- Check both the interval endpoints and the behavior of the integrand on the interval. - A finite interval does not automatically make an integral proper.

1. In a), the upper limit is infinite, so the integral is improper. 2. In b), the integrand is unbounded at the finite endpoint \(x=0\), so the integral is improper. 3. In c), both limits are finite and \(\ln x\) is continuous on \([1,3]\), so the integral is proper.

a) Improper because the interval is unbounded. b) Improper because the integrand is unbounded at \(x=0\). c) Proper.

- Find any point inside the stated interval where the integrand is not defined. - What happens when you approach that point from below compared with from above? - Which endpoint of the left-hand integral must approach the undefined point without reaching it?

1. The integrand is undefined and unbounded at \(x=0\), which lies inside the interval. 2. Split there: \(\lim_{a\to0^-}\int_{-1}^{a}\frac{1}{x^2}\,\text{d}x\) and \(\lim_{b\to0^+}\int_b^1\frac{1}{x^2}\,\text{d}x\). Both must exist finitely for the original integral to converge.

It is improper at \(x=0\). Check \(\lim_{a\to0^-}\int_{-1}^{a}\frac{1}{x^2}\,\text{d}x\) and \(\lim_{b\to0^+}\int_b^1\frac{1}{x^2}\,\text{d}x\) separately.

- The height of the graph approaching zero is not the same as the total area approaching a finite number. - Integrate to a variable upper bound \(b\). - Consider the limit of the logarithm as \(b\) grows.

1. Use a finite upper endpoint: \(\int_1^b\frac1x\,\text{d}x=\ln(b)\). 2. Since \(\ln(b)\to\infty\), the improper integral diverges even though \(f(x)\to0\).

No. \(\int_1^b\frac1x\,\text{d}x=\ln(b)\to\infty\), so the improper integral diverges.

- Picture two nonnegative curves, one entirely above the other. What can you conclude about their accumulated areas? - Can the same known area help in both directions of an inequality? Try to imagine an extreme counterexample.

1. To prove convergence by comparison with a known finite area, the nonnegative function must be no larger than the benchmark. 2. Condition A gives this finite upper bound, so it guarantees convergence. Condition B only bounds the unknown integral from below; the unknown function might still have infinite area.

A. \(0\le f(x)\le\frac{1}{x^2}\) for \(x\ge1\). It bounds the area above by a finite number; B gives only a lower bound.

- Replace each improper endpoint with a variable. - Apply the power rule for integration. - Evaluate the resulting limit.

1. a) \(\int_1^\infty4x^{-3}\,dx =\lim_{b\to\infty}\left[-\frac{2}{x^2}\right]_1^b =2\). 2. b) \(\int_0^1x^{-1/3}\,dx =\lim_{a\to0^+}\left[\frac{3}{2}x^{2/3}\right]_a^1 =\frac{3}{2}\). Both limits are finite, so both integrals converge.

a) Converges to \(2\). b) Converges to \(\frac{3}{2}\).

- Identify the improper endpoint in each integral. - Write each as a limit. - Examine the limiting behavior of the exponential or logarithm.

1. a) \(\int_0^\infty e^{-0.5x}\,dx =\lim_{b\to\infty}\left[-2e^{-0.5x}\right]_0^b=2\). It converges. 2. b) \(\int_1^2\frac{1}{x-1}\,dx =\lim_{a\to1^+}\left[\ln|x-1|\right]_a^2\). As \(a\to1^+\), \(\ln(a-1)\to-\infty\), so the integral diverges.

a) Converges to \(2\). b) Diverges; it has no finite value.

- Replace the infinite upper bound with a finite variable endpoint. - Find an antiderivative of the exponential function. - Use the end behavior of \(e^{-cx}\) when \(c>0\).

