MATH 141 practice midterm with full solutions (McGill)

a) $\Delta x = \frac{2 - 0}{4} = \frac{1}{2}$. The left sum samples $0, \frac{1}{2}, 1, \frac{3}{2}$ and the right sum $\frac{1}{2}, 1, \frac{3}{2}, 2$. The values are $f(0) = 0$, $f(\frac{1}{2}) = \frac{5}{4}$, $f(1) = 3$, $f(\frac{3}{2}) = \frac{21}{4}$, $f(2) = 8$. So $L_4 = \frac{1}{2}\left(0 + \frac{5}{4} + 3 + \frac{21}{4}\right) = \frac{1}{2} \cdot \frac{19}{2} = \frac{19}{4}$ and $R_4 = \frac{1}{2}\left(\frac{5}{4} + 3 + \frac{21}{4} + 8\right) = \frac{1}{2} \cdot \frac{35}{2} = \frac{35}{4}$.

The property that decides is monotonicity, not the picture: $f'(x) = 2x + 2 > 0$ on $[0, 2]$, so $f$ is increasing. On each strip the left end gives the SMALLEST value and the right end the LARGEST, so every left rectangle sits under the curve and every right rectangle sticks out above it. Hence $L_4 \le A \le R_4$, that is $\frac{19}{4} \le A \le \frac{35}{4}$, and $R_4$ is the overestimate. The gap $R_4 - L_4 = \Delta x\left[f(2) - f(0)\right] = 4$ is wide: four strips do not pin $A$ down.

b) With $n$ strips, $\Delta x = \frac{2}{n}$ and $x_i = \frac{2i}{n}$, so $f(x_i) = \frac{4i^2}{n^2} + \frac{4i}{n}$. Then $R_n = \sum_{i=1}^{n} \frac{2}{n}\left(\frac{4i^2}{n^2} + \frac{4i}{n}\right) = \frac{8}{n^3}\sum_{i=1}^{n} i^2 + \frac{8}{n^2}\sum_{i=1}^{n} i$. The sums start at $i = 1$ and end at $n$, so the formulas apply: $R_n = \frac{8}{n^3} \cdot \frac{n(n + 1)(2n + 1)}{6} + \frac{8}{n^2} \cdot \frac{n(n + 1)}{2} = \frac{4(n + 1)(2n + 1)}{3n^2} + \frac{4(n + 1)}{n} = \frac{4(n + 1)(5n + 1)}{3n^2}$.

Check at $n = 4$: $\frac{4 \cdot 5 \cdot 21}{48} = \frac{420}{48} = \frac{35}{4}$, the value of a). Now write $R_n = \frac{4}{3}\left(1 + \frac{1}{n}\right)\left(5 + \frac{1}{n}\right)$ and let $n \to \infty$: $\int_0^2 (x^2 + 2x)\,dx = \frac{4}{3} \cdot 5 = \frac{20}{3}$. It lies in the bracket of a), since $\frac{19}{4} = 4.75 < 6.67 < 8.75$.

c) Factor out $\frac{1}{n}$: $\frac{1}{n + i} = \frac{1}{n} \cdot \frac{1}{1 + \frac{i}{n}}$. Read it in order: $\Delta x = \frac{1}{n}$, so $b - a = 1$; the sample point is $x_i = \frac{i}{n} = 0 + i\,\Delta x$, so $a = 0$ and $b = 1$; the function is $f(x) = \frac{1}{1 + x}$, continuous on $[0, 1]$. The limit is $\int_0^1 \frac{dx}{1 + x} = \Big[\ln(1 + x)\Big]_0^1 = \ln 2$. Another correct reading is $x_i = 1 + \frac{i}{n}$ with $f(x) = \frac{1}{x}$ on $[1, 2]$, which gives $\int_1^2 \frac{dx}{x} = \ln 2$ as well. Sanity check: each of the $n$ terms lies between $\frac{1}{2n}$ and $\frac{1}{n}$, so the sum lies between $\frac{1}{2}$ and $1$, and $\ln 2 \approx 0.69$ does.

d) By linearity, $\int_{-2}^{2} \left(x + \sqrt{4 - x^2}\right)dx = \int_{-2}^{2} x\,dx + \int_{-2}^{2} \sqrt{4 - x^2}\,dx$. The first integral is the signed area between the line $y = x$ and the axis: a triangle of area $2$ below the axis on $[-2, 0]$ and one of area $2$ above it on $[0, 2]$, so it is $-2 + 2 = 0$. The graph of $y = \sqrt{4 - x^2}$ is the upper half of the circle $x^2 + y^2 = 4$, so the second integral is half the area of a disc of radius $2$: $\frac{1}{2}\pi \cdot 2^2 = 2\pi$. The value is $2\pi$. Note that the integrand itself is negative near $x = -2$ (it equals $-2$ there): the integral is a signed area, and nothing here asks for an area.

