Exercise 1: Four integrals, and the move that opens each one

Evaluate each integral exactly. Three points each: the first line must NAME the move (the substitution with its differential and its new bounds, the choice of $u$ and $dv$ , the identity or the algebra used), then the computation follows. No technique is announced.

a) $\displaystyle\int_0^{1} x\ln(1 + x)\,dx$
b) $\displaystyle\int_0^{\pi/2} \frac{\cos^3 x}{1 + \sin x}\,dx$
c) $\displaystyle\int_0^{\ln 3} \frac{dx}{e^x + 3e^{-x}}$
d) $\displaystyle\int_3^{8} \frac{dx}{x\sqrt{x + 1}}$

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Answers

a) $\frac{1}{4}$
b) $\frac{1}{2}$
c) $\frac{\pi}{6\sqrt{3}} = \frac{\pi\sqrt{3}}{18}$
d) $\ln\frac{3}{2}$

a)

Move: integration by parts with $u = \ln(1 + x)$ , $du = \frac{dx}{1 + x}$ , and $dv = x\,dx$ . The logarithm is $u$ because differentiating it turns it into a rational function. For $v$ , ANY antiderivative of $x$ is allowed, and the one that factors is the smart one: $v = \frac{x^2 - 1}{2} = \frac{(x - 1)(x + 1)}{2}$ . Then $\int_0^1 x\ln(1 + x)\,dx = \Big[\frac{x^2 - 1}{2}\ln(1 + x)\Big]_0^1 - \int_0^1 \frac{(x - 1)(x + 1)}{2(1 + x)}\,dx = 0 - 0 - \frac{1}{2}\int_0^1 (x - 1)\,dx = -\frac{1}{2}\left(\frac{1}{2} - 1\right) = \frac{1}{4}$ . The bracket vanishes at both bounds: $x^2 - 1 = 0$ at $x = 1$ and $\ln 1 = 0$ at $x = 0$ .

With the usual $v = \frac{x^2}{2}$ the answer is the same but longer: $\frac{\ln 2}{2} - \frac{1}{2}\int_0^1 \frac{x^2}{1 + x}\,dx$ , and long division gives $\frac{x^2}{1 + x} = x - 1 + \frac{1}{1 + x}$ , so the value is $\frac{\ln 2}{2} - \frac{1}{2}\left(\frac{1}{2} - 1 + \ln 2\right) = \frac{1}{4}$ : the $\ln 2$ cancels. Size check: on $[0, 1]$ , the concave $\ln(1 + x)$ lies between its chord $x\ln 2$ and its tangent $x$ , so the integral lies between $\frac{\ln 2}{3} \approx 0.23$ and $\frac{1}{3}$ , and $0.25$ does.

b)

Move: simplify BEFORE integrating. $\cos^3 x = \cos x\,(1 - \sin^2 x) = \cos x\,(1 - \sin x)(1 + \sin x)$ , and $1 + \sin x > 0$ on $[0, \frac{\pi}{2}]$ , so the fraction reduces to $\cos x\,(1 - \sin x) = \cos x - \sin x\cos x$ . Then $\int_0^{\pi/2} (\cos x - \sin x\cos x)\,dx = \Big[\sin x - \frac{\sin^2 x}{2}\Big]_0^{\pi/2} = 1 - \frac{1}{2} = \frac{1}{2}$ . The same through $u = \sin x$ , $du = \cos x\,dx$ , bounds $0$ and $1$ : $\int_0^1 \frac{1 - u^2}{1 + u}\,du = \int_0^1 (1 - u)\,du = \frac{1}{2}$ . Treating $\cos^3 x$ as an odd power is right, but forgetting to cancel $1 + u$ afterwards sends people into a needless long division.

c)

Move: multiply top and bottom by $e^x$ , so the integrand becomes $\frac{e^x}{e^{2x} + 3}$ , then $u = e^x$ , $du = e^x\,dx$ , bounds $x = 0 \mapsto u = 1$ and $x = \ln 3 \mapsto u = 3$ . Now $\int_1^3 \frac{du}{u^2 + 3} = \frac{1}{\sqrt{3}}\Big[\arctan\frac{u}{\sqrt{3}}\Big]_1^3 = \frac{1}{\sqrt{3}}\left(\arctan\sqrt{3} - \arctan\frac{1}{\sqrt{3}}\right) = \frac{1}{\sqrt{3}}\left(\frac{\pi}{3} - \frac{\pi}{6}\right) = \frac{\pi}{6\sqrt{3}}$ . The factor $\frac{1}{\sqrt{3}}$ in front and the $\frac{u}{\sqrt{3}}$ inside both come from $u^2 + 3 = 3\left(1 + \left(\frac{u}{\sqrt{3}}\right)^2\right)$ ; forgetting either is the classic slip. Size check: $e^x + 3e^{-x} \ge 2\sqrt{3}$ (sum of two positive numbers with product $3$ ) and $\le 4$ on the interval, so the value lies between $\frac{\ln 3}{4}$ and $\frac{\ln 3}{2\sqrt{3}}$ , about $0.27$ and $0.32$ ; $\frac{\pi\sqrt{3}}{18} \approx 0.30$ .

d)

Move: the root blocks everything, so rationalize with $u = \sqrt{x + 1}$ , $x = u^2 - 1$ , $dx = 2u\,du$ , bounds $x = 3 \mapsto u = 2$ and $x = 8 \mapsto u = 3$ . Then $\int_3^8 \frac{dx}{x\sqrt{x + 1}} = \int_2^3 \frac{2u\,du}{(u^2 - 1)u} = \int_2^3 \frac{2\,du}{u^2 - 1}$ . Now partial fractions: $\frac{2}{(u - 1)(u + 1)} = \frac{1}{u - 1} - \frac{1}{u + 1}$ (cover-up at $u = 1$ and $u = -1$ ). The value is $\Big[\ln\frac{u - 1}{u + 1}\Big]_2^3 = \ln\frac{2}{4} - \ln\frac{1}{3} = \ln\frac{3}{2}$ . Two chapters in a row: the rationalizing substitution turns a root into a rational function, and the rational function is then decomposed.