1. Write \(\int_0^\infty g(x)\,\text{d}x=\lim_{b\to\infty}\int_0^b1.5e^{-0.25x}\,\text{d}x\). 2. An antiderivative is \(-6e^{-0.25x}\). 3. Thus \(\lim_{b\to\infty}[-6e^{-0.25x}]_0^b=\lim_{b\to\infty}(6-6e^{-0.25b})=6\). 4. Because the limit is finite, the improper integral converges.

The improper integral converges to \(6\).

- Read the finite boundary of the shaded region from the graph. - Replace the infinite upper endpoint with a variable and use a limit. - Rewrite the reciprocal square as a negative power before integrating.

1. The graph shows that the left boundary is \(x=2\), so the area is \(A=\lim_{b\to\infty}\int_2^b8x^{-2}\,\text{d}x\). 2. An antiderivative is \(-\frac8x\). 3. Thus \(A=\lim_{b\to\infty}\left[-\frac8x\right]_2^b=\lim_{b\to\infty}\left(4-\frac8b\right)=4\). 4. Because the limit is finite, the region has finite area.

The region has finite area \(4\) square units.

- Replace infinity with a variable. - Apply the power rule. - Evaluate the limit of the reciprocal power.

1. Replace the infinite limit: \(\int_1^\infty3x^{-4}\,dx =\lim_{b\to\infty}\int_1^b3x^{-4}\,dx\). 2. An antiderivative is \(-x^{-3}\). 3. Therefore, \(\lim_{b\to\infty}\left[-\frac{1}{x^3}\right]_1^b =\lim_{b\to\infty}\left(1-\frac{1}{b^3}\right)=1\).

The integral converges to \(1\).

- What happens to the exponential term as \(x\) increases without bound? - Integrate the vertical distance between the curve and its asymptote, not the full height of the curve. - Replace the infinite upper endpoint by a variable before taking the limit.

1. As \(x\to\infty\), \(e^{3-x}\to0\). Therefore \(h(x)-5\to0\), so \(h(x)\to5\). The line \(y=5\) is a horizontal asymptote. 2. The vertical distance between the curve and the asymptote is \(h(x)-5=\frac12e^{3-x}\). Thus \(A=\int_0^\infty\frac12e^{3-x}\,\text{d}x\). 3. Therefore \(A=\lim_{b\to\infty}\left[-\frac12e^{3-x}\right]_0^b=\frac{e^3}{2}\), which is finite.

a) Since \(h(x)-5=\frac12e^{3-x}\to0\) as \(x\to\infty\), the horizontal asymptote is \(y=5\). b) \(\frac{e^3}{2}\) square units

- Identify the improper endpoint. - Use the power rule. - Evaluate the relevant limit.

1. a) \(\int_0^4x^{-1/2}\,dx =\lim_{a\to0^+}\left[2\sqrt{x}\right]_a^4=4\). It converges. 2. b) \(\int_4^\infty x^{-1/2}\,dx =\lim_{b\to\infty}(2\sqrt b-4)\), which diverges to infinity.

a) Converges to \(4\). b) Diverges.

- Replace the improper endpoint with a variable. - Apply the power rule carefully. - A convergent improper integral must have a finite limit.

1. a) \(\int_1^\infty2x^{-3}\,dx =\lim_{b\to\infty}\left[-\frac{1}{x^2}\right]_1^b=1\). 2. b) \(\int_0^12x^{-3}\,dx =\lim_{a\to0^+}\left[-\frac{1}{x^2}\right]_a^1\), which diverges to infinity.

a) Converges to \(1\). b) Diverges.

- Rewrite the reciprocal square root as a power. - Keep the lower endpoint \(u\) symbolic when applying the Fundamental Theorem of Calculus. - An unbounded function can still have a finite improper integral.

1. The area is \(A(u)=\int_u^4x^{-1/2}\,\text{d}x\). 2. By the power rule, \(A(u)=[2\sqrt{x}]_u^4=4-2\sqrt{u}\). 3. As \(u\to0^+\), \(2\sqrt{u}\to0\), so \(A(u)\to4\). The function is unbounded near \(0\), but the limiting area is finite.

