s=2\sqrt{4-y}$. An equilateral triangle of side $s$ has area $\frac{\sqrt{3}}{4}{s}^2$, so

   $$A(y)=\frac{\sqrt{3}}{4}\cdot 4(4-y)=\sqrt{3}(4-y).$$

   By the volume-by-slicing formula,

   $$V={\int }_0^4\sqrt{3}(4-y)dy=\sqrt{3}{\left[4y-\frac{y^2}{2}\right]}_0^4=\sqrt{3}(16-8)=8\sqrt{3}.$$

   (b) Slice in $x$. Now the slices are indexed by $x$, with $-2\le x\le 2$. At position $x$ the base runs from $y=0$ to $y=4-x^2$; that length is the diameter of the semicircle, so the radius is $\frac{1}{2}(4-x^2)$ and

   $$A(x)=\frac{1}{2}\pi {\left(\frac{4-x^2}{2}\right)}^2=\frac{\pi }{8}{\left(4-x^2\right)}^2.$$

   The integrand is even, so integrate over $[0,2]$ and double:

   $$V=2\cdot \frac{\pi }{8}{\int }_0^2\left(16-8x^2+x^4\right)dx=\frac{\pi }{4}{\left[16x-\frac{8x^3}{3}+\frac{x^5}{5}\right]}_0^2=\frac{\pi }{4}\left(32-\frac{64}{3}+\frac{32}{5}\right)=\frac{\pi }{4}\cdot \frac{480-320+96}{15}=\frac{64\pi }{15}.$$

   Halving the width to get the radius is the step most often skipped; leaving it out multiplies the answer by $4$.

   (c) Bound by the largest slice. A solid of length $\ell$ whose slices all have area at most $A_{\max }$ has volume at most $A_{\max }\ell$, because the integrand is bounded by $A_{\max }$ throughout.

   In (a) the largest triangle is at $y=0$, with side $4$ and area $4\sqrt{3}$; the solid runs over $0\le y\le 4$, so $V\le 16\sqrt{3}$. Our $8\sqrt{3}$ is exactly half, which is right: $A(y)$ decreases linearly to $0$, so its average is half its maximum.

   In (b) the largest semicircle is at $x=0$, with diameter $4$ and area $\frac{\pi }{8}\cdot 16=2\pi$; the solid runs over $-2\le x\le 2$, so $V\le 8\pi$. Our $\frac{64\pi }{15}\approx 0.53\times 8\pi$ is comfortably inside.

3. 3.Sketch · 8 min

   Let $\mathcal{R}$ be the region enclosed by $y=x^2$ and $y=2x$. A picture without the axis of rotation drawn on it earns nothing here: every radius you write down must be visible as a distance on your sketch.

   1. (a)Sketch $\mathcal{R}$, marking the intersection points. Draw the line $y=4$ on the same axes, then draw one representative vertical slice of $\mathcal{R}$ and the washer it sweeps out when $\mathcal{R}$ is rotated about $y=4$. Label the outer and inner radii as functions of $x$, say which curve supplies which, and why.
   2. (b)Set up and evaluate the volume of the solid obtained by rotating $\mathcal{R}$ about $y=4$.
   3. (c)Redraw the slice for rotation about the $x$-axis instead. Which curve is now the outer one? Compute this volume too, and explain in a sentence why the ordering of the two volumes makes sense.
   Show the solution to question 3

   (a) The region, the axis, and one slice. The curves meet where $x^2=2x$, so at $x=0$ and $x=2$: the points $(0,0)$ and $(2,4)$. Between them the line is on top (at $x=1$, $2>1$), so $\mathcal{R}$ is the thin lens between the straight segment from $(0,0)$ to $(2,4)$ above and the parabolic arc below, widest near $x=1$ where it runs from $y=1$ to $y=2$.

   The line $y=4$ is horizontal, passes through the upper corner $(2,4)$, and otherwise lies above the region, so $\mathcal{R}$ sits entirely below the axis and touches it at one point. Draw it dashed.

   A vertical slice at $x$ runs from the parabola at $(x,x^2)$ up to the line at $(x,2x)$. Rotated about $y=4$ it sweeps a washer whose radii are the distances from the axis to the two ends of the slice. The lower end is farther from the axis, so the parabola gives the outer radius and the line the inner one:

   $$R(x)=4-x^2,r(x)=4-2x.$$

   The rule is "farther from the axis is outer", not "higher is outer"; when the axis is above the region those two disagree. On the sketch, draw the washer as two concentric circles seen edge-on, centred on the dashed line, with $R$ reaching down to the parabola and $r$ reaching down to the line.

   (b) Washers about $y=4$. By the washer method, slicing along the axis of rotation,

   $$V=\pi {\int }_0^2\left[(4-x^2{)}^2-(4-2x{)}^2\right]dx=\pi {\int }_0^2\left[(16-8x^2+x^4)-(16-16x+4x^2)\right]dx=\pi {\int }_0^2\left(x^4-12x^2+16x\right)dx.$$

   Hence

   $$V=\pi {\left[\frac{x^5}{5}-4x^3+8x^2\right]}_0^2=\pi \left(\frac{32}{5}-32+32\right)=\frac{32\pi }{5}.$$

   (c) Washers about the $x$-axis. Now the axis is below the region, so the higher curve is the farther one: the line gives $R=2x$ and the parabola $r=x^2$. Then

   $$V=\pi {\int }_0^2\left(4x^2-x^4\right)dx=\pi {\left[\frac{4x^3}{3}-\frac{x^5}{5}\right]}_0^2=\pi \left(\frac{32}{3}-\frac{32}{5}\right)=\frac{64\pi }{15}.$$

   This is smaller than $\frac{32\pi }{5}=\frac{96\pi }{15}$, and it should be: the bulk of the lens lies low, nearer the $x$-axis than the line $y=4$, so the same area is carried around a shorter circle. The Theorem of Pappus makes this exact — the region's centroid is at height $\frac{8}{5}$, which is $\frac{8}{5}$ from the $x$-axis and $\frac{12}{5}$ from $y=4$, and the ratio of the volumes is $\frac{12/5}{8/5}=\frac{3}{2}=\frac{96/15}{64/15}$.