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Exercise 2: Improper integrals: two values, two verdicts

For each integral, say first WHERE it is improper and write the limit or limits. Evaluate a) and b); decide c) and d) by comparison, without an antiderivative. The figure shows the integrand of b).

a) $\displaystyle\int_1^{\infty} \frac{dx}{x^2(x + 1)}$ . A classmate splits the integrand into partial fractions and writes one limit per fraction. What goes wrong?
b) $\displaystyle\int_0^{4} \frac{dx}{\sqrt{4x - x^2}}$
c) $\displaystyle\int_1^{\infty} \frac{x\,dx}{\sqrt{x^5 + 1}}$ . Show that it converges, and prove that its value $I$ satisfies $\sqrt{2} \le I \le 2$ .
d) $\displaystyle\int_0^{\pi/2} \frac{dx}{\sin x}$

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Answers

a) Converges to $1 - \ln 2$ ; taken one fraction at a time, it is $-\infty + \infty$
b) Improper at $0$ and at $4$ ; converges to $\frac{\pi}{2} + \frac{\pi}{2} = \pi$
c) $\frac{1}{\sqrt{2}}x^{-3/2} \le \frac{x}{\sqrt{x^5 + 1}} \le x^{-3/2}$ on $[1, \infty)$ : converges, $\sqrt{2} \le I \le 2$
d) Improper at $0$ ; $\frac{1}{\sin x} \ge \frac{1}{x}$ there: diverges

a)

Improper at $\infty$ only: the integrand is continuous on $[1, \infty)$ . Decompose, with one term per power of the repeated factor: $\frac{1}{x^2(x + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1}$ , so $1 = Ax(x + 1) + B(x + 1) + Cx^2$ . At $x = 0$ : $B = 1$ . At $x = -1$ : $C = 1$ . The $x^2$ coefficient: $0 = A + C$ , so $A = -1$ . An antiderivative is $-\ln x - \frac{1}{x} + \ln(x + 1) = \ln\frac{x + 1}{x} - \frac{1}{x}$ , with the two logarithms COMBINED before any limit is taken.

Then $\int_1^t \frac{dx}{x^2(x + 1)} = \ln\left(1 + \frac{1}{t}\right) - \frac{1}{t} - (\ln 2 - 1)$ , and as $t \to \infty$ the first two terms tend to $0$ : the integral converges to $1 - \ln 2 \approx 0.31$ . It is below $\int_1^\infty \frac{dx}{x^3} = \frac{1}{2}$ , as it must be. The classmate writes $-\int_1^\infty \frac{dx}{x} + \int_1^\infty \frac{dx}{x^2} + \int_1^\infty \frac{dx}{x + 1}$ : the first and the last are divergent, $-\infty$ and $+\infty$ , and $\infty - \infty$ is not a number. Splitting a convergent improper integral is legal only into pieces that each converge.

b)

Complete the square: $4x - x^2 = 4 - (x - 2)^2$ , which vanishes at $x = 0$ and at $x = 4$ . The integrand blows up at BOTH endpoints (the figure shows the two branches going up), so split at an interior point, $2$ is natural, and take one limit on each side: $\int_0^4 = \lim_{s \to 0^+}\int_s^2 + \lim_{t \to 4^-}\int_2^t$ . With $x - 2 = 2\sin\theta$ , $dx = 2\cos\theta\,d\theta$ and $\sqrt{4 - 4\sin^2\theta} = 2\cos\theta$ (positive for $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ ), the integrand becomes $d\theta$ : an antiderivative is $\arcsin\frac{x - 2}{2}$ .

Left piece: $\arcsin 0 - \lim_{s \to 0^+}\arcsin\frac{s - 2}{2} = 0 - \arcsin(-1) = \frac{\pi}{2}$ . Right piece: $\lim_{t \to 4^-}\arcsin\frac{t - 2}{2} - 0 = \frac{\pi}{2}$ . Both converge, so the integral converges to $\pi$ . The unbounded region of the figure has a finite area because the branches blow up like $\frac{1}{2\sqrt{x}}$ , an exponent $\frac{1}{2} < 1$ on $(0, 1]$ .

c)

Improper at $\infty$ only, and the integrand is positive. For $x \ge 1$ : $x^5 < x^5 + 1 \le 2x^5$ , so $\sqrt{x^5} < \sqrt{x^5 + 1} \le \sqrt{2}\,x^{5/2}$ , and dividing $x$ by these, $\frac{x^{-3/2}}{\sqrt{2}} \le \frac{x}{\sqrt{x^5 + 1}} < x^{-3/2}$ . The upper bound is a $p$ -integral with $p = \frac{3}{2} > 1$ : $\int_1^\infty x^{-3/2}\,dx = \lim_{t \to \infty}\left(2 - \frac{2}{\sqrt{t}}\right) = 2$ . By the comparison test the integral converges, and $I \le 2$ . Integrating the lower bound: $I \ge \frac{2}{\sqrt{2}} = \sqrt{2}$ . The direction matters: the upper bound proves convergence, the lower bound only gives the bracket.

d)

Improper at $0$ , where $\sin x = 0$ ; the integrand is positive on $(0, \frac{\pi}{2}]$ . For $0 < x \le \frac{\pi}{2}$ , $0 < \sin x \le x$ (the graph of $\sin$ lies under its tangent at $0$ ), so $\frac{1}{\sin x} \ge \frac{1}{x}$ . And $\int_0^{\pi/2} \frac{dx}{x} = \lim_{s \to 0^+}\left(\ln\frac{\pi}{2} - \ln s\right) = \infty$ , the $p$ -integral with $p = 1$ on $(0, a]$ . An integral ABOVE a divergent one diverges: $\int_0^{\pi/2}\frac{dx}{\sin x}$ diverges. Confirmation: $\ln\left(\tan\frac{x}{2}\right)$ is an antiderivative, and it tends to $-\infty$ as $x \to 0^+$ .

Exercise 3: One region: its area, its centroid, and the volume Pappus gives for free

Let $R$ be the region enclosed by $y = \sqrt{x}$ and $y = x^3$ , shown in the figure with its centroid.

a) Find the points where the curves meet and the area $A$ of $R$ .
b) Find the centroid $(\bar{x}, \bar{y})$ of $R$ . Check that the red point of the figure is plausible.
c) $R$ is rotated about the line $y = -1$ . Use the theorem of Pappus to find the volume of the solid, then check it with washers.