\(A(u)=4-2\sqrt{u}\), and \(\lim_{u\to0^+}A(u)=4\).

- Integrate the exponential while accounting for the coefficient in its exponent. - Keep the upper limit \(k\) symbolic during endpoint evaluation. - Use the long-term behavior of exponential decay to evaluate the limit.

1. An antiderivative of \(4e^{-0.5x}\) is \(-8e^{-0.5x}\). 2. Therefore \(I(k)=[-8e^{-0.5x}]_0^k=8-8e^{-0.5k}\). 3. As \(k\to\infty\), \(e^{-0.5k}\to0\), so \(\lim_{k\to\infty}I(k)=8\).

\(I(k)=8-8e^{-0.5k}\), and \(\lim_{k\to\infty}I(k)=8\).

- Use the function's long-term value to identify the horizontal asymptote. - Replace the infinite upper endpoint with a variable. - Rewrite the denominator using exponent \(-2\) before integrating.

1. Since \(6/(x+1)^2\to0\), the horizontal asymptote is \(y=0\). 2. The area is \(\lim_{b\to\infty}\int_0^b\frac6{(x+1)^2}\,\text{d}x\). 3. An antiderivative is \(-6/(x+1)\), so the area is \(\lim_{b\to\infty}\left(6-\frac6{b+1}\right)=6\).

a) \(y=0\) b) \(6\) square units

- Read the finite vertical boundary from the displayed graph. - Rewrite the shifted power before applying the reverse power rule. - Replace the infinite endpoint with a variable and evaluate the limit.

1. The graph shows that the finite boundary is \(x=2\). Thus \(A=\lim_{b\to\infty}\int_2^b\frac4{(x+2)^{3/2}}\,\text{d}x\). 2. An antiderivative is \(-8/\sqrt{x+2}\). 3. Therefore \(A=\lim_{b\to\infty}\left(4-\frac8{\sqrt{b+2}}\right)=4\), so the region has finite area.

The region has finite area \(4\) square units.

- Rewrite the integrand with a negative exponent. - Replace infinity with a variable before applying the power rule. - Evaluate the limit of the reciprocal square-root term.

1. Replace the infinite upper endpoint with \(b\): \(\int_1^\infty x^{-3/2}\,dx=\lim_{b\to\infty}\int_1^b x^{-3/2}\,dx\). 2. An antiderivative is \(-2x^{-1/2}\). Therefore, \(\lim_{b\to\infty}[-2x^{-1/2}]_1^b=\lim_{b\to\infty}\left(2-\frac{2}{\sqrt{b}}\right)=2\). The improper integral converges.

The integral converges to \(2\).

- Compare \(\sqrt{x}\) with \(1\) on the given domain. - Compare positive reciprocals after comparing the denominators. - Evaluate the benchmark integral up to a variable upper bound, then decide what the comparison tells you.

1. For \(x\ge1\), \(1\le\sqrt{x}\). Hence \(\sqrt{x}+1\le2\sqrt{x}\), so \(g(x)\ge\frac{1}{2\sqrt{x}}\ge0\). 2. The benchmark integral is \(\int_1^b\frac{1}{2\sqrt{x}}\,\text{d}x=[\sqrt{x}]_1^b=\sqrt{b}-1\), which tends to infinity. 3. Because \(g\) is at least this large, its improper integral also diverges by the comparison test.

Diverges: \(g(x)\ge\frac{1}{2\sqrt{x}}\) for \(x\ge1\), and the lower benchmark has an infinite integral.

- Compare the denominator \(1+x\) with \(1\) on the integration interval. - The numerator is nonnegative and has a familiar convergent integral. - Set up the comparison of the two accumulated areas before sending the upper limit to infinity.

1. For \(x\ge0\), \(1+x\ge1\), so \(0\le\frac{e^{-x}}{1+x}\le e^{-x}\). 2. The comparison integral \(\int_0^\infty e^{-x}\,\text{d}x=1\) converges. 3. By the comparison test, the original nonnegative improper integral also converges.

