Riemann Integration Questions and Answers

Answer: C

\(f\) is uniformly continuous on both \([0, 1)\) and \((0, \infty)\).

By the fundamental theorem of calculus, \(f'(x) = \sin^{2}(x^{2})\), so \(0 \le f'(x) \le 1\) for all \(x \ge 0\). By the mean value theorem,

\[ |f(x) - f(y)| \le |x - y| \]

so \(f\) is Lipschitz with constant 1 on the whole of \([0, \infty)\), and therefore uniformly continuous on every subset of it, including both \([0,1)\) and \((0, \infty)\).

The options denying uniform continuity on either set are ruled out by this bound; the growth of \(x^{2}\) inside the sine does not matter because the derivative stays bounded.

Answer: D

The true statement is \(\lim_{n\to\infty}\)\(\int_{[-\pi,\pi]\setminus[-\delta,\delta]}f(x)P_{n,\delta}(x)dx=0\).

Putting x = 0 in the hypothesis gives f(0) \(\gt\) f(0)/2, so f(0) \(\gt\) 0 and f is positive on (\(-\delta, \delta\)). Look at the base \(1 + \cos x - \cos\delta\):

• for |x| \(\lt\) \(\delta\), cos x \(\gt\) cos \(\delta\), so the base exceeds 1 and its powers grow without bound;
• for \(\delta\) \(\lt\) |x| ≤ \(\pi\), \(0 \lt \cos\delta - \cos x \lt 2\), so the base lies strictly between -1 and 1 and its powers tend to 0.

Outside [\(-\delta, \delta\)] the integrand is bounded by max|f| and tends to 0 at every point except the two endpoints, so by dominated convergence its integral tends to 0.

The other three integrals all contain an interval next to 0, for example [0, \(\delta\)/2], where f ≥ f(0)/2 \(\gt\) 0 and the base is at least a fixed number q \(\gt\) 1. Those integrals grow like \(q^{n}\) and do not tend to 0. This is the idea behind the Landau kernels used to prove the Weierstrass theorem.

Answer: C

The limit exists and equals 3.

Let \(g(x) = \cos(3x) + x^{2} + \sqrt{x + 4}\) and substitute \(x = \varepsilon t\):

\[ \begin{aligned} \frac{1}{\varepsilon}\int_{0}^{\infty}e^{-x/\varepsilon}g(x)\,dx \\ = \int_{0}^{\infty}e^{-t}g(\varepsilon t)\,dt \end{aligned} \]

As \(\varepsilon \to 0\), \(g(\varepsilon t) \to g(0)\) for each \(t\), and \(e^{-t}|g(\varepsilon t)|\) is dominated by an integrable function for \(\varepsilon \le 1\), so the limit is

\[ \begin{aligned} g(0)\int_{0}^{\infty}e^{-t}\,dt &= \cos 0 + 0 + \sqrt{4} \\ &= 1 + 2 = 3. \end{aligned} \]

The kernel \(\frac{1}{\varepsilon}e^{-x/\varepsilon}\) concentrates its unit mass at 0, so the answer is the value at 0, not 0 or \(\pi\).

Answer: A

The integral equals 0.

The integrand \(f(x) = x\,e^{-2x^{2}}\) is odd: \(f(-x) = -x\,e^{-2x^{2}} = -f(x)\). The integral of an odd function over limits symmetric about zero vanishes, because the area for negative x exactly cancels the area for positive x.

Directly, with \(u = -2x^{2}\):

\[ \int x\,e^{-2x^{2}}\,dx = -\tfrac{1}{4}e^{-2x^{2}} \]

which takes the same value at x = +2 and x = −2, so the difference is zero. The value 1/2 would come from integrating only from 0 to infinity and doubling, which ignores the sign of x.

Answer: D

The area is \(\frac{7}{3}e^{8}\).

The function is positive on \([1, 2]\), so the area is the definite integral. Substitute \(u = x^{3}\):

\[ \begin{aligned} du &= 3x^{2}\,dx \\ x^{5}\,dx &= x^{3}\cdot x^{2}\,dx = \tfrac{1}{3}u\,du \end{aligned} \]

The limits become \(u = 1\) and \(u = 8\). Integrating by parts, \(\int u e^{u}\,du = (u - 1)e^{u}\):

\[ \begin{aligned} A &= \frac{1}{3}\int_{1}^{8} u e^{u}\,du \\ &= \frac{1}{3}\Big[(u - 1)e^{u}\Big]_{1}^{8} \\ &= \frac{1}{3}\left(7e^{8} - 0\right) = \frac{7}{3}e^{8} \end{aligned} \]

The lower limit contributes nothing because \(u - 1 = 0\) at \(u = 1\), which is why the answer has no stray \(e\) term. The coefficients 1/3, 4/3 and 5/3 come from slips in the substitution or the by-parts step; at the upper limit the factor \(u - 1\) is 7.

Answer: A

The limit is \(\frac{e^{\pi} + 1}{2(e^{\pi} - 1)}\).