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Answers

a) $x = 0$ and $x = 1$ ; $A = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}$
b) $\bar{x} = \frac{12}{25}$ , $\bar{y} = \frac{3}{7}$
c) $V = 2\pi\left(\frac{3}{7} + 1\right)\frac{5}{12} = \frac{25\pi}{21}$ , confirmed by washers

a)

$\sqrt{x} = x^3$ gives $x = x^6$ for $x \ge 0$ , so $x(x^5 - 1) = 0$ : $x = 0$ or $x = 1$ , at the points $(0, 0)$ and $(1, 1)$ . On $(0, 1)$ , $\sqrt{x} > x^3$ (at $x = \frac{1}{4}$ : $\frac{1}{2}$ against $\frac{1}{64}$ ), so the top is $\sqrt{x}$ . $A = \int_0^1 \left(x^{1/2} - x^3\right)dx = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}$ .

b)

A vertical strip at $x$ has area $(\sqrt{x} - x^3)\,dx$ , its arm about the $y$ -axis is $x$ , and its midpoint sits at height $\frac{\sqrt{x} + x^3}{2}$ . So $M_y = \int_0^1 x\left(x^{1/2} - x^3\right)dx = \frac{2}{5} - \frac{1}{5} = \frac{1}{5}$ and $M_x = \int_0^1 \frac{1}{2}\left((\sqrt{x})^2 - (x^3)^2\right)dx = \frac{1}{2}\int_0^1 (x - x^6)\,dx = \frac{1}{2}\left(\frac{1}{2} - \frac{1}{7}\right) = \frac{5}{28}$ . Divide by the area: $\bar{x} = \frac{1/5}{5/12} = \frac{12}{25} = 0.48$ and $\bar{y} = \frac{5/28}{5/12} = \frac{12}{28} = \frac{3}{7} \approx 0.43$ .

The trap in $M_x$ is to square the difference, $\frac{1}{2}(\sqrt{x} - x^3)^2$ , instead of taking the difference of the squares: the strip is a washer-like band from $x^3$ to $\sqrt{x}$ , and its moment is $\frac{1}{2}(f^2 - g^2)$ . Plausibility: the red point $(0.48, 0.43)$ lies inside $R$ , as it must, since $R$ is convex (it lies under the concave curve $\sqrt{x}$ and over the convex curve $x^3$ ) and the centroid of a convex region is inside it. Its abscissa is also close to where the strip height $\sqrt{x} - x^3$ is largest: the derivative $\frac{1}{2\sqrt{x}} - 3x^2$ vanishes when $x^{5/2} = \frac{1}{6}$ , a little below $\frac{1}{2}$ .

c)

Pappus: the volume is the area times the distance travelled by the centroid. The centroid is at height $\frac{3}{7}$ , so its distance to the axis $y = -1$ is $\frac{3}{7} + 1 = \frac{10}{7}$ , and it travels $2\pi \cdot \frac{10}{7}$ . $V = 2\pi \cdot \frac{10}{7} \cdot \frac{5}{12} = \frac{100\pi}{84} = \frac{25\pi}{21}$ .

Check with washers perpendicular to the $x$ -axis: the radii are measured FROM the axis $y = -1$ , so $R_{\text{out}} = \sqrt{x} + 1$ and $R_{\text{in}} = x^3 + 1$ . $V = \pi\int_0^1 \left[(\sqrt{x} + 1)^2 - (x^3 + 1)^2\right]dx = \pi\int_0^1 \left(x + 2\sqrt{x} - x^6 - 2x^3\right)dx = \pi\left(\frac{1}{2} + \frac{4}{3} - \frac{1}{7} - \frac{1}{2}\right) = \frac{25\pi}{21}$ . The two routes agree; with the radii $\sqrt{x}$ and $x^3$ , forgetting the shift, one would get the volume about the $x$ -axis, $\frac{5\pi}{14}$ .

Exercise 4: One solid, two methods: shells, then washers that force an inverse

Let $R$ be the region between the parabolas $y = x^2$ and $y = 4x - x^2$ . The figure shows $R$ , one vertical strip, and the axis $x = 3$ . The solid $S$ is obtained by rotating $R$ about the line $x = 3$ .

a) Find the volume of $S$ by cylindrical shells.
b) Set up and evaluate the volume of $S$ with washers. Explain why the inner and outer radii require solving $y = 4x - x^2$ for $x$ , and which root to keep.
c) Find the volume of the solid obtained by rotating $R$ about the line $x = -1$ instead. Explain the result with a symmetry of $R$ .

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Answers

a) $V = 2\pi\int_0^2 (3 - x)(4x - 2x^2)\,dx = \frac{32\pi}{3}$
b) $x = 2 - \sqrt{4 - y}$ on the left, $x = \sqrt{y}$ on the right; $V = \pi\int_0^4 \left[(1 + \sqrt{4 - y})^2 - (3 - \sqrt{y})^2\right]dy = \frac{32\pi}{3}$
c) $\frac{32\pi}{3}$ again: the height $4x - 2x^2$ is symmetric about $x = 1$ , which is at distance $2$ from both axes

a)

Intersections: $x^2 = 4x - x^2$ gives $2x(x - 2) = 0$ , so $x = 0$ and $x = 2$ , at $(0, 0)$ and $(2, 4)$ ; between them $4x - x^2 \ge x^2$ (at $x = 1$ : $3$ against $1$ ). A vertical strip at $x$ , parallel to the axis, sweeps a shell of height $h(x) = (4x - x^2) - x^2 = 4x - 2x^2$ and radius equal to its distance to the axis. The region lies to the LEFT of $x = 3$ , so the radius is $3 - x$ , not $x$ and not $x - 3$ . $V = 2\pi\int_0^2 (3 - x)(4x - 2x^2)\,dx = 2\pi\int_0^2 \left(12x - 10x^2 + 2x^3\right)dx = 2\pi\left(24 - \frac{80}{3} + 8\right) = 2\pi \cdot \frac{16}{3} = \frac{32\pi}{3}$ .

b)

Washers need slices PERPENDICULAR to the axis, so horizontal slices at height $y$ , $0 \le y \le 4$ . A horizontal line at height $y$ crosses $R$ between the right branch of $y = 4x - x^2$ on the left and $y = x^2$ on the right, so both curves must be solved for $x$ . From $y = x^2$ with $x \ge 0$ : $x = \sqrt{y}$ . From $x^2 - 4x + y = 0$ : $x = 2 \pm \sqrt{4 - y}$ , and on $R$ the $x$ runs over $[0, 2]$ , so the root to keep is $x = 2 - \sqrt{4 - y}$ (the $+$ root is the other half of the parabola, for $2 \le x \le 4$ , outside $R$ ). Check at $y = 3$ : the slice runs from $x = 1$ to $x = \sqrt{3}$ , and indeed $4 \cdot 1 - 1 = 3$ and $(\sqrt{3})^2 = 3$ .