4. 4.Computation · 7 min

   Let $\mathcal{R}$ be the region under $y=4x-x^2$ and above the $x$-axis. For each rotation, write down the radius, the height and the thickness of a typical cylindrical shell before you integrate, and give the volume exactly.

   1. (a)Rotation about the $y$-axis.
   2. (b)Rotation about the line $x=5$.
   3. (c)Explain concretely what makes (a) awkward with washers, then carry the washer computation through anyway and confirm your answer.
   Show the solution to question 4

   (a) Shells about the $y$-axis. $4x-x^2=x(4-x)$ is non-negative exactly on $[0,4]$, so that is the region's extent. The axis is vertical, so use vertical strips. The strip at $x$ has distance $x$ from the axis, height $4x-x^2$, and thickness $dx$; rotated, it is a shell of volume $2\pi x(4x-x^2)dx$. By the shell method,

   $$V={\int }_0^{4}2\pi x\left(4x-x^2\right)dx=2\pi {\int }_0^4\left(4x^2-x^3\right)dx=2\pi {\left[\frac{4x^3}{3}-\frac{x^4}{4}\right]}_0^4=2\pi \left(\frac{256}{3}-64\right)=\frac{128\pi }{3}.$$

   (b) Shells about $x=5$. The axis is again vertical and lies to the right of the region, so the strip at $x$ is at distance $5-x$ from it (positive throughout $[0,4]$, as a radius must be). Height and thickness are unchanged:

   $$V={\int }_0^{4}2\pi (5-x)\left(4x-x^2\right)dx=2\pi {\int }_0^4\left(20x-9x^2+x^3\right)dx=2\pi {\left[10x^2-3x^3+\frac{x^4}{4}\right]}_0^4=2\pi (160-192+64)=64\pi ,$$

   using $(5-x)(4x-x^2)=20x-5x^2-4x^2+x^3$. Sanity check: the region is symmetric about $x=2$, so it balances at distance $2$ from the $y$-axis and $3$ from $x=5$; the ratio of the volumes is $\frac{64\pi }{128\pi /3}=\frac{3}{2}$, exactly the ratio of those distances, as the Theorem of Pappus predicts.

   (c) Washers, the hard way. For washers about a vertical axis the slices are horizontal, indexed by $y$ from $0$ to $4$, and each radius is an $x$-value on the curve. That means solving $y=4x-x^2$ for $x$: the quadratic $x^2-4x+y=0$ gives

   $$x=2\pm \sqrt{4-y}.$$

   A horizontal line at height $y$ crosses the arch twice, and the slice of $\mathcal{R}$ runs between the two crossings, so the washer's radii come from the two different branches: outer $R=2+\sqrt{4-y}$, inner $r=2-\sqrt{4-y}$. That is what makes it awkward: an inversion, two branches, and square roots. Carrying on,

   $$R^2-r^2=(R-r)(R+r)=2\sqrt{4-y}\cdot 4=8\sqrt{4-y},$$

   so

   $$V=\pi {\int }_0^{4}8\sqrt{4-y}dy=8\pi {\left[-\frac{2}{3}(4-y{)}^{3/2}\right]}_0^4=8\pi \cdot \frac{2}{3}\cdot 8=\frac{128\pi }{3},$$

   in agreement with (a). Shells and washers always agree; the choice is only about which cut leaves the algebra simplest.

5. 5.Critique · 8 min

   Let $\mathcal{R}$ be the region bounded by $y=\sqrt{x}$, the $x$-axis and the line $x=4$. A student is asked for the volume of the solid obtained by rotating $\mathcal{R}$ about the line $y=3$ and submits the following.

   Step 1. The axis is horizontal, so slice vertically. The region lies under the curve $y=\sqrt{x}$, so this is the disk method.

   Step 2. The radius is the distance from the curve to the axis, which is $3-\sqrt{x}$.

   Step 3. $V=\pi {\int }_0^4{\left(3-\sqrt{x}\right)}^{2}dx=\pi {\int }_0^4\left(9-6\sqrt{x}+x\right)dx=\pi {\left[9x-4x^{3/2}+\frac{x^2}{2}\right]}_0^4=\pi (36-32+8)=12\pi .$

   The arithmetic in Step 3 is correct.

   1. (a)Identify the step that is wrong and explain exactly what the student has misunderstood. The number $12\pi$ is the volume of a genuine solid of revolution about $y=3$: describe the region that produces it.
   2. (b)Find the correct volume with washers. Draw the slice and state both radii.
   3. (c)Confirm your answer to (b) with cylindrical shells.
   Show the solution to question 5

   (a) Step 1 is wrong; Step 2 inherits the error. The disk method applies when the slice of the region runs from the axis of rotation out to the curve, so that rotating it fills a solid disk. Here the vertical slice at $x$ runs from $(x,0)$ up to $(x,\sqrt{x})$, and the axis $y=3$ lies above both ends. The slice never reaches the axis, so rotating it sweeps a ring with a hole, and the method needed is washers. "The region lies under a curve" describes the region's relation to the $x$-axis, not to the axis of rotation, and it is the axis of rotation that decides disk versus washer.

   Given that, Step 2's quantity $3-\sqrt{x}$ is a correct distance but the wrong role: it is the distance from the axis to the nearer end of the slice, so it is the inner radius. The outer radius is the distance from the axis to the farther end, the point on the $x$-axis, which is $3-0=3$.

   The student's integral $\pi {\int }_0^4(3-\sqrt{x}{)}^{2}dx$ is a correct disk-method computation for a different region: the one whose slice at $x$ runs from the curve $y=\sqrt{x}$ up to the axis $y=3$. That is the region between $y=\sqrt{x}$ and $y=3$ for $0\le x\le 4$, the complement of $\mathcal{R}$ inside the rectangle $[0,4]\times [0,3]$. So $12\pi$ is the volume of the solid generated by everything in the rectangle except $\mathcal{R}$.