a) Let $g(t) = \frac{\sin t}{t}$, continuous on $(0, \infty)$. For $x > 0$ both bounds $\sqrt{x}$ and $x^2$ lie in $(0, \infty)$, so the integral can be split at the constant $1$: $F(x) = \int_1^{x^2} g(t)\,dt - \int_1^{\sqrt{x}} g(t)\,dt$. FTC 1 and the chain rule on each piece: $F'(x) = g(x^2) \cdot 2x - g(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\sin(x^2)}{x^2} \cdot 2x - \frac{\sin\sqrt{x}}{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{2\sin(x^2)}{x} - \frac{\sin\sqrt{x}}{2x}$. Looking for an antiderivative first would be a dead end: there is none in elementary terms, and FTC 1 does not need one.

At $x = 1$ both bounds equal $1$, so $F(1) = \int_1^1 g(t)\,dt = 0$, and $F'(1) = 2\sin 1 - \frac{1}{2}\sin 1 = \frac{3}{2}\sin 1$. The tangent line is $y = \frac{3}{2}(\sin 1)(x - 1)$. The two traps: forgetting the chain factors $2x$ and $\frac{1}{2\sqrt{x}}$, and forgetting the minus sign of the lower bound.

b) The factor $\cos x$ is the derivative of $1 + \sin x$: set $u = 1 + \sin x$, $du = \cos x\,dx$. Carry the bounds: $x = 0$ gives $u = 1$ and $x = \frac{\pi}{2}$ gives $u = 2$. The integral becomes $\int_1^2 u^{-1/2}\,du = \Big[2\sqrt{u}\Big]_1^2 = 2\sqrt{2} - 2$. Size check: the integrand lies between $0$ and $1$ on an interval of length $\frac{\pi}{2}$, and $2\sqrt{2} - 2 \approx 0.83$ does.

c) With $u = 1 + x^3$, $du = 3x^2\,dx$ absorbs only $x^2$. Split $x^5 = x^3 \cdot x^2$: the $x^2\,dx$ becomes $\frac{du}{3}$, and the remaining $x^3$ is NOT a constant, it is rewritten from $u = 1 + x^3$ as $x^3 = u - 1$. The bounds: $x = 0$ gives $u = 1$, $x = 1$ gives $u = 2$. So $\int_0^1 \frac{x^5}{\sqrt{1 + x^3}}\,dx = \frac{1}{3}\int_1^2 \frac{u - 1}{\sqrt{u}}\,du = \frac{1}{3}\int_1^2 \left(u^{1/2} - u^{-1/2}\right)du = \frac{1}{3}\Big[\frac{2}{3}u^{3/2} - 2u^{1/2}\Big]_1^2$.

At $u = 2$: $\frac{2}{3} \cdot 2\sqrt{2} - 2\sqrt{2} = -\frac{2\sqrt{2}}{3}$; at $u = 1$: $\frac{2}{3} - 2 = -\frac{4}{3}$. The difference is $\frac{4 - 2\sqrt{2}}{3}$, and the integral is $\frac{4 - 2\sqrt{2}}{9}$.

d) She obtains $\frac{1}{3}\Big[\frac{2}{3}u^{3/2} - 2u^{1/2}\Big]_0^1 = \frac{1}{3}\left(\frac{2}{3} - 2\right) = -\frac{4}{9}$. The integrand $\frac{x^5}{\sqrt{1 + x^3}}$ is nonnegative on $[0, 1]$, so its integral cannot be negative: the sign alone condemns the answer. Her integrand is in $u$ but her bounds are still in $x$, whereas $u$ runs from $1$ to $2$.