Both radii are measured from $x = 3$ : the far edge is the left one, $R_{\text{out}} = 3 - (2 - \sqrt{4 - y}) = 1 + \sqrt{4 - y}$ , and $R_{\text{in}} = 3 - \sqrt{y}$ . Expanding, $R_{\text{out}}^2 - R_{\text{in}}^2 = (5 - y + 2\sqrt{4 - y}) - (9 - 6\sqrt{y} + y) = -4 - 2y + 2\sqrt{4 - y} + 6\sqrt{y}$ . Then $V = \pi\int_0^4 \left(-4 - 2y + 2(4 - y)^{1/2} + 6y^{1/2}\right)dy = \pi\left(-16 - 16 + \frac{4}{3} \cdot 8 + 4 \cdot 8\right) = \pi\left(\frac{32}{3}\right)$ , since $\int_0^4 (4 - y)^{1/2}dy = \int_0^4 y^{1/2}dy = \frac{16}{3}$ . Same volume, twice the work: this is why shells are the method of choice when the axis is vertical and the curves are given as $y = f(x)$ .

c)

Shells again, with radius $x - (-1) = x + 1$ since $R$ is to the RIGHT of $x = -1$ : $V = 2\pi\int_0^2 (x + 1)(4x - 2x^2)\,dx = 2\pi\int_0^2 \left(4x + 2x^2 - 2x^3\right)dx = 2\pi\left(8 + \frac{16}{3} - 8\right) = \frac{32\pi}{3}$ . The same value is no accident: the height $h(x) = 2x(2 - x)$ is symmetric about $x = 1$ , so the centroid of $R$ has $\bar{x} = 1$ , at distance $2$ from both $x = 3$ and $x = -1$ . With $A = \int_0^2 (4x - 2x^2)\,dx = \frac{8}{3}$ , Pappus gives $2\pi \cdot 2 \cdot \frac{8}{3} = \frac{32\pi}{3}$ for both axes, although the two solids have different shapes.

Exercise 5: Pumping out a bowl shaped like a paraboloid

A tank is the surface obtained by rotating the parabola $y = x^2$ , $0 \le x \le 2$ , about the $y$ -axis (lengths in metres). It is full of water, of density $1000$ kg/m $^3$ ; take $g = 9.8$ m/s $^2$ . The figure shows a cross-section and one horizontal layer of water. Give exact answers, in terms of $\pi$ .

a) Find the work needed to pump all the water out over the rim.
b) Find the work needed to pump out only enough water to lower the level from $4$ m to $2$ m. Compare the fraction of the work and the fraction of the water.
c) The water must in fact be pumped to a spout $1$ m above the rim. Find the new total work.

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Answers

a) $W = 9800\pi\int_0^4 y(4 - y)\,dy = \frac{313600\pi}{3}$ J
b) $9800\pi \cdot \frac{16}{3} = \frac{156800\pi}{3}$ J: half the work for three quarters of the water
c) $9800\pi\left(\frac{32}{3} + 8\right) = \frac{548800\pi}{3}$ J

a)

Measure $y$ upward from the bottom of the bowl. The layer at height $y$ , of thickness $dy$ , is a disc of radius $x = \sqrt{y}$ (the tank wall is $y = x^2$ ), so its volume is $\pi(\sqrt{y})^2\,dy = \pi y\,dy$ , its mass $1000\pi y\,dy$ kg and its weight $9800\pi y\,dy$ N. It must rise from height $y$ to the rim at height $4$ : a distance $4 - y$ , not $y$ . So $W = \int_0^4 9800\pi\,y(4 - y)\,dy = 9800\pi\Big[2y^2 - \frac{y^3}{3}\Big]_0^4 = 9800\pi\left(32 - \frac{64}{3}\right) = 9800\pi \cdot \frac{32}{3} = \frac{313600\pi}{3}$ J, about $3.3 \times 10^5$ J.

b)

Only the layers between $y = 2$ and $y = 4$ leave, each still lifted to the rim: $W_{\text{top}} = 9800\pi\int_2^4 y(4 - y)\,dy = 9800\pi\left(\frac{32}{3} - \left(8 - \frac{8}{3}\right)\right) = 9800\pi \cdot \frac{16}{3} = \frac{156800\pi}{3}$ J, exactly HALF of a). The water removed is $\pi\int_2^4 y\,dy = 6\pi$ m $^3$ out of $\pi\int_0^4 y\,dy = 8\pi$ m $^3$ : three quarters of the water for half of the work. The bowl is wide at the top, so most of the water is near the rim, where it has little distance to travel; the last quarter, at the narrow bottom, costs the other half. Treating the bowl as a cylinder, with a constant layer area, would miss exactly this.

c)

Every layer now travels $5 - y$ instead of $4 - y$ : one extra metre for each. $W = 9800\pi\int_0^4 y(5 - y)\,dy = 9800\pi\left(40 - \frac{64}{3}\right) = 9800\pi \cdot \frac{56}{3} = \frac{548800\pi}{3}$ J. The difference with a) is $9800\pi \cdot 8$ J, the total weight of the water, $9800 \cdot 8\pi$ N, times the extra metre: the extra lift is the same for every layer, so it simply multiplies the total weight.

Exercise 6: A curve whose length element opens, and the surface it sweeps

Let $C$ be the curve $y = \frac{x^4}{8} + \frac{1}{4x^2}$ for $1 \le x \le 2$ .

The figure below shows the situation.

a) Show that $1 + \left(\frac{dy}{dx}\right)^2$ is a perfect square, and find the length of $C$ exactly. Compare with the length of the chord joining its endpoints.
b) $C$ is rotated about the $y$ -axis. Find the area of the surface generated.

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Answers

a) $1 + (y')^2 = \left(\frac{x^3}{2} + \frac{1}{2x^3}\right)^2$ ; $L = \frac{33}{16}$ , longer than the chord $\frac{\sqrt{985}}{16}$
b) $S = 2\pi\int_1^2 x\left(\frac{x^3}{2} + \frac{1}{2x^3}\right)dx = \frac{67\pi}{10}$

a)

$\frac{dy}{dx} = \frac{x^3}{2} - \frac{1}{2x^3}$ . Write it $a - b$ with $a = \frac{x^3}{2}$ and $b = \frac{1}{2x^3}$ , whose product is $ab = \frac{1}{4}$ . Then $1 + (a - b)^2 = a^2 - 2ab + b^2 + 1 = a^2 + \frac{1}{2} + b^2 = (a + b)^2$ , because $1 - 2ab = \frac{1}{2} = 2ab$ . So $\sqrt{1 + (y')^2} = \frac{x^3}{2} + \frac{1}{2x^3}$ , positive on $[1, 2]$ , and $L = \int_1^2 \left(\frac{x^3}{2} + \frac{x^{-3}}{2}\right)dx = \Big[\frac{x^4}{8} - \frac{1}{4x^2}\Big]_1^2 = \left(2 - \frac{1}{16}\right) - \left(\frac{1}{8} - \frac{1}{4}\right) = \frac{31}{16} + \frac{1}{8} = \frac{33}{16}$ .