Converges, since \(0\le\frac{e^{-x}}{1+x}\le e^{-x}\) on \([0,\infty)\) and \(\int_0^\infty e^{-x}\,\text{d}x=1\).

- Read the left edge of the shaded region using the x-axis labels. - Compare the curve heights, then consider how to justify the same inequality beyond the graph. - What happens to positive reciprocals as their denominators increase?

1. The shaded region starts at \(x=1\). For \(x\ge1\), \(x^2+1\ge x^2>0\), so \(0<f(x)\le h(x)\). 2. The benchmark is \(\int_1^\infty\frac{1}{x^2}\,\text{d}x=\lim_{b\to\infty}\left[-\frac1x\right]_1^b=1\). 3. Therefore \(\int_1^\infty f(x)\,\text{d}x\) converges, and its value lies between \(0\) and \(1\).

The shaded region starts at \(x=1\). The full improper integral converges because \(0<f(x)\le\frac1{x^2}\) for \(x\ge1\), and the upper comparison has area \(1\). Therefore the area under \(f\) is less than \(1\) square unit.

- Examine the difference between the function and the proposed asymptote. - Read the finite vertical boundary of the shaded region from the graph. - Express the unbounded area as an improper integral of the vertical difference.

1. The vertical difference is \(f(x)-(4-x)=2e^{0.5x-1}\), which approaches \(0\) as \(x\to-\infty\). Therefore \(y=4-x\) is a slant asymptote. 2. The graph shows that the finite boundary is \(x=2\). Since \(f\) lies above the asymptote, the area is \(A=\int_{-\infty}^{2}2e^{0.5x-1}\,\text{d}x\). 3. Therefore \(A=\lim_{a\to-\infty}[4e^{0.5x-1}]_a^2=4\).

a) The difference \(f(x)-(4-x)\) approaches \(0\) as \(x\to-\infty\). b) The area is finite and equals \(4\) square units.

- Subtract the lower function from the upper function. - Split the integral at the interior singularity and treat the two sides separately. - Rewrite the cube-root expression as a power when finding an antiderivative.

1. The vertical difference is \(f(x)-g(x)=\frac{1}{\sqrt[3]{x^2}}\). 2. Split the improper integral at \(x=0\): \(A=\int_{-1}^{0}\frac{1}{\sqrt[3]{x^2}}\,\text{d}x+\int_0^8\frac{1}{\sqrt[3]{x^2}}\,\text{d}x\). 3. Using the real cube root, an antiderivative on each side is \(3\sqrt[3]{x}\). The left integral converges to \(3\), and the right integral converges to \(6\). 4. Both one-sided improper integrals converge, so the total area is \(9\) square units.

The region has finite area \(9\) square units.

- Form the vertical difference between the two functions. - Identify the interior singularity before applying the Fundamental Theorem. - It is enough to show that one required one-sided improper integral diverges.

1. The vertical difference is \(f(x)-g(x)=\frac{2}{(x-1)^2}\). 2. The area integral is improper at \(x=1\). Consider the left part: \(\int_0^1\frac{2}{(x-1)^2}\,\text{d}x=\lim_{t\to1^-}\left[-\frac{2}{x-1}\right]_0^t\). 3. As \(t\to1^-\), this expression grows without bound. Therefore the left one-sided integral diverges, so the entire region has infinite area.

The region has no finite area; the improper integral diverges.

- Write the improper integral as a limit. - Apply the power rule after shifting by \(2\). - Set the resulting expression equal to \(4\).

1. An antiderivative is \(-\frac{k}{x+2}\). 2. Therefore, \(\int_0^\infty\frac{k}{(x+2)^2}\,dx =\lim_{b\to\infty}\left[-\frac{k}{x+2}\right]_0^b =\frac{k}{2}\). 3. Set \(\frac{k}{2}=4\). Thus, \(k=8\).

\(k=8\)

- Integrate the exponential function carefully. - Use \(e^{-2x}\to0\) as \(x\to\infty\). - Set the resulting limit equal to \(1\).

