MATH 141 practice final exam with full solutions (McGill)

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.

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 $x \to 0^+$.

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, $\frac{5\pi}{14}$.

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.

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.

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.

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.