The endpoints are $\left(1, \frac{3}{8}\right)$ and $\left(2, \frac{33}{16}\right)$ , a rise of $\frac{27}{16}$ over a run of $1$ , so the chord is $\sqrt{1 + \frac{729}{256}} = \frac{\sqrt{985}}{16}$ . Since $985 < 33^2 = 1089$ , the chord is shorter than $\frac{33}{16}$ , as a chord must be. Notice the antiderivative: $\frac{x^4}{8} - \frac{1}{4x^2}$ is $y$ with one sign changed, the signature of these designed curves.

b)

About the $y$ -axis the radius of the circle described by a point of $C$ is its distance to that axis, $x$ , not $y$ . So $S = \int_1^2 2\pi x\,ds = 2\pi\int_1^2 x\left(\frac{x^3}{2} + \frac{1}{2x^3}\right)dx = 2\pi\int_1^2 \left(\frac{x^4}{2} + \frac{1}{2x^2}\right)dx = 2\pi\Big[\frac{x^5}{10} - \frac{1}{2x}\Big]_1^2 = 2\pi\left(\frac{32}{10} - \frac{1}{4} - \frac{1}{10} + \frac{1}{2}\right) = 2\pi \cdot \frac{67}{20} = \frac{67\pi}{10}$ . Two errors cost the mark here: writing $2\pi x\,dx$ instead of $2\pi x\,ds$ , which gives the area of a flat ring, and taking the radius $y$ , which rotates about the other axis.

Exercise 7: How far from the target does the drone land?

A delivery drone aims at a target on the ground. In a model of its precision, the distance $X$ , in metres, between the target and the landing point has the density $f(x) = \frac{kx}{(1 + x^2)^2}$ for $x \ge 0$ , and $f(x) = 0$ for $x < 0$ . The figure shows the graph of $f$ for the value of $k$ that makes it a density.

a) Find $k$ .
b) Find the cumulative distribution function $F$ , the probability that the drone lands more than $2$ m from the target, and the median distance.
c) Find the mean distance, and explain why it is larger than the median.

Show the solution

Answers

a) $\int_0^\infty \frac{x\,dx}{(1 + x^2)^2} = \frac{1}{2}$ , so $k = 2$
b) $F(x) = 1 - \frac{1}{1 + x^2}$ for $x \ge 0$ ; $P(X > 2) = \frac{1}{5}$ ; median $1$ m
c) $\mu = \int_0^\infty \frac{2x^2}{(1 + x^2)^2}\,dx = \frac{\pi}{2}$ m, about $1.57$ m

a)

A density must be nonnegative, which holds for $k > 0$ and $x \ge 0$ , and have total area $1$ . With $u = 1 + x^2$ , $du = 2x\,dx$ : $\int_0^t \frac{x\,dx}{(1 + x^2)^2} = \frac{1}{2}\Big[-\frac{1}{u}\Big]_1^{1 + t^2} = \frac{1}{2}\left(1 - \frac{1}{1 + t^2}\right) \to \frac{1}{2}$ as $t \to \infty$ . So $k \cdot \frac{1}{2} = 1$ and $k = 2$ .

b)

For $x \ge 0$ , $F(x) = \int_0^x \frac{2t\,dt}{(1 + t^2)^2} = 1 - \frac{1}{1 + x^2}$ , and $F(x) = 0$ for $x < 0$ . Then $P(X > 2) = 1 - F(2) = \frac{1}{1 + 4} = \frac{1}{5}$ : one landing in five misses by more than $2$ m, the shaded tail of the figure. The median $m$ solves $F(m) = \frac{1}{2}$ : $\frac{1}{1 + m^2} = \frac{1}{2}$ , so $m^2 = 1$ and $m = 1$ m, the positive root since $X \ge 0$ . The median is NOT the peak of the density, which is at $x = \frac{1}{\sqrt{3}}$ : it is the point that splits the AREA in two.

c)

$\mu = \int_0^\infty x f(x)\,dx = \int_0^\infty \frac{2x^2}{(1 + x^2)^2}\,dx$ , an improper integral; it converges because the integrand behaves like $\frac{2}{x^2}$ at infinity. By parts with $u = x$ and $dv = \frac{2x\,dx}{(1 + x^2)^2}$ , $v = -\frac{1}{1 + x^2}$ : $\int_0^t \frac{2x^2\,dx}{(1 + x^2)^2} = \Big[-\frac{x}{1 + x^2}\Big]_0^t + \int_0^t \frac{dx}{1 + x^2} = -\frac{t}{1 + t^2} + \arctan t \to 0 + \frac{\pi}{2}$ . So $\mu = \frac{\pi}{2} \approx 1.57$ m. (With $x = \tan\theta$ the integrand becomes $2\sin^2\theta\,d\theta$ on $[0, \frac{\pi}{2}]$ , the same $\frac{\pi}{2}$ .) The mean exceeds the median $1$ because the density has a long right tail: the rare landings far from the target pull the average up, but not the middle value.

Exercise 8: Five series, five verdicts, five tests

Decide whether each series converges or diverges. Two points each: name the test, CHECK its hypotheses, and state the conclusion. A verdict without its test earns nothing.

a) $\displaystyle\sum_{n=1}^{\infty} \frac{n + 1}{n\sqrt{n + 3}}$
b) $\displaystyle\sum_{n=1}^{\infty} \frac{(2n)!}{n!\,n^n}$
c) $\displaystyle\sum_{n=1}^{\infty} \left(\frac{n}{n + 1}\right)^{n^2}$
d) $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n\sqrt{\ln n}}$
e) $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{n} - \sin\frac{1}{n}\right)$

Show the solution

Answers

a) Limit comparison with $\frac{1}{\sqrt{n}}$ , limit $1$ : diverges
b) Ratio test, $L = \frac{4}{e} > 1$ : diverges
c) Root test, $L = \frac{1}{e} < 1$ : converges
d) Integral test, $\int_2^\infty \frac{dx}{x\sqrt{\ln x}} = \infty$ : diverges
e) Terms positive; limit comparison with $\frac{1}{n^3}$ , limit $\frac{1}{6}$ : converges

a)

Positive terms. For large $n$ the numerator behaves like $n$ and the denominator like $n \cdot n^{1/2}$ , so compare with $b_n = \frac{1}{\sqrt{n}}$ : $\frac{a_n}{b_n} = \frac{(n + 1)\sqrt{n}}{n\sqrt{n + 3}} = \frac{1 + \frac{1}{n}}{\sqrt{1 + \frac{3}{n}}} \to 1$ , a finite positive limit. $\sum \frac{1}{\sqrt{n}}$ is a $p$ -series with $p = \frac{1}{2} \le 1$ , divergent, so the series diverges by the limit comparison test. The terms do tend to $0$ : the divergence test says nothing here.