1. An antiderivative is \(-\frac{k-1}{2}e^{-2x}\). 2. Therefore, \(\int_0^\infty(k-1)e^{-2x}\,dx =\frac{k-1}{2}\). 3. Set \(\frac{k-1}{2}=1\). Thus, \(k=3\).

\(k=3\)

- Replace the infinite endpoint with a finite variable endpoint. - The denominator suggests using its inner quadratic expression when finding an antiderivative. - Evaluate the rational term's limit as the endpoint grows.

1. Write \(\int_0^\infty f(x)\,\text{d}x=\lim_{b\to\infty}\int_0^b\frac{10x}{(x^2+1)^2}\,\text{d}x\). 2. With \(u=x^2+1\), an antiderivative is \(-\frac{5}{x^2+1}\). 3. Therefore \(\lim_{b\to\infty}\left[-\frac{5}{x^2+1}\right]_0^b=5\). 4. The finite limit shows that the improper integral converges.

The improper integral converges to \(5\).

- Form the vertical difference between the function and the horizontal line. - Use integration by parts on the resulting linear-times-exponential integrand. - Compare the exponential decay with the linear factor when taking the limit.

1. The vertical difference is \(xe^{-0.5x}\), so the area is \(A=\int_0^\infty xe^{-0.5x}\,\text{d}x\). 2. Integration by parts gives \(\int xe^{-0.5x}\,\text{d}x=(-2x-4)e^{-0.5x}+C\). 3. Since \((2x+4)e^{-0.5x}\to0\), \(A=\lim_{b\to\infty}[(-2x-4)e^{-0.5x}]_0^b=4\).

The region has finite area \(4\) square units.

- Rewrite the integrand using the negative power \(r^{-2}\). - Use the same antiderivative for the finite and infinite upper bounds. - Compare the intervals and the limiting reciprocal term when explaining the two values.

1. An antiderivative is \(-\frac{24}{r}\). 2. For part a), \(\int_4^{12}\frac{24}{r^2}\,\text{d}r=24\left(\frac{1}{4}-\frac{1}{12}\right)=4\). 3. For part b), \(\int_4^\infty\frac{24}{r^2}\,\text{d}r=\lim_{b\to\infty}24\left(\frac{1}{4}-\frac{1}{b}\right)=6\). 4. The improper integral includes all of \([4,12]\) plus the positive tail beyond \(12\), so it is larger. It remains finite because \(1/b\to0\).

a) \(4\) b) \(6\) c) The improper integral includes an additional positive tail, but that tail has finite accumulation because the reciprocal term approaches \(0\).

- Replace the infinite upper limit with a finite variable endpoint. - Find the logarithmic antiderivative of the reciprocal linear function. - Decide whether the resulting logarithmic expression approaches a finite limit.

1. The area is \(A=\int_0^\infty\frac4{x+2}\,\text{d}x=\lim_{b\to\infty}\int_0^b\frac4{x+2}\,\text{d}x\). 2. An antiderivative is \(4\ln(x+2)\), so \(A=\lim_{b\to\infty}[4\ln(x+2)]_0^b\). 3. This equals \(\lim_{b\to\infty}(4\ln(b+2)-4\ln2)\), which grows without bound. 4. Therefore the improper integral diverges and the region has infinite area.

The improper integral diverges, so the region has infinite area.

- Identify the endpoint where the denominator is zero. - Replace that endpoint with a variable. - Use reverse chain rule to integrate.

1. The integrand is unbounded at \(x=2\), so use a one-sided limit. 2. An antiderivative is \(-\sqrt{4-2x}\). 3. Thus, \(\int_0^2\frac{1}{\sqrt{4-2x}}\,dx =\lim_{b\to2^-}\left[-\sqrt{4-2x}\right]_0^b=2\). The integral converges.

The integral converges to \(2\).