   (b) Washers. Slice vertically at $x$, $0\le x\le 4$; outer radius $R=3$, inner radius $r=3-\sqrt{x}$. By the washer method,

   $$V=\pi {\int }_0^4\left[3^2-{\left(3-\sqrt{x}\right)}^2\right]dx=\pi {\int }_0^4\left(6\sqrt{x}-x\right)dx=\pi {\left[4x^{3/2}-\frac{x^2}{2}\right]}_0^4=\pi (32-8)=24\pi .$$

   A check that also explains the student's number: the whole rectangle $[0,4]\times [0,3]$ rotated about $y=3$ is a cylinder of radius $3$ and length $4$, of volume $36\pi$, and $\mathcal{R}$ and its complement fill it between them: $24\pi +12\pi =36\pi$.

   (c) Shells. For a horizontal axis, shells come from horizontal strips, indexed by $y$ from $0$ to $2$ (the curve reaches $y=2$ at $x=4$). The strip at height $y$ runs from the curve, where $x=y^2$, to the line $x=4$, so its length is $4-y^2$; its distance from the axis $y=3$ is $3-y$. By the shell method,

   $$V={\int }_0^{2}2\pi (3-y)\left(4-y^2\right)dy=2\pi {\int }_0^2\left(12-4y-3y^2+y^3\right)dy=2\pi {\left[12y-2y^2-y^3+\frac{y^4}{4}\right]}_0^2=2\pi (24-8-8+4)=24\pi .$$

   The two methods agree, and neither of them is the student's $12\pi$.

6. 6.Computation · 9 min

   Take $g=9.8{\text{m/s}}^2$, and for water $\rho g=9800{\text{N/m}}^3$. In each part say what is being sliced — the motion, the rope, or the liquid — and write down the work contributed by one slice before integrating.

   1. (a)A force of $12$ N holds a spring stretched $5$ cm beyond its natural length. How much work is done in stretching it from $5$ cm to $15$ cm beyond its natural length?
   2. (b)A bucket holding $10$ kg of water is raised $20$ m at constant speed by a rope of mass $0.5$ kg/m that hangs from a winch and is wound in as the bucket rises. The bucket leaks at a steady rate and is empty at the instant it reaches the top. Ignoring the mass of the empty bucket, find the total work done.
   3. (c)A trough $6$ m long has a vertical cross-section that is an isosceles triangle, $2$ m across the top and $2$ m deep, with the vertex at the bottom. It is full of water. Find the work required to pump all the water out through a spout $1$ m above the top of the trough.
   Show the solution to question 6

   (a) Slice the motion; find $k$ first. Hooke's Law says $F(x)=kx$ with $x$ the extension beyond the natural length. One measurement fixes $k$: $12=k\cdot 0.05$, so $k=240$ N/m. Over a short step $dx$ at extension $x$ the force is nearly constant, contributing $240xdx$, so

   $$W={\int }_{0.05}^{0.15}240xdx=120{\left[x^2\right]}_{0.05}^{0.15}=120(0.0225-0.0025)=2.4\text{J}.$$

   Sanity check: the force runs from $12$ N to $36$ N over $0.1$ m, so the work lies between $1.2$ J and $3.6$ J; because the force is linear it is the midpoint, $2.4$ J.

   (b) Two contributions, sliced two different ways. Let $h$ be the height of the bucket above its starting point, $0\le h\le 20$.

   The water: slice the motion. The water drains linearly from $10$ kg to $0$ kg over the lift, so at height $h$ its mass is $10-\frac{h}{2}$ and its weight is $9.8\left(10-\frac{h}{2}\right)$ N. A step $dh$ contributes that weight times $dh$:

   $$W_{\text{water}}={\int }_0^{20}9.8\left(10-\frac{h}{2}\right)dh=9.8{\left[10h-\frac{h^2}{4}\right]}_0^{20}=9.8(200-100)=980\text{J}.$$

   The rope: slice the rope. The rope weighs $0.5\times 9.8=4.9$ N per metre. The piece of rope initially $x$ metres below the winch, of length $dx$, weighs $4.9dx$ and is lifted exactly $x$ metres, contributing $4.9xdx$:

   $$W_{\text{rope}}={\int }_0^{20}4.9xdx=4.9\cdot \frac{400}{2}=980\text{J}.$$

   (Slicing the motion instead gives the same thing: when the bucket is at height $h$ the hanging rope is $20-h$ long, and ${\int }_0^{20}4.9(20-h)dh=980$.)

   Total: $W=980+980=1960$ J.

   Sanity check: a non-leaking bucket alone would take $10\times 9.8\times 20=1960$ J; leaking to empty halves that, since the average mass carried is $5$ kg. The $10$ kg of rope rises through its centre of mass, $10$ m: $98\times 10=980$ J.

   (c) Slice the liquid into horizontal layers. Put the origin at the bottom vertex with $y$ measured upward, so the water occupies $0\le y\le 2$ and the spout is at $y=3$. By similar triangles the width of the triangle at height $y$ is $w=y$ (it is $2$ at $y=2$), so a layer at height $y$ has horizontal area $A(y)=6y$ and volume $6ydy$, hence weight $9800\cdot 6ydy$. It must be lifted $d(y)=3-y$. Check the extremes: the top layer ($y=2$) rises $1$ m and the bottom layer rises $3$ m. Then

   $$W={\int }_0^{2}9800\cdot 6y(3-y)dy=58,800{\int }_0^2\left(3y-y^2\right)dy=58,800{\left[\frac{3y^2}{2}-\frac{y^3}{3}\right]}_0^2=58,800\left(6-\frac{8}{3}\right)=58,800\cdot \frac{10}{3}=196,000\text{J}.$$

   Sanity check: the water's volume is $\frac{1}{2}\cdot 2\cdot 2\cdot 6=12{\text{m}}^3$, weight $117,600$ N. A triangle's centroid is two thirds of the way from the vertex to the base, at $y=\frac{4}{3}$, so the whole mass rises on average $3-\frac{4}{3}=\frac{5}{3}$ m, and $117,600\times \frac{5}{3}=196,000$ J.