\(|\sin t|\) has period \(\pi\). Splitting \([0, \infty)\) into intervals \([k\pi, (k+1)\pi]\) and putting \(t = k\pi + s\) pulls out a factor \(e^{-k\pi}\) from each piece:

\[ \begin{aligned} I &= \sum_{k=0}^{\infty} e^{-k\pi} \int_{0}^{\pi} e^{-s}\sin s\,ds \\ \int_{0}^{\pi} e^{-s}\sin s\,ds &= \frac{1 + e^{-\pi}}{2} \end{aligned} \]

The geometric series sums to \(\frac{1}{1 - e^{-\pi}}\), so

\[ \begin{aligned} I &= \frac{1 + e^{-\pi}}{2(1 - e^{-\pi})} \\ &= \frac{e^{\pi} + 1}{2(e^{\pi} - 1)} \approx 0.545 \end{aligned} \]

Two bounds rule out the rest without any working. Since \(|\sin t| \leq 1\), \(I \leq \int_{0}^{\infty} e^{-t}\,dt = 1\), which excludes \(\frac{2(e^{\pi} + 1)}{e^{\pi} - 1} \approx 2.18\) and \(\frac{2(e^{\pi} - 1)}{e^{\pi} + 1} \approx 1.83\). Since \(|\sin t| \geq \sin t\), \(I \geq \int_{0}^{\infty} e^{-t}\sin t\,dt = \frac{1}{2}\), which excludes \(\frac{e^{\pi} - 1}{2(e^{\pi} + 1)} \approx 0.46\).

Answer: A

The area traced out is \(6\pi\).

Squaring \(y = \sqrt{6x - 5 - x^{2}}\) gives \((x - 3)^{2} + y^{2} = 4\) with \(y \geq 0\): the upper half of the circle of radius 2 centred at \((3,0)\), running from \((1,0)\) to \((5,0)\). Its distance from the origin ranges over \([1,5]\).

Each circle \(x^{2} + y^{2} = r^{2}\) with \(1 \leq r \leq 5\) meets the semi-circle at exactly one point, where \(x = \frac{r^{2} + 5}{6}\). A rotation about the origin keeps every point on its own circle, so that point sweeps an arc of angle \(\frac{\pi}{2}\) and length \(\frac{\pi}{2}r\). Adding up the arcs:

\[ \begin{aligned} \text{Area} &= \int_{1}^{5} \frac{\pi}{2}\,r\,dr \\ &= \frac{\pi}{2} \cdot \frac{25 - 1}{2} = 6\pi \end{aligned} \]

This is the area of a quarter of the annulus between radii 1 and 5. \(2\pi\) is the area of the half-disc under the semi-circle, and \(4\pi\) is the area of the whole disc of radius 2; neither accounts for the sweep.

Answer: B

Only one statement is true: \(f(x)\) is continuous at \(\frac{\pi}{2}\). The final key accepts this option alone.

At \(\frac{\pi}{2}\), which is irrational, \(f\left(\frac{\pi}{2}\right) = 0\). For every \(x\), \(|f(x)| \leq |\cos x|\), and \(\cos x \to 0\) as \(x \to \frac{\pi}{2}\), so \(f(x) \to 0\): continuous.

At 0, which is rational, \(f(0) = \cos 0 = 1\), but irrational points arbitrarily close to 0 give \(f = 0\). So \(f\) is not continuous at 0.

On \([0,1]\), \(\cos x \geq \cos 1 \approx 0.54 \gt 0\). Every subinterval contains rationals, where \(f \geq \cos 1\), and irrationals, where \(f = 0\). So every upper Darboux sum is at least \(\cos 1\) while every lower sum is 0, and \(f\) is not Riemann integrable on \([0,1]\). Both integrability statements fail, whatever value they claim.

Answer: B

P is NOT correct and Q is correct.

P: \(f(0) = 0\). The left derivative at 0 is \(c\). The right derivative is

\[ \begin{aligned} &\lim_{x \to 0^{+}} \frac{x g(x) + x^{2} h(x)}{x} \\ &= \lim_{x \to 0^{+}} \left(g(x) + x h(x)\right) \end{aligned} \]

Because \(h\) is bounded, \(x h(x) \to 0\), but \(g\) is only bounded, not continuous, so \(g(x)\) need not tend to \(g(0)\) or to anything. Knowing \(c = g(0) = h(0)\) does not make the two one-sided derivatives agree, so P fails.

Q: now \(f, g, h\) are smooth. For \(x \gt 0\), \(f'(x) = g + xg' + 2xh + x^{2}h'\), so \(f'(0) = g(0) = 0\), and \(f''(0) = 2g'(0) + 2h(0) \gt 0\). Since \(f\) is twice differentiable at 0, the left side must match (\(c = 0\), \(2d = f''(0)\)). With \(f'(0) = 0\) and \(f''(0) \gt 0\), the second derivative test gives a local minimum.

Answer: A

The volume is \(\frac{88}{5}\pi\).

The curves meet where \(x^{2} + 2 = 4x - 1\), that is \(x^{2} - 4x + 3 = 0\), so \(x = 1\) and \(x = 3\). Between them the line lies above the parabola (at \(x = 2\): 7 against 6), and both are positive, so the washer method gives

\[ \begin{aligned} V = \pi\int_{1}^{3} \Big[ &(4x - 1)^{2} \\ &- (x^{2} + 2)^{2} \Big]\, dx \end{aligned} \]

\[ \begin{aligned} &= \pi\int_{1}^{3} (-x^{4} + 12x^{2} - 8x - 3)\,dx \\ &= \pi\left(-\frac{242}{5} + 104 - 32 - 6\right) \\ &= \pi\left(-\frac{242}{5} + 66\right) = \frac{88}{5}\pi \end{aligned} \]

The \(x^{4}\) term is the one to watch: it contributes \(-\frac{3^{5} - 1}{5} = -\frac{242}{5}\). The other options, \(\frac{84}{5}\pi\), \(\frac{82}{5}\pi\) and \(\frac{92}{5}\pi\), do not match the integral; note also that revolving the region needs the difference of the squares, not the square of the difference.