- Use the product rule and collect the coefficients multiplying \(e^x\). - Evaluate the proposed antiderivative at \(-1\) and at the variable lower limit. - Separate the polynomial–exponential product into terms such as \(k^2e^k\). - Compare polynomial growth with exponential decay as \(k\to-\infty\).

1. a) By the product rule, \(G'(x)=(2x-2)e^x+(x^2-2x-1)e^x=(x^2-3)e^x=g(x)\). 2. b) By the Fundamental Theorem of Calculus, \(I(k)=G(-1)-G(k)=\frac{2}{e}-(k^2-2k-1)e^k\). 3. c) As \(k\to-\infty\), each of \(k^2e^k\), \(ke^k\), and \(e^k\) approaches \(0\). Therefore, \(\lim_{k\to-\infty}I(k)=\frac{2}{e}\). 4. d) Exponential decay dominates polynomial growth, so the product \((k^2-2k-1)e^k\) approaches \(0\).

a) \(G'(x)=g(x)\) b) \(I(k)=\frac{2}{e}-(k^2-2k-1)e^k\) c) \(\lim_{k\to-\infty}I(k)=\frac{2}{e}\approx0.736\) d) The exponential factor approaches \(0\) faster than the polynomial factor grows.

- Replace the infinite upper endpoint with a variable when finding the total area. - Compare the finite accumulated area with the total by forming a ratio. - Translate \(90\%\) of the total into an equation involving \(k\).

1. The total area is \(\int_0^\infty24e^{-0.2x}\,\text{d}x=\lim_{b\to\infty}[-120e^{-0.2x}]_0^b=120\). 2. On \([0,5]\), the area is \(120(1-e^{-1})\). 3. As a percentage of the total, this is \(100(1-e^{-1})\%\approx63.21\%\). 4. For \(90\%\) of the total, solve \(120(1-e^{-0.2k})=0.9(120)\). This gives \(e^{-0.2k}=0.1\), so \(k=5\ln(10)\approx11.51\).

a) \(120\) square units b) \(120(1-e^{-1})\) square units, or approximately \(63.21\%\) of the total c) \(k=5\ln(10)\approx11.51\)

- Use exponential decay for the first limit. - Factor out the always-positive exponential term to find the zero and determine the sign. - Evaluate an antiderivative at the variable upper endpoint, then take the limit.

1. Both exponential terms approach \(0\), so \(\lim_{x\to\infty}f(x)=0\). 2. Factor \(f(x)=e^{-x}(3-e^{-x})\). Since \(e^{-x}>0\), the zero satisfies \(e^{-x}=3\), so \(x_0=-\ln3\). 3. For \(x\ge x_0\), the factor \(3-e^{-x}\ge0\), so the area is the signed integral. An antiderivative is \(F(x)=-3e^{-x}+\frac12e^{-2x}\). 4. Therefore \(A(k)=F(k)-F(-\ln3)=\frac92-3e^{-k}+\frac12e^{-2k}\), and \(\lim_{k\to\infty}A(k)=\frac92\).

a) \(0\) b) \(x_0=-\ln3\) c) \(A(k)=\frac92-3e^{-k}+\frac12e^{-2k}\), and \(\lim_{k\to\infty}A(k)=\frac92\)

- Use the signs of the numerator and denominator for part a). - The denominator contains an inner expression whose derivative is a multiple of the numerator. - Find the finite accumulated area first, then let the endpoint grow without bound.

1. For \(x>0\), the numerator \(4x\) and denominator \((x^2+1)^2\) are both positive, so \(g(x)>0\). 2. Using \(u=x^2+1\), an antiderivative is \(-\frac{2}{x^2+1}\). Thus \(A(z)=2-\frac{2}{z^2+1}\). 3. Taking \(z\to\infty\) gives \(\lim_{z\to\infty}A(z)=2\). Therefore the improper integral converges to \(2\).

a) \(g(x)>0\) for \(x>0\) b) \(A(z)=2-\frac{2}{z^2+1}\) c) The improper integral converges to \(2\).