7. 7.Proof · 10 min

   Throughout, $f_{\text{avg}}$ denotes the average value $\frac{1}{b-a}{\int }_a^{b}f(x)dx$ on $[a,b]$, with $a<b$. You may use the comparison property of the integral (if $u\le v$ on $[a,b]$ then ${\int }_a^{b}u\le {\int }_a^{b}v$) and additivity over adjacent intervals.

   1. (a)Let $f$ and $g$ be continuous on $[a,b]$ with $f(x)\le g(x)$ for every $x$, and suppose $f_{\text{avg}}=g_{\text{avg}}$. Prove that $f(x)=g(x)$ for every $x$ in $[a,b]$.
   2. (b)Show by an explicit example that (a) fails if continuity is dropped, and point to the line of your proof that used continuity.
   3. (c)The Mean Value Theorem for Integrals provides $c$ in the closed interval $[a,b]$ with $f(c)=f_{\text{avg}}$. Prove that when $f$ is continuous on $[a,b]$, such a $c$ can always be found in the open interval $(a,b)$.
   Show the solution to question 7

   (a) A non-negative continuous function with zero integral is zero. Put $h=g-f$. Then $h$ is continuous, $h\ge 0$ on $[a,b]$, and

   $${\int }_a^{b}hdx={\int }_a^{b}gdx-{\int }_a^{b}fdx=(b-a)\left(g_{\text{avg}}-f_{\text{avg}}\right)=0.$$

   Suppose, for contradiction, that $h(c)>0$ for some $c\in [a,b]$. Continuity of $h$ at $c$ with $\epsilon =\frac{1}{2}h(c)$ gives $\delta >0$ such that $|h(x)-h(c)|<\frac{1}{2}h(c)$, hence $h(x)>\frac{1}{2}h(c)$, for every $x\in [a,b]$ with $|x-c|<\delta$. Let $[p,q]=\left[\max \left(a,c-\frac{\delta }{2}\right),\min \left(b,c+\frac{\delta }{2}\right)\right]$; this is a subinterval of $[a,b]$ of positive length on which $h>\frac{1}{2}h(c)$. By additivity and the comparison property (using $h\ge 0$ on the two outer pieces and $h>\frac{1}{2}h(c)$ on the middle one),

   $${\int }_a^{b}hdx={\int }_a^{p}hdx+{\int }_p^{q}hdx+{\int }_q^{b}hdx\ge 0+\frac{h(c)}{2}(q-p)+0>0,$$

   contradicting ${\int }_a^{b}h=0$. Hence $h(x)=0$, that is $f(x)=g(x)$, for every $x\in [a,b]$. $■$

   (b) One point is not enough without continuity. On $[0,1]$ let $f\equiv 0$ and let $g(x)=1$ if $x=\frac{1}{2}$, $g(x)=0$ otherwise. Then $f\le g$ everywhere, and ${\int }_0^{1}gdx=0$: every Riemann sum for $g$ has at most one nonzero term, of size at most $\Delta x$, which tends to $0$. So $f_{\text{avg}}=g_{\text{avg}}=0$, yet $g\left(\frac{1}{2}\right)\ne f\left(\frac{1}{2}\right)$.

   The proof breaks at the sentence "continuity of $h$ at $c$ gives $\delta >0$ such that $h(x)>\frac{1}{2}h(c)$ on a whole interval around $c$". That is the only place continuity was used, and it is essential: it converts a positive value at a point, which has no length and contributes nothing to an integral, into a positive value on an interval, which does. For the step function $h=g$ there is no such interval.

   (c) Use the Mean Value Theorem for derivatives. Define $F(x)={\int }_a^{x}f(t)dt$ for $x\in [a,b]$. Since $f$ is continuous, the Fundamental Theorem of Calculus says $F$ is differentiable on $[a,b]$ with $F^{\prime }(x)=f(x)$; in particular $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$, which are exactly the hypotheses of the Mean Value Theorem. That theorem supplies $c\in (a,b)$ — the open interval — with

   $$F(b)-F(a)=F^{\prime }(c)(b-a),\text{that is}{\int }_a^{b}f(x)dx=f(c)(b-a).$$

   Dividing by $b-a$ gives $f(c)=f_{\text{avg}}$ with $c$ strictly between $a$ and $b$. $■$

   The closed-interval proof, via the Extreme and Intermediate Value Theorems, cannot promise this: the IVT is silent about where its point lies. Routing through $F$ trades that for the MVT's guarantee of an interior point.

8. 8.Computation · 9 min

   In each part show the algebra that turns $1+(\text{derivative}{)}^2$ into a perfect square before you take the root, and say why the root equals the positive expression on the interval concerned.

   1. (a)Find the length of $y=\frac{x^4}{8}+\frac{1}{4x^2}$ for $1\le x\le 2$.
   2. (b)Find the length of $x=\frac{y^3}{3}+\frac{1}{4y}$ for $1\le y\le 3$.
   3. (c)For the curve in (a), find the arc length function $s(x)$ measured from the point where $x=1$. Compute $s^{\prime }(1)$ and explain the value you get in terms of the tangent line at that point.
   Show the solution to question 8

   (a) Complete the square. $y^{\prime }=\frac{x^3}{2}-\frac{1}{2x^3}$, so

   $$(y^{\prime }{)}^2=\frac{x^6}{4}-\frac{1}{2}+\frac{1}{4x^6},1+(y^{\prime }{)}^2=\frac{x^6}{4}+\frac{1}{2}+\frac{1}{4x^6}={\left(\frac{x^3}{2}+\frac{1}{2x^3}\right)}^2.$$

   The added $1$ flips the sign of the cross term, which is the whole trick. For $x>0$ both $\frac{x^3}{2}$ and $\frac{1}{2x^3}$ are positive, so the square root of the square is the expression itself, not its negative. By the arc length formula,

   $$L={\int }_1^2\left(\frac{x^3}{2}+\frac{1}{2x^3}\right)dx={\left[\frac{x^4}{8}-\frac{1}{4x^2}\right]}_1^2=\left(2-\frac{1}{16}\right)-\left(\frac{1}{8}-\frac{1}{4}\right)=\frac{31}{16}+\frac{1}{8}=\frac{33}{16}.$$

   Sanity check: the endpoints are $\left(1,\frac{3}{8}\right)$ and $\left(2,\frac{33}{16}\right)$, at straight-line distance $\sqrt{1+{\left(\frac{27}{16}\right)}^2}=\frac{\sqrt{985}}{16}\approx 1.96$; the curve, at $\frac{33}{16}\approx 2.06$, is a little longer, as it must be.