b)

Factorials: the ratio test. $\frac{a_{n+1}}{a_n} = \frac{(2n + 2)!}{(2n)!} \cdot \frac{n!}{(n + 1)!} \cdot \frac{n^n}{(n + 1)^{n+1}} = \frac{(2n + 2)(2n + 1)}{n + 1} \cdot \frac{n^n}{(n + 1)^{n+1}} = \frac{2(2n + 1)}{n + 1} \cdot \left(\frac{n}{n + 1}\right)^n$ . The first factor tends to $4$ ; the second is $\frac{1}{(1 + 1/n)^n} \to \frac{1}{e}$ . So $L = \frac{4}{e}$ , and $L > 1$ because $e < 4$ : the series diverges, its terms even tend to infinity. The two traps: writing $\frac{(2n + 2)!}{(2n)!} = 2n + 2$ , forgetting the factor $2n + 1$ , and reading $\left(\frac{n}{n + 1}\right)^n$ as $1$ because its base tends to $1$ .

c)

The whole term is an $n$ -th power, $a_n = \left[\left(\frac{n}{n + 1}\right)^n\right]^n$ : the root test. $\sqrt[n]{a_n} = \left(\frac{n}{n + 1}\right)^n = \frac{1}{\left(1 + \frac{1}{n}\right)^n} \to \frac{1}{e} < 1$ , so the series converges. Same trap as in b), one level up: the base $\frac{n}{n + 1}$ tends to $1$ , but $1^\infty$ is an indeterminate form, and here the exponent $n^2$ wins.

d)

Let $f(x) = \frac{1}{x\sqrt{\ln x}}$ on $[2, \infty)$ . Hypotheses: $f$ is continuous and positive there ( $\ln x \ge \ln 2 > 0$ ), and decreasing, since $x$ and $\sqrt{\ln x}$ are positive and increasing, so their product increases. With $u = \ln x$ , $du = \frac{dx}{x}$ : $\int_2^t \frac{dx}{x\sqrt{\ln x}} = \Big[2\sqrt{\ln x}\Big]_2^t = 2\sqrt{\ln t} - 2\sqrt{\ln 2} \to \infty$ . The integral diverges, so the series diverges by the integral test, however slowly. Comparing with $\frac{1}{n}$ is useless here: $\frac{1}{n\sqrt{\ln n}} < \frac{1}{n}$ for $n \ge 3$ , and being smaller than a divergent series proves nothing.

e)

Sign first: for $0 < t \le 1$ , $\sin t < t$ , so every term $\frac{1}{n} - \sin\frac{1}{n}$ is positive. Size: with $t = \frac{1}{n}$ and the Maclaurin series $\sin t = t - \frac{t^3}{6} + \frac{t^5}{120} - \cdots$ , $t - \sin t = \frac{t^3}{6} - \frac{t^5}{120} + \cdots$ , so compare with $b_n = \frac{1}{n^3}$ : $\frac{a_n}{b_n} = \frac{t - \sin t}{t^3} = \frac{1}{6} - \frac{t^2}{120} + \cdots \to \frac{1}{6}$ as $n \to \infty$ . The limit is finite and positive and $\sum \frac{1}{n^3}$ converges ( $p = 3$ ), so the series converges. Comparing with $\frac{1}{n}$ , the natural first guess, gives the limit $0$ , which decides nothing for a divergent benchmark.

Exercise 9: Two alternating series: a late decrease, then an error to control

Parts a) and b) are independent. The figure shows the partial sums $S_1, \dots, S_8$ of the series of b), with its sum dashed.

a) Show that $\displaystyle\sum_{n=1}^{\infty} (-1)^n\frac{n^2}{n^3 + 4}$ converges, checking that its terms decrease in absolute value from some rank on (give the rank). Does it converge absolutely?
b) Let $S = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n\,3^n}$ . How many terms guarantee an error of at most $10^{-3}$ ? Give that partial sum as a fraction, say whether it is too large or too small, and give an interval that contains $S$ .
c) Find $S$ exactly, and check it against b).

Show the solution

Answers

a) $f'(x) = \frac{x(8 - x^3)}{(x^3 + 4)^2} < 0$ for $x > 2$ : decreasing from $n = 2$ ; converges, conditionally ( $\frac{n^2}{n^3 + 4} \sim \frac{1}{n}$ )
b) $b_5 = \frac{1}{1215} < 10^{-3} < b_4 = \frac{1}{324}$ : four terms; $S_4 = \frac{31}{108}$ , too small; $\frac{31}{108} < S < \frac{31}{108} + \frac{1}{1215}$
c) $S = \ln\left(1 + \frac{1}{3}\right) = \ln\frac{4}{3} \approx 0.2877$

a)

Here $b_n = \frac{n^2}{n^3 + 4} > 0$ and $b_n = \frac{1/n}{1 + 4/n^3} \to 0$ . The decrease is not obvious, numerator and denominator both grow, so study $f(x) = \frac{x^2}{x^3 + 4}$ : $f'(x) = \frac{2x(x^3 + 4) - x^2 \cdot 3x^2}{(x^3 + 4)^2} = \frac{8x - x^4}{(x^3 + 4)^2} = \frac{x(8 - x^3)}{(x^3 + 4)^2}$ , negative exactly when $x^3 > 8$ , that is $x > 2$ . So $(b_n)$ decreases from $n = 2$ on: $b_1 = \frac{1}{5} < b_2 = \frac{1}{3}$ , then $b_3 = \frac{9}{31} < \frac{1}{3}$ , and so on. The alternating series test only needs the decrease EVENTUALLY, since the first term does not affect convergence: the series converges.

Absolutely? $\sum b_n$ against $\frac{1}{n}$ : $\frac{b_n}{1/n} = \frac{n^3}{n^3 + 4} \to 1$ , and the harmonic series diverges, so $\sum b_n$ diverges by limit comparison. The series converges conditionally, not absolutely.

b)

$b_n = \frac{1}{n\,3^n}$ is positive, decreasing (both $n$ and $3^n$ increase) and tends to $0$ , so the alternating series estimate applies: $|S - S_N| \le b_{N+1}$ . The first omitted term is the one to test: $b_4 = \frac{1}{4 \cdot 81} = \frac{1}{324} > 10^{-3}$ , while $b_5 = \frac{1}{5 \cdot 243} = \frac{1}{1215} < 10^{-3}$ . So $N = 4$ terms suffice (and three do not guarantee it). $S_4 = \frac{1}{3} - \frac{1}{18} + \frac{1}{81} - \frac{1}{324} = \frac{108 - 18 + 4 - 1}{324} = \frac{93}{324} = \frac{31}{108}$ .

