MATH 141 practice midterm with full solutions (McGill)

$\Delta x = \frac{2 - 0}{4} = \frac{1}{2}$. 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 $R_4 = \frac{1}{2}\left(\frac{5}{4} + 3 + \frac{21}{4} + 8\right) = \frac{1}{2} \cdot \frac{35}{2} = \frac{35}{4}$.

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: $R_n = \frac{8}{n^3} \cdot \frac{n(n + 1)(2n + 1)}{6} + \frac{8}{n^2} \cdot \frac{n(n + 1)}{2} = \frac{4(n + 1)(2n + 1)}{3n^2} + \frac{4(n + 1)}{n} = \frac{4(n + 1)(5n + 1)}{3n^2}$.

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.

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.

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.

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.

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 $\int_0^1 \frac{x^5}{\sqrt{1 + x^3}}\,dx = \frac{1}{3}\int_1^2 \frac{u - 1}{\sqrt{u}}\,du = \frac{1}{3}\int_1^2 \left(u^{1/2} - u^{-1/2}\right)du = \frac{1}{3}\Big[\frac{2}{3}u^{3/2} - 2u^{1/2}\Big]_1^2$.

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 $2$.

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 $\int_0^{\pi/3} \frac{\sin^2 x}{\cos^2 x}\sin x\,dx = \int_1^{1/2} \frac{1 - u^2}{u^2}(-du) = \int_{1/2}^{1} \left(u^{-2} - 1\right)du = \Big[-\frac{1}{u} - u\Big]_{1/2}^{1} = (-2) - \left(-\frac{5}{2}\right) = \frac{1}{2}$.

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.

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 $\sin 2\pi = 0$.

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 $2\Big[\ln(u + 1)\Big]_1^3 = 2(\ln 4 - \ln 2) = 2\ln 2$.

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.

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 $(0, 1)$.

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 $6$.

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.

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 $\frac{4}{3}e^3 - \frac{4}{9}e^3 + \frac{4}{9} = \frac{8e^3 + 4}{9}$.

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 $\cos^2 x = 1 - \sin^2 x$: $\int \sin^n x\,dx = -\sin^{n-1} x\cos x + (n - 1)\int \sin^{n-2} x\,dx - (n - 1)\int \sin^n x\,dx$.

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.

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 $\pi \approx 3.14$).

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 $u = \sin x$, $\int_0^1 (1 - u^2)^2\,du = 1 - \frac{2}{3} + \frac{1}{5} = \frac{8}{15}$.