   (b) The same trick in $y$. The curve is given as $x$ in terms of $y$, so use $L=\int \sqrt{1+(dx/dy{)}^2}dy$. Here $\frac{dx}{dy}=y^2-\frac{1}{4y^2}$, and

   $$1+{\left(\frac{dx}{dy}\right)}^2=1+y^4-\frac{1}{2}+\frac{1}{16y^4}=y^4+\frac{1}{2}+\frac{1}{16y^4}={\left(y^2+\frac{1}{4y^2}\right)}^2,$$

   with the bracket positive for $y>0$. Hence

   $$L={\int }_1^3\left(y^2+\frac{1}{4y^2}\right)dy={\left[\frac{y^3}{3}-\frac{1}{4y}\right]}_1^3=\left(9-\frac{1}{12}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{107}{12}-\frac{1}{12}=\frac{53}{6}.$$

   Sanity check: the endpoints are $\left(\frac{7}{12},1\right)$ and $\left(\frac{109}{12},3\right)$, whose chord has length $\sqrt{{\left(\frac{17}{2}\right)}^2+4}=\frac{\sqrt{305}}{2}\approx 8.73$, slightly under $\frac{53}{6}\approx 8.83$.

   (c) The arc length function. With the root already removed in (a),

   $$s(x)={\int }_1^x\left(\frac{t^3}{2}+\frac{1}{2t^3}\right)dt={\left[\frac{t^4}{8}-\frac{1}{4t^2}\right]}_1^x=\frac{x^4}{8}-\frac{1}{4x^2}+\frac{1}{8},$$

   and as a check $s(2)=2-\frac{1}{16}+\frac{1}{8}=\frac{33}{16}$, the answer to (a). By the Fundamental Theorem of Calculus, $s^{\prime }(x)=\frac{x^3}{2}+\frac{1}{2x^3}$, so $s^{\prime }(1)=\frac{1}{2}+\frac{1}{2}=1$.

   The meaning: $s^{\prime }(x)=\sqrt{1+(y^{\prime }{)}^2}\ge 1$ always, with equality exactly where $y^{\prime }=0$. And indeed $y^{\prime }(1)=\frac{1}{2}-\frac{1}{2}=0$: the tangent at $x=1$ is horizontal, so a small horizontal step $dx$ there produces a step $ds$ of the same length along the curve; at that instant the curve is no longer than its shadow on the $x$-axis. Since $y^{\prime }<0$ for $0<x<1$ and $y^{\prime }>0$ for $x>1$, this is the lowest point of the curve, $\left(1,\frac{3}{8}\right)$.

9. 9.Computation · 9 min

   Write each surface area as $\int 2\pi (\text{radius})ds$, saying what the radius is and which variable $ds$ is written in, before you compute. All answers must be exact.

   1. (a)The surface obtained by rotating $y=\frac{x^3}{3}$, $0\le x\le 1$, about the $x$-axis.
   2. (b)The surface obtained by rotating $y=\frac{x^3}{3}+\frac{1}{4x}$, $1\le x\le 2$, about the $x$-axis.
   3. (c)A zone of a sphere of radius $a$ is the part of its surface lying between two parallel planes that both cut the sphere, at distance $h$ apart. Show that the zone's area is $2\pi ah$, whatever the position of the planes.
   Show the solution to question 9

   (a) A substitution kills the root. Rotation about the $x$-axis, so the radius is $y=\frac{x^3}{3}$, and with $y^{\prime }=x^2$ the arc length element in $x$ is $ds=\sqrt{1+x^4}dx$. By the surface area formula,

   $$S={\int }_0^{1}2\pi \cdot \frac{x^3}{3}\sqrt{1+x^4}dx.$$

   The factor $x^3$ is, up to a constant, the derivative of $1+x^4$. Substitute $u=1+x^4$, $du=4x^{3}dx$; the limits $x=0,1$ become $u=1,2$:

   $$S=\frac{2\pi }{3}\cdot \frac{1}{4}{\int }_1^2\sqrt{u}du=\frac{\pi }{6}\cdot \frac{2}{3}[u^{3/2}{]}_1^2=\frac{\pi }{9}\left(2\sqrt{2}-1\right).$$

   Sanity check: on $[0,1]$ the root lies between $1$ and $\sqrt{2}$, so $S$ lies between $2\pi {\int }_0^1\frac{x^3}{3}dx=\frac{\pi }{6}\approx 0.52$ and $\sqrt{2}$ times that, $\approx 0.74$; our $\frac{\pi }{9}(2\sqrt{2}-1)\approx 0.64$ is between them.

