Tests edge cases, boundary conditions, and common misconceptions for integrals.

Question:

A student encounters the limit:

$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^2}{n^3} \sqrt{4 - \frac{k^2}{n^2}}$

(a) Express this limit as a definite integral. (b) Identify which type of Riemann sum (left, right, midpoint, or trapezoidal) this represents, or state if it is ambiguous. (c) Evaluate the definite integral.

Solution:

(a) Factor out $\frac{1}{n}$:

$\sum_{k=1}^{n} \frac{k^2}{n^3}\sqrt{4 - \frac{k^2}{n^2}} = \sum_{k=1}^{n} \frac{1}{n} \cdot \left(\frac{k}{n}\right)^2 \sqrt{4 - \left(\frac{k}{n}\right)^2}$

With $\Delta x = \frac{1}{n}$ and $x_k = \frac{k}{n}$ (right endpoint of the $k$-th subinterval on $[0, 1]$):

$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n} \cdot x_k^2 \sqrt{4 - x_k^2} = \int_0^1 x^2\sqrt{4 - x^2}\,dx$

(b) Since $x_k = \frac{k}{n}$ uses the right endpoint of each subinterval, this is a right Riemann sum.

(c) Use the substitution $x = 2\sin\theta$, $dx = 2\cos\theta\,d\theta$:

$\int_0^1 x^2\sqrt{4-x^2}\,dx = \int_0^{\arcsin(1/2)} 4\sin^2\theta \cdot 2\cos\theta \cdot 2\cos\theta\,d\theta$

$= 16\int_0^{\pi/6} \sin^2\theta\cos^2\theta\,d\theta = 4\int_0^{\pi/6} \sin^2(2\theta)\,d\theta$

Using $\sin^2(2\theta) = \frac{1 - \cos(4\theta)}{2}$:

$= 4\int_0^{\pi/6}\frac{1 - \cos(4\theta)}{2}\,d\theta = 2\left[\theta - \frac{\sin(4\theta)}{4}\right]_0^{\pi/6}$

$= 2\left(\frac{\pi}{6} - \frac{\sin(2\pi/3)}{4}\right) = 2\left(\frac{\pi}{6} - \frac{\sqrt{3}/2}{4}\right) = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$

Question:

Evaluate $\displaystyle\int e^{2x}\sin(3x)\,dx$.

A student chooses $u = \sin(3x)$ and $dv = e^{2x}\,dx$ for the first application of integration by parts. Show that this choice works but leads to a longer computation than choosing $u = e^{2x}$. Evaluate the integral completely using the more efficient choice and explain why “LIATE” alone does not settle this choice.

Solution:

LIATE ranks: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Both $e^{2x}$ (Exponential) and $\sin(3x)$ (Trigonometric) are in the LIATE list. Trigonometric comes before Exponential, so LIATE suggests $u = \sin(3x)$. But for products of exponentials and trig functions, either choice works, and the key is to apply integration by parts twice and solve algebraically.

Using $u = e^{2x}$, $dv = \sin(3x)\,dx$:

$du = 2e^{2x}\,dx$, $v = -\frac{1}{3}\cos(3x)$.

$\int e^{2x}\sin(3x)\,dx = -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{3}\int e^{2x}\cos(3x)\,dx$

Apply parts again to the remaining integral. Let $u = e^{2x}$, $dv = \cos(3x)\,dx$:

$du = 2e^{2x}\,dx$, $v = \frac{1}{3}\sin(3x)$.

$\int e^{2x}\cos(3x)\,dx = \frac{1}{3}e^{2x}\sin(3x) - \frac{2}{3}\int e^{2x}\sin(3x)\,dx$

Substituting back:

$I = -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{3}\left(\frac{1}{3}e^{2x}\sin(3x) - \frac{2}{3}I\right)$

$I = -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{9}e^{2x}\sin(3x) - \frac{4}{9}I$

$\frac{13}{9}I = e^{2x}\left(-\frac{1}{3}\cos(3x) + \frac{2}{9}\sin(3x)\right)$

$I = \frac{e^{2x}}{13}(2\sin(3x) - 3\cos(3x)) + C$

The misconception: students sometimes stop after one application of parts, or they choose a different assignment for the second integration by parts (switching $u$ and $dv$), which creates a circular argument instead of solving for $I$.

Question:

Find the total area of the region bounded by $y = x^3 - 4x$ and $y = x^2 - 4$.

A student computes $\displaystyle\int_{-2}^{2}\left[(x^3 - 4x) - (x^2 - 4)\right]\,dx$ and gets the wrong answer. Explain the error and compute the correct total area.

Solution:

First find intersection points: $x^3 - 4x = x^2 - 4 \implies x^3 - x^2 - 4x + 4 = 0$.

By inspection $x = 1$ is a root: $1 - 1 - 4 + 4 = 0$. Factoring:

$(x-1)(x^2 - 4) = (x-1)(x-2)(x+2) = 0$

Intersection points: $x = -2, 1, 2$.