The error $S - S_4$ has the sign of the first omitted term, $+\frac{1}{1215}$ : $S_4$ is too SMALL. Hence $\frac{31}{108} < S < \frac{31}{108} + \frac{1}{1215}$ . The figure shows it: the odd partial sums sit above the dashed line, the even ones below, and $S_4$ is an even one.

c)

For $-1 < x \le 1$ , $\ln(1 + x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n}$ . At $x = \frac{1}{3}$ , inside the interval, this is exactly our series: $S = \ln\frac{4}{3}$ . Check: $\ln\frac{4}{3} = 2\ln 2 - \ln 3 \approx 1.3863 - 1.0986 = 0.2877$ , while $\frac{31}{108} \approx 0.2870$ : the difference, about $0.0006$ , is positive and below $\frac{1}{1215} \approx 0.0008$ , as b) promised.

Exercise 10: An interval with its two endpoints, and a series found by differentiating

Parts a) and b) are independent.

a) Find the centre, the radius and the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x + 1)^n}{3^n\sqrt{n^2 + 1}}$ . Test each endpoint with its own test.
b) Starting from the geometric series of $\frac{1}{1 + x}$ , find a power series for $\frac{1}{(1 + x)^3}$ centred at $0$ , with its radius. Deduce the exact sum of $\displaystyle\sum_{n=0}^{\infty} (-1)^n\frac{(n + 1)(n + 2)}{3^n}$ .

Show the solution

Answers

a) Centre $-1$ , radius $3$ ; diverges at $x = 2$ , converges conditionally at $x = -4$ : interval $[-4, 2)$
b) $\frac{1}{(1 + x)^3} = \sum_{n=0}^{\infty} (-1)^n\frac{(n + 1)(n + 2)}{2}x^n$ , $R = 1$ ; the sum is $\frac{2}{(4/3)^3} = \frac{27}{32}$

a)

Centre $a = -1$ , read in the power $(x - (-1))^n$ . Ratio test on the absolute values, $x \ne -1$ fixed: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x + 1|}{3} \cdot \frac{\sqrt{n^2 + 1}}{\sqrt{(n + 1)^2 + 1}} \to \frac{|x + 1|}{3}$ . The series converges absolutely when $|x + 1| < 3$ and diverges when $|x + 1| > 3$ : radius $R = 3$ , and the open interval is $(-4, 2)$ . The ratio test is silent at the two endpoints, where the limit is exactly $1$ , so each is tested separately.

At $x = 2$ : $x + 1 = 3$ and the series is $\sum \frac{1}{\sqrt{n^2 + 1}}$ , positive terms. Limit comparison with $\frac{1}{n}$ : $\frac{n}{\sqrt{n^2 + 1}} = \frac{1}{\sqrt{1 + 1/n^2}} \to 1$ , and the harmonic series diverges, so the series diverges at $x = 2$ . At $x = -4$ : $x + 1 = -3$ and the series is $\sum \frac{(-1)^n}{\sqrt{n^2 + 1}}$ . Alternating series test: $b_n = \frac{1}{\sqrt{n^2 + 1}}$ is positive, decreasing because $n^2 + 1$ increases, and tends to $0$ : it converges, conditionally by the computation at $x = 2$ . The interval of convergence is $[-4, 2)$ .

b)

For $|x| < 1$ , $\frac{1}{1 + x} = \sum_{n=0}^{\infty} (-1)^n x^n$ . Differentiate twice, term by term, which keeps the radius $1$ : $\frac{-1}{(1 + x)^2} = \sum_{n=1}^{\infty} (-1)^n n x^{n-1}$ and $\frac{2}{(1 + x)^3} = \sum_{n=2}^{\infty} (-1)^n n(n - 1)x^{n-2}$ . Shift the index with $m = n - 2$ , so $(-1)^n = (-1)^m$ : $\frac{2}{(1 + x)^3} = \sum_{m=0}^{\infty} (-1)^m (m + 2)(m + 1)x^m$ , hence $\frac{1}{(1 + x)^3} = \sum_{m=0}^{\infty} (-1)^m\frac{(m + 1)(m + 2)}{2}x^m$ for $|x| < 1$ . Check the first terms against the binomial series: $1 - 3x + 6x^2 - \cdots$ .

The numerical series is $\sum (-1)^n (n + 1)(n + 2)\left(\frac{1}{3}\right)^n = 2 \cdot \frac{1}{(1 + 1/3)^3}$ , evaluated at $x = \frac{1}{3}$ , which is inside the interval $|x| < 1$ . So the sum is $2 \cdot \frac{27}{64} = \frac{27}{32}$ . Forgetting the factor $2$ of the second derivative, or evaluating at $x = -\frac{1}{3}$ because of the sign $(-1)^n$ , are the two ways to lose this mark.

Exercise 11: A Maclaurin series from the table, read three ways

Let $f(x) = x\ln(1 + x^2)$ . Use the Maclaurin series of the table; no derivative of $f$ is to be computed by hand.

a) Find the Maclaurin series of $f$ , its general term, and its interval of convergence.
b) Find $f^{(7)}(0)$ and $f^{(8)}(0)$ .
c) Compute $\displaystyle\lim_{x \to 0} \frac{x\ln(1 + x^2) - x\sin^2 x}{x^5}$ .

Show the solution

Answers

a) $f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{2n+1}}{n} = x^3 - \frac{x^5}{2} + \frac{x^7}{3} - \cdots$ on $[-1, 1]$
b) $f^{(7)}(0) = \frac{7!}{3} = 1680$ ; $f^{(8)}(0) = 0$
c) $-\frac{1}{2} + \frac{1}{3} = -\frac{1}{6}$

a)

The table gives $\ln(1 + u) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}u^n}{n}$ for $-1 < u \le 1$ . Substitute $u = x^2$ : $\ln(1 + x^2) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{2n}}{n}$ , then multiply by $x$ : $f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{2n+1}}{n} = x^3 - \frac{x^5}{2} + \frac{x^7}{3} - \cdots$ . The interval moves with the variable: the condition is $-1 < x^2 \le 1$ , and $x^2 > -1$ always holds, so it is $|x| \le 1$ . At $x = \pm 1$ the series is $\pm\sum \frac{(-1)^{n-1}}{n}$ , the alternating harmonic series, convergent, which confirms both endpoints. For $|x| > 1$ the terms $\frac{|x|^{2n+1}}{n}$ tend to infinity. Interval: $[-1, 1]$ .