   (b) The perfect-square family. Radius $y=\frac{x^3}{3}+\frac{1}{4x}$, and $y^{\prime }=x^2-\frac{1}{4x^2}$ has the shape $u-\frac{1}{4u}$ with $u=x^2$, so

   $$1+(y^{\prime }{)}^2=x^4+\frac{1}{2}+\frac{1}{16x^4}={\left(x^2+\frac{1}{4x^2}\right)}^2,ds=\left(x^2+\frac{1}{4x^2}\right)dx$$

   (the bracket is positive for $x>0$). Multiply out the integrand:

   $$\left(\frac{x^3}{3}+\frac{1}{4x}\right)\left(x^2+\frac{1}{4x^2}\right)=\frac{x^5}{3}+\frac{x}{12}+\frac{x}{4}+\frac{1}{16x^3}=\frac{x^5}{3}+\frac{x}{3}+\frac{1}{16x^3}.$$

   Hence

   $$S=2\pi {\int }_1^2\left(\frac{x^5}{3}+\frac{x}{3}+\frac{1}{16x^3}\right)dx=2\pi {\left[\frac{x^6}{18}+\frac{x^2}{6}-\frac{1}{32x^2}\right]}_1^2.$$

   At $x=2$ the bracket is $\frac{32}{9}+\frac{2}{3}-\frac{1}{128}$; at $x=1$ it is $\frac{1}{18}+\frac{1}{6}-\frac{1}{32}$. Subtracting term by term: $\frac{32}{9}-\frac{1}{18}=\frac{63}{18}=\frac{7}{2}$, $\frac{2}{3}-\frac{1}{6}=\frac{1}{2}$, $\frac{1}{32}-\frac{1}{128}=\frac{3}{128}$. The difference is $4+\frac{3}{128}=\frac{515}{128}$, so

   $$S=2\pi \cdot \frac{515}{128}=\frac{515\pi }{64}.$$

   Sanity check: the arc has length ${\left[\frac{x^3}{3}-\frac{1}{4x}\right]}_1^2=\frac{59}{24}\approx 2.46$ and its radii run from $\frac{7}{12}$ to $\frac{67}{24}$. A single frustum with those radii and that slant height has area $\pi \left(\frac{7}{12}+\frac{67}{24}\right)\frac{59}{24}\approx 8.30\pi$, a shade above our $\frac{515}{64}\pi \approx 8.05\pi$, as expected for a curve that sags below its chord.

   (c) The zone: the root cancels. Generate the sphere by rotating $y=\sqrt{a^2-x^2}$ about the $x$-axis, and let the planes be $x=c$ and $x=c+h$ with $-a\le c<c+h\le a$. The radius is $y$, and

   $$y^{\prime }=\frac{-x}{\sqrt{a^2-x^2}},1+(y^{\prime }{)}^2=\frac{a^2}{a^2-x^2},ds=\frac{a}{\sqrt{a^2-x^2}}dx.$$

   The root in the radius cancels the root in $ds$ exactly:

   $$S={\int }_c^{c+h}2\pi \sqrt{a^2-x^2}\cdot \frac{a}{\sqrt{a^2-x^2}}dx=2\pi a{\int }_c^{c+h}dx=2\pi ah.$$

   The answer contains no $c$, which is the claim: the integrand is the constant $2\pi a$, so every slab of thickness $dx$ carries the same surface area wherever it is cut. If one plane is tangent at a pole ($c=-a$ or $c+h=a$) the derivative $y^{\prime }$ blows up there, but the integrand simplified to a constant before the endpoint was reached, so the improper integral is harmless. As a check, $h=2a$ gives the full sphere, $4\pi a^2$.

10. 10.Computation · 11 min

    Use $\rho g=9800{\text{N/m}}^3$ for water. In every part choose a vertical coordinate, write down the depth $d(y)$ and the width $w(y)$ of a horizontal strip as two separate functions, check both against the top strip and the bottom strip, and only then integrate.

    1. (a)The face of a dam is a vertical trapezoid, $20$ m wide along the top, $10$ m wide along the bottom and $6$ m tall, and the water reaches the top. Find the force on the face.
    2. (b)A circular window of radius $1$ m in the vertical wall of an aquarium has its centre $5$ m below the surface. Find the force on it. Evaluate any integral that appears by recognising it as an area or by symmetry, not by antidifferentiation.
    3. (c)Prove the shortcut $F=\rho g\bar{d}A$, where $A$ is the plate's area and $\bar{d}$ is the depth of its centroid, from the strip formula and the definition of the centroid. Use it to check (b), and then to find the depth of the centroid of the trapezoid in (a) from your answer there.
    Show the solution to question 10

    (a) Horizontal strips down the trapezoid. Let $y$ be the depth below the surface, $0\le y\le 6$; since the water reaches the top of the face, $d(y)=y$. The width shrinks linearly from $20$ to $10$ over $6$ m:

    $$w(y)=20-\frac{10}{6}y=20-\frac{5y}{3},w(0)=20,w(6)=10.$$

    The strip at depth $y$ has area $w(y)dy$ and feels pressure $\rho gy$, so by the hydrostatic force formula

    $$F={\int }_0^{6}9800y\left(20-\frac{5y}{3}\right)dy=9800{\left[10y^2-\frac{5y^3}{9}\right]}_0^6=9800(360-120)=9800\times 240=2,352,000\text{N},$$

    about $2.35\times 10^6$ N.

    (b) Strips across the window. Let $y$ be the signed vertical distance below the window's centre, $-1\le y\le 1$. Then $d(y)=5+y$: the top strip ($y=-1$) is at depth $4$ and the bottom strip at depth $6$. The width is the chord of the unit circle, $w(y)=2\sqrt{1-y^2}$. So

    $$F=9800{\int }_{-1}^1(5+y)\cdot 2\sqrt{1-y^2}dy=9800\left[5{\int }_{-1}^{1}2\sqrt{1-y^2}dy+{\int }_{-1}^{1}2y\sqrt{1-y^2}dy\right].$$

    The first integral adds up the widths of the horizontal chords of the unit disc, so it is the disc's area, $\pi$. The second has an odd integrand on a symmetric interval, so it is $0$: the extra pressure on the lower half is exactly matched by the shortfall on the upper half. Hence

    $$F=9800\times 5\pi =49,000\pi \text{N}\approx 1.54\times 10^5\text{N}.$$

    (c) The centroid shortcut. Take $y$ to be depth below the surface, with the plate occupying $a\le y\le b$, so $d(y)=y$. The strip at depth $y$ has area $w(y)dy$, and its own centre lies at depth $y$, so its moment about the surface line is $yw(y)dy$. Summing strips, the plate's area and its moment about the surface are

    $$A={\int }_a^{b}w(y)dy,M={\int }_a^{b}yw(y)dy,$$

    and by definition the centroid's depth is the area-weighted average depth, $\bar{d}=M/A$. (This is the centroid formula with the strips cut horizontally rather than vertically; the derivation is the same, since a thin strip balances at its centre.) Therefore

    $$F=\rho g{\int }_a^{b}yw(y)dy=\rho gM=\rho g\bar{d}A.■$$

    The shortcut is exact, not an approximation, because the pressure is linear in depth.