The student”s error: they integrated from $-2$ to $2$ without accounting for the curve crossing at $x = 1$. On $[-2, 1]$We must determine which curve is on top; on $[1, 2]$The other may be on top.

Test point $x = 0$: $x^3 - 4x = 0$, $x^2 - 4 = -4$. So $x^3 - 4x \gt x^2 - 4$ on $[-2, 1]$.

Test point $x = 1.5$: $x^3 - 4x = 3.375 - 6 = -2.625$, $x^2 - 4 = 2.25 - 4 = -1.75$. So $x^2 - 4 \gt x^3 - 4x$ on $[1, 2]$.

\text{Total area = \int_{-2}^{1}\left[(x^3 - 4x) - (x^2 - 4)\right]\,dx + \int_1^{2}\left[(x^2 - 4) - (x^3 - 4x)\right]\,dx

$= \int_{-2}^{1}(x^3 - x^2 - 4x + 4)\,dx + \int_1^{2}(-x^3 + x^2 + 4x - 4)\,dx$

First integral:

$\left[\frac{x^4}{4} - \frac{x^3}{3} - 2x^2 + 4x\right]_{-2}^{1} = \left(\frac{1}{4} - \frac{1}{3} - 2 + 4\right) - \left(4 + \frac{8}{3} - 8 - 8\right)$

$= \frac{25}{12} - \left(-\frac{20}{3}\right) = \frac{25}{12} + \frac{80}{12} = \frac{105}{12} = \frac{35}{4}$

Second integral:

$\left[-\frac{x^4}{4} + \frac{x^3}{3} + 2x^2 - 4x\right]_1^{2} = \left(-4 + \frac{8}{3} + 8 - 8\right) - \left(-\frac{1}{4} + \frac{1}{3} + 2 - 4\right)$

$= -\frac{4}{3} - \left(-\frac{23}{12}\right) = -\frac{16}{12} + \frac{23}{12} = \frac{7}{12}$

\text{Total area = \frac{35}{4} + \frac{7}{12} = \frac{105 + 7}{12} = \frac{112}{12} = \frac{28}{3}

The student’s integral gives $\displaystyle\int_{-2}^{2}(x^3 - x^2 - 4x + 4)\,dx = \frac{28}{3}$ only because the areas happen to be positive in both subintervals when separated correctly. The student’s single integral from $-2$ to $2$ actually evaluates to:

$\int_{-2}^{2}(x^3 - x^2 - 4x + 4)\,dx = \left[\frac{x^4}{4} - \frac{x^3}{3} - 2x^2 + 4x\right]_{-2}^{2}$

This gives $\frac{28}{3}$ here too, but this is coincidental. The fundamental error is not splitting at intersection points, which would give wrong answers .

Tests synthesis of integrals with other topics.

Question:

Let $R$ be the region bounded by $y = \sqrt{x}$, $y = 0$And $x = 4$.

(a) Find the volume generated when $R$ is revolved about the line $x = 6$ using the shell method. (b) Verify your answer using the washer method. (c) A student claims the volume about $x = 6$ should be the same as the volume about the $y$-axis because “it’s just a translation.” Explain why this is false, and compute the volume about the $y$-axis for comparison.

Solution:

(a) Shell method (parallel to axis of revolution): use horizontal shells.

A shell at height $y$ has radius $r = 6 - x = 6 - y^2$ and height $h = dy$ (thin strip). For shells, we integrate along the axis perpendicular to the axis of revolution.

Since we revolve about $x = 6$ (vertical line), shells are vertical: radius $= 6 - x$Height $= \sqrt{x} - 0 = \sqrt{x}$.

$V = 2\pi\int_0^4 (6-x)\sqrt{x}\,dx = 2\pi\int_0^4 (6x^{1/2} - x^{3/2})\,dx$

$= 2\pi\left[4x^{3/2} - \frac{2}{5}x^{5/2}\right]_0^4 = 2\pi\left(4 \cdot 8 - \frac{2}{5} \cdot 32\right) = 2\pi\left(32 - \frac{64}{5}\right) = 2\pi \cdot \frac{96}{5} = \frac{192\pi}{5}$

(b) Washer method: washers perpendicular to $x = 6$So we integrate with respect to $y$.

Outer radius: $R = 6 - 0 = 6$ (from $x = 6$ to the $y$-axis). More precisely, for washers perpendicular to the axis $x = 6$: at height $y$The region extends from $x = y^2$ to $x = 4$. Revolved about $x = 6$:

• Outer radius: $6 - y^2$ (from axis to the left edge of region at $x = y^2$)
• Inner radius: $6 - 4 = 2$ (from axis to the right edge of region at $x = 4$)

$V = \pi\int_0^2\left[(6-y^2)^2 - 2^2\right]\,dy = \pi\int_0^2(36 - 12y^2 + y^4 - 4)\,dy$

$= \pi\int_0^2(32 - 12y^2 + y^4)\,dy = \pi\left[32y - 4y^3 + \frac{y^5}{5}\right]_0^2$

$= \pi\left(64 - 32 + \frac{32}{5}\right) = \pi\left(32 + \frac{32}{5}\right) = \frac{192\pi}{5}$

(c) Volume about the $y$-axis (disk method):

$V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi \cdot 8 = 8\pi$

The volumes are different ($\frac{192\pi}{5} \approx 120.6$ vs $8\pi \approx 25.1$). Revolving about a different axis changes the radius of every point, so the volume changes. The student’s “translation” argument is wrong because the region itself does not translate — only the axis does, which changes the distance from every point to the axis.