Bracket: on $[0, 1]$, $1 \le 1 + x^3 \le 2$, so $\frac{x^5}{\sqrt{2}} \le \frac{x^5}{\sqrt{1 + x^3}} \le x^5$, and integrating, $\frac{1}{6\sqrt{2}} \le I \le \frac{1}{6}$. With $\sqrt{2} \approx 1.41$ that is $0.118 \le I \le 0.167$, and $\frac{4 - 2\sqrt{2}}{9} \approx \frac{4 - 2.83}{9} \approx 0.13$ fits.

a) Move: an ODD power of sine, so detach one $\sin x\,dx$ and set $u = \cos x$, $du = -\sin x\,dx$, converting $\sin^2 x = 1 - u^2$. The integrand is continuous on $[0, \frac{\pi}{3}]$, where $\cos x \ge \frac{1}{2}$. Bounds: $x = 0$ gives $u = 1$, $x = \frac{\pi}{3}$ gives $u = \frac{1}{2}$. Then $\int_0^{\pi/3} \frac{\sin^2 x}{\cos^2 x}\sin x\,dx = \int_1^{1/2} \frac{1 - u^2}{u^2}(-du) = \int_{1/2}^{1} \left(u^{-2} - 1\right)du = \Big[-\frac{1}{u} - u\Big]_{1/2}^{1} = (-2) - \left(-\frac{5}{2}\right) = \frac{1}{2}$.

Second route, as a check: $\frac{\sin^3 x}{\cos^2 x} = \frac{\sin x(1 - \cos^2 x)}{\cos^2 x} = \sec x\tan x - \sin x$, whose antiderivative is $\sec x + \cos x$; from $0$ to $\frac{\pi}{3}$ it gives $\left(2 + \frac{1}{2}\right) - (1 + 1) = \frac{1}{2}$.

b) Move: integration by parts with $u = x$, $du = dx$, $dv = \sec^2 x\,dx$, $v = \tan x$. The polynomial is $u$ because differentiating it removes it, and $\sec^2 x$ is $dv$ because it integrates at sight; the opposite choice would need $v = \frac{x^2}{2}$ and raise the power. Then $\int_0^{\pi/4} x\sec^2 x\,dx = \Big[x\tan x\Big]_0^{\pi/4} - \int_0^{\pi/4} \tan x\,dx = \frac{\pi}{4} - \Big[\ln|\sec x|\Big]_0^{\pi/4} = \frac{\pi}{4} - \ln\sqrt{2} = \frac{\pi}{4} - \frac{1}{2}\ln 2$. Size check: about $0.79 - 0.35 = 0.44$, positive as it must be.

c) Move: only EVEN powers, so no factor can be saved; lower the degree. First $\sin x\cos x = \frac{1}{2}\sin 2x$, so $\sin^2 x\cos^2 x = \frac{1}{4}\sin^2 2x$, then the half-angle formula $\sin^2 2x = \frac{1 - \cos 4x}{2}$: the integrand is $\frac{1}{8}(1 - \cos 4x)$. Then $\int_0^{\pi/2} \frac{1}{8}(1 - \cos 4x)\,dx = \frac{1}{8}\Big[x - \frac{\sin 4x}{4}\Big]_0^{\pi/2} = \frac{1}{8} \cdot \frac{\pi}{2} = \frac{\pi}{16}$, since $\sin 2\pi = 0$.

d) Move: the root blocks everything, so rationalize with $u = \sqrt{x}$, $x = u^2$, $dx = 2u\,du$; the WHOLE of $dx$ is converted. Bounds: $x = 1$ gives $u = 1$, $x = 9$ gives $u = 3$. Then $\int_1^9 \frac{dx}{x + \sqrt{x}} = \int_1^3 \frac{2u\,du}{u^2 + u} = \int_1^3 \frac{2\,du}{u + 1}$, because $u^2 + u = u(u + 1)$ and the factor $u$ cancels: no partial fractions are needed once the fraction is simplified. The value is $2\Big[\ln(u + 1)\Big]_1^3 = 2(\ln 4 - \ln 2) = 2\ln 2$.

a) Improper at $\infty$ only: $x^2 + 4x + 5 = (x + 2)^2 + 1 > 0$, so the integrand is continuous on $[0, \infty)$. Completing the square gives the arctangent: $\int_0^t \frac{dx}{(x + 2)^2 + 1} = \arctan(t + 2) - \arctan 2$, and as $t \to \infty$, $\arctan(t + 2) \to \frac{\pi}{2}$. The integral converges to $\frac{\pi}{2} - \arctan 2$, which is also $\arctan\frac{1}{2}$ since $\arctan a + \arctan\frac{1}{a} = \frac{\pi}{2}$ for $a > 0$. It is an exact answer: no calculator is needed, nor wanted.