    Check of (b). A disc's centroid is its centre, at depth $5$, and its area is $\pi \cdot 1^2=\pi$, so $F=9800\times 5\times \pi =49,000\pi$ N, as before.

    Centroid of the trapezoid. Its area is $A=\frac{20+10}{2}\cdot 6=90{\text{m}}^2$, so from (a)

    $$\bar{d}=\frac{F}{\rho gA}=\frac{240}{90}=\frac{8}{3}\text{m}.$$

    Plausibility: a $6$ m rectangle would balance at depth $3$; a triangle with its $20$ m side at the top and its vertex at the bottom would balance at depth $2$; the trapezoid, wider at the top than a rectangle would be relative to its bottom, sits between them at $\frac{8}{3}$.

11. 11.Proof · 12 min

    Name the symmetry principle and the Theorem of Pappus wherever you use them, and each time you use Pappus check its hypothesis: the region must lie on one side of the axis (touching it is allowed).

    1. (a)Let $b,h>0$. Prove that the centroid of the triangle with vertices $(0,0)$, $(b,0)$ and $(0,h)$ is $\left(\frac{b}{3},\frac{h}{3}\right)$.
    2. (b)Use (a) and the Theorem of Pappus to find the volume of the solid obtained by rotating this triangle about the $y$-axis, and then about the $x$-axis. Identify each as a classical formula.
    3. (c)Let $\mathcal{Q}$ be the quarter disc $x^2+y^2\le r^2$, $x\ge 0$, $y\ge 0$. Using the symmetry principle, the Theorem of Pappus and the volume of a sphere, find the centroid of $\mathcal{Q}$ without evaluating any integral. Then confirm $\bar{x}$ by direct integration.
    Show the solution to question 11

    (a) Centroid of the triangle. The hypotenuse joins $(b,0)$ to $(0,h)$, so the triangle is the region under $f(x)=h\left(1-\frac{x}{b}\right)$, above the $x$-axis, for $0\le x\le b$, and its area is $A=\frac{1}{2}bh$ (or, from the integral, ${\int }_0^{b}h\left(1-\frac{x}{b}\right)dx=h\left(b-\frac{b}{2}\right)=\frac{1}{2}bh$).

    By the centroid formulas,

    $$\bar{x}=\frac{1}{A}{\int }_0^{b}xh\left(1-\frac{x}{b}\right)dx=\frac{2}{bh}\cdot h{\left[\frac{x^2}{2}-\frac{x^3}{3b}\right]}_0^b=\frac{2}{b}\left(\frac{b^2}{2}-\frac{b^2}{3}\right)=\frac{2}{b}\cdot \frac{b^2}{6}=\frac{b}{3}.$$

    For $\bar{y}$, substitute $u=1-\frac{x}{b}$, $dx=-bdu$, so that ${\int }_0^b{\left(1-\frac{x}{b}\right)}^{2}dx=b{\int }_0^1{u}^{2}du=\frac{b}{3}$. Then

    $$\bar{y}=\frac{1}{A}{\int }_0^b\frac{1}{2}{\left[h\left(1-\frac{x}{b}\right)\right]}^{2}dx=\frac{2}{bh}\cdot \frac{h^2}{2}\cdot \frac{b}{3}=\frac{h}{3}.$$

    So the centroid is $\left(\frac{b}{3},\frac{h}{3}\right)$. $■$ As a check, the median from $(0,h)$ to the midpoint $\left(\frac{b}{2},0\right)$ of the opposite side passes through this point (two thirds of the way along it), and a triangle's centroid is where its medians meet.

    (b) Two cones by Pappus. The triangle lies in $x\ge 0$, touching the $y$-axis along one leg, so the hypothesis of the Theorem of Pappus holds for that axis; the distance from the centroid to the axis is $d=\bar{x}=\frac{b}{3}$. Hence

    $$V=2\pi dA=2\pi \cdot \frac{b}{3}\cdot \frac{bh}{2}=\frac{1}{3}\pi b^{2}h,$$

    the volume of a cone of base radius $b$ and height $h$: one third of base area times height.

    Likewise the triangle lies in $y\ge 0$, touching the $x$-axis along its other leg, and $d=\bar{y}=\frac{h}{3}$, so

    $$V=2\pi \cdot \frac{h}{3}\cdot \frac{bh}{2}=\frac{1}{3}\pi h^{2}b,$$

    the cone of base radius $h$ and height $b$. Swapping which leg is the axis swaps the roles of $b$ and $h$, as it should.

    (c) Pappus run backwards. Swapping $x$ and $y$ leaves the description $x^2+y^2\le r^2$, $x\ge 0$, $y\ge 0$ unchanged, so $\mathcal{Q}$ is symmetric about the line $y=x$, and by the symmetry principle its centroid lies on that line: $\bar{x}=\bar{y}$.

    Now rotate $\mathcal{Q}$ about the $y$-axis. The region lies in $x\ge 0$ and touches the axis along a radius, so Pappus applies; the solid swept out is a hemisphere of radius $r$, of volume $\frac{1}{2}\cdot \frac{4}{3}\pi r^3=\frac{2}{3}\pi r^3$, and the area of $\mathcal{Q}$ is $\frac{1}{4}\pi r^2$. The Theorem of Pappus gives

    $$\frac{2}{3}\pi r^3=2\pi \bar{x}\cdot \frac{\pi r^2}{4}=\frac{{\pi }^2{r}^2}{2}\bar{x},\text{so}\bar{x}=\frac{2\pi r^3/3}{{\pi }^2{r}^2/2}=\frac{4r}{3\pi },$$

    and the centroid is $\left(\frac{4r}{3\pi },\frac{4r}{3\pi }\right)$.