Question:

For what values of $p$ does the improper integral $\displaystyle\int_0^{\infty} \frac{x^{p-1}}{1 + x}\,dx$ converge?

Hint: Split the integral at $x = 1$ and analyze the behavior of each part separately using comparison with $p$-integrals.

Solution:

Split at $x = 1$: $\displaystyle\int_0^{\infty} \frac{x^{p-1}}{1+x}\,dx = \int_0^1 \frac{x^{p-1}}{1+x}\,dx + \int_1^{\infty} \frac{x^{p-1}}{1+x}\,dx$

Near $x = 0$ (first integral): When $x \in (0, 1)$, $\frac{1}{2} \lt \frac{1}{1+x} \lt 1$So:

$\frac{x^{p-1}}{2} \lt \frac{x^{p-1}}{1+x} \lt x^{p-1}$

By comparison with $\displaystyle\int_0^1 x^{p-1}\,dx$ (converges iff $p - 1 > -1$I.e., $p > 0$):

The first integral converges iff $p > 0$.

As $x \to \infty$ (second integral): When $x > 1$, $\frac{1}{2} \lt \frac{1}{1+x} \lt \frac{1}{x}$So:

$\frac{x^{p-1}}{2} \lt \frac{x^{p-1}}{1+x} \lt x^{p-2}$

By comparison with $\displaystyle\int_1^{\infty} x^{p-2}\,dx$ (converges iff $p - 2 \lt -1$I.e., $p \lt 1$):

The second integral converges iff $p \lt 1$.

Conclusion: The full integral converges iff both parts converge, i.e., $0 \lt p \lt 1$.

Note: this integral equals $B(p, 1-p) = \pi \csc(\pi p)$ (the beta function), which has the same domain of convergence.

Question:

Let f(x) = \begin{cases} 2x & \text{if 0 \leq x \lt 2 \\ 8 - 2x & \text{if 2 \leq x \leq 4 \end{cases}

Let $\displaystyle F(x) = \int_0^x f(t)\,dt$.

(a) Find and graph $F(x)$. (b) Is $F$ differentiable at $x = 2$? Compute $F'(2)$ from the definition of the derivative and explain. (c) A student claims that since $f$ has a corner at $x = 2$, $F$ must also have a corner at $x = 2$. Is this correct?

Solution:

(a) For $0 \leq x \lt 2$:

$F(x) = \int_0^x 2t\,dt = t^2\Big|_0^x = x^2$

For $2 \leq x \leq 4$:

$F(x) = \int_0^2 2t\,dt + \int_2^x (8 - 2t)\,dt = 4 + \left[8t - t^2\right]_2^x = 4 + (8x - x^2 - 12) = -x^2 + 8x - 8$

So F(x) = \begin{cases} x^2 & \text{if 0 \leq x \lt 2 \\ -x^2 + 8x - 8 & \text{if 2 \leq x \leq 4 \end{cases}

(Note: $4 + (8x - x^2 - 12) = -x^2 + 8x - 8$Not $(x-4)^2 = x^2 - 8x + 16$. The two expressions differ.)

(b) Check differentiability at $x = 2$:

$F(2) = -4 + 16 - 8 = 4$

Left-hand derivative: $F'_{-}(2) = \lim_{h \to 0^-}\frac{F(2+h) - F(2)}{h} = \lim_{h \to 0^-}\frac{(2+h)^2 - 4}{h} = \lim_{h \to 0^-}\frac{4h + h^2}{h} = 4$.

Right-hand derivative: $F'_{+}(2) = \lim_{h \to 0^+}\frac{F(2+h) - F(2)}{h} = \lim_{h \to 0^+}\frac{-(2+h)^2 + 8(2+h) - 8 - 4}{h}$

$= \lim_{h \to 0^+}\frac{-4 - 4h - h^2 + 16 + 8h - 12}{h} = \lim_{h \to 0^+}\frac{-h^2 + 4h}{h} = \lim_{h \to 0^+}(-h + 4) = 4$.

Since $4 = 4$, $F$ is differentiable at $x = 2$With $F'(2) = 4 = f(2)$.

(c) The student is incorrect. Although $f$ has a corner at $x = 2$ ($f$ changes from slope $+2$ to slope $-2$), $f$ is continuous at $x = 2$ ($f(2^-) = 4 = f(2^+)$). By FTC part 1, since $f$ is continuous at $x = 2$, $F$ is differentiable at $x = 2$ with $F'(2) = f(2) = 4$. Integration “smooths” the corner: $F$ is continuously differentiable even though $f$ is not.

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

• Confusing terminology or concepts that appear similar but have distinct meanings.
• Overlooking key assumptions or boundary conditions that limit applicability.