b) Improper at $0$: $\frac{\ln x}{\sqrt{x}} \to -\infty$ as $x \to 0^+$. By parts with $u = \ln x$, $dv = x^{-1/2}\,dx$, $v = 2\sqrt{x}$: $\int \frac{\ln x}{\sqrt{x}}\,dx = 2\sqrt{x}\ln x - \int \frac{2\sqrt{x}}{x}\,dx = 2\sqrt{x}\ln x - 4\sqrt{x} + C$. So $\int_t^1 \frac{\ln x}{\sqrt{x}}\,dx = -4 - 2\sqrt{t}\ln t + 4\sqrt{t}$. The limit $\lim_{t \to 0^+} \sqrt{t}\ln t$ is of the form $0 \cdot (-\infty)$: write it $\frac{\ln t}{t^{-1/2}}$, of the form $\frac{-\infty}{\infty}$, and L'Hôpital's rule gives $\frac{1/t}{-\frac{1}{2}t^{-3/2}} = -2\sqrt{t} \to 0$. The integral converges to $-4$, negative as it must be, since $\ln x < 0$ on $(0, 1)$.

c) Both endpoints are harmless ($\ln x - 1 = -1$ at $x = 1$ and $1$ at $x = e^2$), but $\ln x - 1 = 0$ at $x = e$, which lies INSIDE $[1, e^2]$: the integrand blows up there. Split at $e$, each piece a limit: $\int_1^{e^2} = \lim_{s \to e^-}\int_1^s + \lim_{r \to e^+}\int_r^{e^2}$. With $u = \ln x - 1$, $du = \frac{dx}{x}$, an antiderivative is $3\sqrt[3]{u} = 3\sqrt[3]{\ln x - 1}$, valid on each side of $e$ (the cube root of a negative number is real). Left piece: $\lim_{s \to e^-}\left(3\sqrt[3]{\ln s - 1}\right) - 3\sqrt[3]{-1} = 0 + 3 = 3$. Right piece: $3\sqrt[3]{1} - \lim_{r \to e^+} 3\sqrt[3]{\ln r - 1} = 3$. Both converge, so the integral converges to $6$.

Evaluating $3\sqrt[3]{\ln x - 1}$ from $1$ to $e^2$ in one line also prints $6$, by luck: that line uses FTC 2 across a point where the integrand is not continuous. The marks go to the split and the two limits, because the same shortcut prints a finite number for integrals that diverge.

d) Look at the endpoints: $x^2 - 4x + 3 = (x - 1)(x - 3)$ vanishes at $x = 1$, an endpoint, so the integral is improper at $1$. Partial fractions: $\frac{1}{(x - 1)(x - 3)} = \frac{1}{2}\left(\frac{1}{x - 3} - \frac{1}{x - 1}\right)$. Then $\int_0^t \frac{dx}{(x - 1)(x - 3)} = \frac{1}{2}\Big[\ln|x - 3| - \ln|x - 1|\Big]_0^t = \frac{1}{2}\left(\ln(3 - t) - \ln(1 - t) - \ln 3\right)$. As $t \to 1^-$, $-\ln(1 - t) \to +\infty$: the integral diverges. Evaluating from $0$ to $1$ directly would ask for $\ln 0$, which is exactly the warning that a limit was needed.

a) Take $u = \ln x$, $du = \frac{dx}{x}$, and $dv = x^{1/2}\,dx$, $v = \frac{2}{3}x^{3/2}$. The logarithm is $u$ because its derivative is algebraic and kills the logarithm; the opposite choice, $dv = \ln x\,dx$, would need $v = x\ln x - x$ and leave $\int \frac{x\ln x - x}{2\sqrt{x}}\,dx$, no simpler than the start. Then $\int_1^{e^2} \sqrt{x}\ln x\,dx = \Big[\frac{2}{3}x^{3/2}\ln x\Big]_1^{e^2} - \frac{2}{3}\int_1^{e^2} x^{1/2}\,dx$. At $x = e^2$, $x^{3/2} = e^3$ and $\ln x = 2$, so the bracket is $\frac{4}{3}e^3 - 0$. The new integral is $\frac{2}{3} \cdot \frac{2}{3}\Big[x^{3/2}\Big]_1^{e^2} = \frac{4}{9}(e^3 - 1)$. The value is $\frac{4}{3}e^3 - \frac{4}{9}e^3 + \frac{4}{9} = \frac{8e^3 + 4}{9}$.