    Direct confirmation. $\mathcal{Q}$ is the region under $f(x)=\sqrt{r^2-x^2}$ on $[0,r]$, so

    $$\bar{x}=\frac{1}{A}{\int }_0^{r}x\sqrt{r^2-x^2}dx=\frac{4}{\pi r^2}{\left[-\frac{1}{3}{\left(r^2-x^2\right)}^{3/2}\right]}_0^r=\frac{4}{\pi r^2}\cdot \frac{r^3}{3}=\frac{4r}{3\pi },$$

    in agreement. Sanity check: $\frac{4}{3\pi }\approx 0.42$, so the centroid sits a little less than halfway out along each leg and well inside the quarter disc, pulled toward the corner where the region is widest.

12. 12.Computation · 10 min

    Every probability must be written as an integral before it is evaluated, and every improper integral must be handled with an explicit limit. Leave answers in terms of $e$ and $\ln 2$ where they arise.

    1. (a)Let $f(x)=\frac{c}{x^2}$ for $x\ge 1$ and $f(x)=0$ for $x<1$. Find the value of $c$ for which $f$ is a probability density function, find the median, and show that the mean does not exist.
    2. (b)The lifetime $T$, in years, of a component is exponentially distributed with mean $5$. Find $P(T>10)$; find the time $t_0$ by which three quarters of all components have failed; and find the probability that a component fails after the median lifetime but before the mean lifetime.
    3. (c)Let $f(x)=x{e}^{-x}$ for $x\ge 0$ and $f(x)=0$ otherwise. Verify that $f$ is a density, find its mean $\mu$, compute $P(X>\mu )$, and deduce — without finding it — whether the median is smaller or larger than the mean.
    Show the solution to question 12

    (a) Normalise, then look for the median and the mean. For $c\ge 0$ we have $f\ge 0$, so only the total area is in question:

    $${\int }_{-\infty }^{\infty }f(x)dx={\int }_1^{\infty }\frac{c}{x^2}dx=\underset{b\to \infty }{\lim }c{\left[-\frac{1}{x}\right]}_1^b=\underset{b\to \infty }{\lim }c\left(1-\frac{1}{b}\right)=c,$$

    so $c=1$. The median $m$ satisfies ${\int }_{-\infty }^{m}f=\frac{1}{2}$, and for $m\ge 1$ that is

    $${\int }_1^m\frac{dx}{x^2}=1-\frac{1}{m}=\frac{1}{2},\text{so }m=2.$$

    For the mean the integrand is $xf(x)=\frac{1}{x}$ on $[1,\infty )$, and

    $${\int }_1^{\infty }\frac{dx}{x}=\underset{b\to \infty }{\lim }\ln b=\infty .$$

    The improper integral diverges, so the mean does not exist. Half the probability lies in $[1,2]$, yet the tail is heavy enough that the balance point runs off to infinity: a sheet of metal cut in this shape has no point at which it balances.

    (b) Exponential lifetimes. Mean $5$ means $\lambda =\frac{1}{5}$, so $f(t)=\frac{1}{5}{e}^{-t/5}$ for $t\ge 0$. Then

    $$P(T>10)={\int }_{10}^{\infty }\frac{1}{5}{e}^{-t/5}dt=\underset{b\to \infty }{\lim }{\left[-e^{-t/5}\right]}_{10}^b=e^{-2}.$$

    Three quarters have failed by $t_0$ when $P(T\le t_0)={\int }_0^{t_0}\frac{1}{5}{e}^{-t/5}dt=1-e^{-t_0/5}=\frac{3}{4}$, that is $e^{-t_0/5}=\frac{1}{4}$, so

    $$t_0=5\ln 4=10\ln 2\approx 6.93\text{years}.$$

    The median is $m=\frac{\ln 2}{\lambda }=5\ln 2$ and the mean is $5$; since $\ln 2<1$ the median comes first, and

    $$P(5\ln 2<T<5)={\int }_{5\ln 2}^5\frac{1}{5}{e}^{-t/5}dt={\left[-e^{-t/5}\right]}_{5\ln 2}^5=e^{-\ln 2}-e^{-1}=\frac{1}{2}-\frac{1}{e}\approx \mathrm{0.13.}$$

    (c) A density with a tail. Clearly $f\ge 0$. For the total area integrate by parts with $u=x$, $dv=e^{-x}dx$, $v=-e^{-x}$:

    $${\int }_0^{\infty }x{e}^{-x}dx=\underset{b\to \infty }{\lim }\left({\left[-x{e}^{-x}\right]}_0^b+{\int }_0^b{e}^{-x}dx\right)=\underset{b\to \infty }{\lim }\left(-b{e}^{-b}+1-e^{-b}\right)=1,$$

    using $b{e}^{-b}\to 0$ (the exponential beats the polynomial). So $f$ is a density. For the mean, parts again with $u=x^2$, $dv=e^{-x}dx$:

    $$\mu ={\int }_0^{\infty }{x}^2{e}^{-x}dx=\underset{b\to \infty }{\lim }\left({\left[-x^2{e}^{-x}\right]}_0^b+2{\int }_0^{b}x{e}^{-x}dx\right)=0+2\cdot 1=2.$$

    For the tail, an antiderivative of $x{e}^{-x}$ is $-(x+1)e^{-x}$ (differentiate to check: $-e^{-x}+(x+1)e^{-x}=x{e}^{-x}$), so

    $$P(X>\mu )={\int }_2^{\infty }x{e}^{-x}dx=\underset{b\to \infty }{\lim }{\left[-(x+1)e^{-x}\right]}_2^b=3e^{-2}.$$

    Since $e>2.7$ we have $e^2>7.29>6$, hence $3e^{-2}<\frac{1}{2}$. So less than half of the probability lies above the mean, which means $P(X\le 2)>\frac{1}{2}$: the cumulative probability has already passed $\frac{1}{2}$ by $x=2$, and the median, the point where it first reaches $\frac{1}{2}$, lies to the left of the mean. Median $<$ mean, as expected for a density whose long tail is on the right: the tail drags the balance point outward but moves the halfway-area point much less.