b)

The series of $f$ at $0$ IS its Taylor series, so the coefficient of $x^k$ is $\frac{f^{(k)}(0)}{k!}$ . The coefficient of $x^7$ is the term $n = 3$ , $\frac{(-1)^2}{3} = \frac{1}{3}$ , so $f^{(7)}(0) = \frac{7!}{3} = \frac{5040}{3} = 1680$ . There is no $x^8$ : every power is odd, as it must be for the odd function $f$ , so $f^{(8)}(0) = 0$ . Seven differentiations of $x\ln(1 + x^2)$ by the product and chain rules would take a page.

c)

Expand both terms up to $x^5$ . From a): $x\ln(1 + x^2) = x^3 - \frac{x^5}{2} + O(x^7)$ . For $\sin^2 x$ , use $\sin^2 x = \frac{1 - \cos 2x}{2}$ and the series of cosine: $\cos 2x = 1 - 2x^2 + \frac{2x^4}{3} - \cdots$ , so $\sin^2 x = x^2 - \frac{x^4}{3} + O(x^6)$ and $x\sin^2 x = x^3 - \frac{x^5}{3} + O(x^7)$ . The $x^3$ terms cancel, which is why the expansion must go one step further: the numerator is $-\frac{x^5}{2} + \frac{x^5}{3} + O(x^7) = -\frac{x^5}{6} + O(x^7)$ , and the limit is $-\frac{1}{6}$ . Stopping at $x^3$ gives $\frac{0}{x^5}$ and no answer; squaring only the first term of $\sin x$ loses the $-\frac{x^4}{3}$ .

Exercise 12: Taylor's inequality away from zero: arctan 1.1 by hand

The figure shows $y = \arctan x$ and its Taylor polynomial $T_2$ of degree $2$ centred at $a = 1$ (dashed).

a) Compute $T_2(x)$ for $f(x) = \arctan x$ at $a = 1$ .
b) Use it to approximate $\arctan 1.1$ , then bound the error with Taylor's inequality, justifying your $M$ without a calculator.
c) Is the approximation too large or too small? Give an interval that contains $\arctan 1.1$ .
d) Why not simply use the Maclaurin series $x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$ at $x = 1.1$ ?

Show the solution

Answers

a) $T_2(x) = \frac{\pi}{4} + \frac{x - 1}{2} - \frac{(x - 1)^2}{4}$
b) $\arctan 1.1 \approx \frac{\pi}{4} + \frac{19}{400}$ ; $f''' \le 1$ on $[1, 1.1]$ , $|R_2| \le \frac{(0.1)^3}{6} = \frac{1}{6000}$
c) $f''' > 0$ : too small; $\frac{\pi}{4} + \frac{19}{400} < \arctan 1.1 < \frac{\pi}{4} + \frac{19}{400} + \frac{1}{6000}$
d) That series has radius $1$ and diverges at $1.1$

a)

$f(x) = \arctan x$ , $f'(x) = \frac{1}{1 + x^2}$ , $f''(x) = -\frac{2x}{(1 + x^2)^2}$ . At $a = 1$ : $f(1) = \frac{\pi}{4}$ , $f'(1) = \frac{1}{2}$ , $f''(1) = -\frac{2}{4} = -\frac{1}{2}$ . So $T_2(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 = \frac{\pi}{4} + \frac{x - 1}{2} - \frac{(x - 1)^2}{4}$ . The division by $2!$ is the usual slip: without it, the last coefficient would be $-\frac{1}{2}$ .

b)

With $x = 1.1$ , $x - 1 = 0.1$ : $T_2(1.1) = \frac{\pi}{4} + 0.05 - 0.0025 = \frac{\pi}{4} + \frac{19}{400}$ . The answer keeps $\pi$ : it is exact up to the Taylor error, and no calculator is needed. For the bound, $f'''(x) = \frac{6x^2 - 2}{(1 + x^2)^3}$ . On the WHOLE interval $[1, 1.1]$ : the numerator is at most $6(1.21) - 2 = 5.26$ and the denominator at least $2^3 = 8$ , so $0 < f'''(x) \le \frac{5.26}{8} < 1$ . Take $M = 1$ : $|R_2(1.1)| \le \frac{M}{3!}|1.1 - 1|^3 = \frac{1}{6} \cdot \frac{1}{1000} = \frac{1}{6000}$ .

c)

Taylor's formula with the Lagrange remainder writes $R_2(1.1) = \frac{f'''(c)}{6}(0.1)^3$ for some $c$ between $1$ and $1.1$ , and $f'''(c) > 0$ there since $6c^2 - 2 \ge 4$ . So the error is positive: $T_2(1.1)$ is too SMALL, and $\frac{\pi}{4} + \frac{19}{400} < \arctan 1.1 < \frac{\pi}{4} + \frac{19}{400} + \frac{1}{6000}$ . With $\pi \approx 3.1416$ that is between $0.8328$ and $0.8331$ . The figure agrees: just to the right of $1$ the dashed parabola passes under the curve. No alternating series is in sight here, so Taylor's inequality, with the sign of $f'''$ , is the tool that gives both the bound and the side.

d)

The series $\sum \frac{(-1)^n x^{2n+1}}{2n + 1}$ has radius $1$ : at $x = 1.1$ its terms $\frac{(1.1)^{2n+1}}{2n + 1}$ tend to infinity, so it DIVERGES, although $\arctan 1.1$ is a perfectly good number. A series centred at $0$ says nothing beyond its radius. This is why the polynomial is centred at $a = 1$ , the nearest point where $\arctan$ and its derivatives are known exactly, and why the error bound shrinks with $|x - a|^3 = 10^{-3}$ .

MATH 141 practice final exam with full solutions (McGill)

Part A: integrals and techniques (/24)

Exercise 1: Four integrals, and the move that opens each one

Exercise 2: Improper integrals: two values, two verdicts

Part B: applications of the integral (/38)

Exercise 3: One region: its area, its centroid, and the volume Pappus gives for free

Exercise 4: One solid, two methods: shells, then washers that force an inverse

Exercise 5: Pumping out a bowl shaped like a paraboloid

Exercise 6: A curve whose length element opens, and the surface it sweeps

Exercise 7: How far from the target does the drone land?

Part C: sequences and series (/38)

Exercise 8: Five series, five verdicts, five tests

Exercise 9: Two alternating series: a late decrease, then an error to control

Exercise 10: An interval with its two endpoints, and a series found by differentiating

Exercise 11: A Maclaurin series from the table, read three ways

Exercise 12: Taylor's inequality away from zero: arctan 1.1 by hand

See also

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