Size check: on $[1, e^2]$ the integrand grows from $0$ to $2e$, over an interval of length $e^2 - 1$; so the value lies between $0$ and $2e(e^2 - 1) \approx 5.4 \times 6.4 \approx 35$, and $\frac{8e^3 + 4}{9} \approx \frac{8 \times 20 + 4}{9} \approx 18$ does (with $e^3 \approx 20$).

b) Write $\sin^n x = \sin^{n-1} x \cdot \sin x$ and take $u = \sin^{n-1} x$, $du = (n - 1)\sin^{n-2} x\cos x\,dx$, $dv = \sin x\,dx$, $v = -\cos x$. Then $\int \sin^n x\,dx = -\sin^{n-1} x\cos x + (n - 1)\int \sin^{n-2} x\cos^2 x\,dx$. Convert $\cos^2 x = 1 - \sin^2 x$: $\int \sin^n x\,dx = -\sin^{n-1} x\cos x + (n - 1)\int \sin^{n-2} x\,dx - (n - 1)\int \sin^n x\,dx$.

The integral we started with reappears on the right with the coefficient $-(n - 1)$. Treat it as an unknown and bring it to the left: $n\int \sin^n x\,dx = -\sin^{n-1} x\cos x + (n - 1)\int \sin^{n-2} x\,dx$. Dividing by $n$ gives the formula. The hypothesis $n \ge 2$ makes $\sin^{n-2} x$ an honest power, so the right-hand side is a genuine integral of lower degree.

c) On $\left[0, \frac{\pi}{2}\right]$ the bracket vanishes at both ends: $\sin^{n-1} x\cos x$ is $0$ at $x = \frac{\pi}{2}$ because $\cos\frac{\pi}{2} = 0$, and $0$ at $x = 0$ because $\sin 0 = 0$ and $n - 1 \ge 1$. So $W_n = \frac{n - 1}{n}W_{n-2}$. The two starting values are $W_0 = \int_0^{\pi/2} dx = \frac{\pi}{2}$ and $W_1 = \Big[-\cos x\Big]_0^{\pi/2} = 1$. Even indices come from $W_0$: $W_2 = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$, $W_4 = \frac{3}{4} \cdot \frac{\pi}{4} = \frac{3\pi}{16}$. Odd indices come from $W_1$: $W_3 = \frac{2}{3}$, $W_5 = \frac{4}{5} \cdot \frac{2}{3} = \frac{8}{15}$. Note that an even $n$ always ends on $\pi$ and an odd $n$ never does.

d) On $\left[0, \frac{\pi}{2}\right]$, $0 \le \sin x \le 1$, so multiplying $\sin^n x$ by $\sin x$ can only lower it: $\sin^{n+1} x \le \sin^n x$, which is the order of the curves on the figure. Integrating, $W_{n+1} \le W_n$. The values agree: $W_5 = \frac{8}{15} \approx 0.53$, $W_4 = \frac{3\pi}{16} \approx 0.59$, $W_3 = \frac{2}{3} \approx 0.67$ (with $\pi \approx 3.14$).

Half-angle check: $\sin^4 x = \left(\frac{1 - \cos 2x}{2}\right)^2 = \frac{1}{4} - \frac{1}{2}\cos 2x + \frac{1}{4}\cos^2 2x = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$. On $\left[0, \frac{\pi}{2}\right]$ the two cosines integrate to $0$ ($\sin \pi = \sin 2\pi = 0$), so $W_4 = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$. Same value, by a longer road.

e) Substitute $x = \frac{\pi}{2} - t$, $dx = -dt$: $x = 0$ gives $t = \frac{\pi}{2}$, $x = \frac{\pi}{2}$ gives $t = 0$, and $\cos\left(\frac{\pi}{2} - t\right) = \sin t$. So $\int_0^{\pi/2} \cos^5 x\,dx = \int_{\pi/2}^{0} \sin^5 t\,(-dt) = W_5 = \frac{8}{15}$. The odd-power method agrees: with $u = \sin x$, $\int_0^1 (1 - u^2)^2\,du = 1 - \frac{2}{3} + \frac{1}{5} = \frac{8}{15}$.

a) The radicand has the form $x^2 - a^2$ with $a = 1$ and $x \ge 1$: set $x = \sec\theta$ with $0 \le \theta < \frac{\pi}{2}$. Then $dx = \sec\theta\tan\theta\,d\theta$ and $\sqrt{x^2 - 1} = \sqrt{\tan^2\theta} = |\tan\theta| = \tan\theta$, since $\tan\theta \ge 0$ on that interval. Bounds: $\sec\theta = \sqrt 2$ gives $\theta = \frac{\pi}{4}$, and $\sec\theta = 2$ gives $\theta = \frac{\pi}{3}$. The integrand times $dx$ becomes $\frac{\sec\theta\tan\theta\,d\theta}{\sec^3\theta\tan\theta} = \cos^2\theta\,d\theta$.