- Treat \(K\) and \(r_0\) as positive constants. - Use the antiderivative of \(r^{-2}\) and evaluate the limit at infinity. - Write the tail beginning at \(R\) using the formula from part a. - Compare \(K/(2r_0)\) with \(K/r_0\).

1. a) Replace the infinite endpoint with \(b\): \(T(r_0)=\lim_{b\to\infty}\int_{r_0}^b Kr^{-2}\,dr=\lim_{b\to\infty}\left[-\frac{K}{r}\right]_{r_0}^b=\frac{K}{r_0}\). 2. b) The tail beginning at \(R\) is \(K/R\). Require \(\frac{K}{R}=\frac12\cdot\frac{K}{r_0}\). Since \(K>0\), this gives \(R=2r_0\). 3. c) Doubling the lower endpoint changes the tail area from \(K/r_0\) to \(K/(2r_0)\), so the area is halved.

a) \(T(r_0)=\frac{K}{r_0}\) b) \(R=2r_0\) c) Doubling the lower endpoint halves the remaining tail area.

- Use the same antiderivative for the accumulated area and the tail. - The two pieces must add to the total area for every \(b\ge3\). - Set the two area formulas equal to locate the halfway point.

1. An antiderivative is \(-18/x\). Thus \(A(b)=6-\frac{18}{b}\) and \(T(b)=\frac{18}{b}\). 2. The total area is \(A(b)+T(b)=6\), independent of \(b\). 3. Set \(6-\frac{18}{b}=\frac{18}{b}\). Then \(b=6\). At \(x=6\), the accumulated area and the remaining tail are each \(3\) square units.

a) \(A(b)=6-\frac{18}{b}\) and \(T(b)=\frac{18}{b}\) b) \(6\) square units c) \(b=6\); this point divides the total unbounded area into two equal parts.

- Identify where the integrand is undefined before applying the Fundamental Theorem. - A continuous antiderivative formula at the two endpoints is not enough when the interval contains a singularity. - Split an improper integral at an interior singularity and test both one-sided limits separately.

1. The interval \([4,6]\) does not contain the singularity, so \(\int_4^6\frac1{x-3}\,\text{d}x=[\ln|x-3|]_4^6=\ln3\). 2. The interval \([2,4]\) contains the singularity \(x=3\), so the Fundamental Theorem cannot be applied across the whole interval. 3. Splitting at \(x=3\), the left one-sided integral diverges to \(-\infty\) and the right one-sided integral diverges to \(+\infty\). Therefore the improper integral diverges; endpoint antiderivative values cannot bypass an interior singularity.

a) \(\ln3\) b) The interval contains the singularity \(x=3\); the split improper integral diverges.

- Evaluate each total improper integral using its natural antiderivative. - Equal total areas do not imply equal heights or equal partial areas. - For a tail area, use \(4\) as the lower limit and infinity as the upper limit.

1. \(\int_0^\infty2e^{-0.5x}\,\text{d}x=4\), and \(\int_0^\infty\frac4{(x+1)^2}\,\text{d}x=4\). 2. The functions can distribute area differently over the interval while still having the same total improper area. 3. The tail areas are \(4e^{-2}\approx0.541\) for \(f\) and \(\frac45=0.8\) for \(g\). Therefore \(g\) has the larger tail beyond \(x=4\).

a) Both total areas are \(4\) square units. b) Different nonnegative functions can distribute the same total area differently over an infinite interval. c) \(f\): \(4e^{-2}\approx0.541\); \(g\): \(\frac45=0.8\). The function \(g\) has the larger tail area.

- Keep the scale factor \(c\) symbolic while evaluating the improper integral. - The total-area formula immediately determines the value required in part b). - Write separate formulas for the accumulated area and the remaining tail. - The common positive factor \(c\) cancels when the two areas are set equal.

1. a) An antiderivative is \(-c/(x+1)\). Therefore, \(\int_0^\infty\frac{c}{(x+1)^2}\,dx=\lim_{b\to\infty}\left[-\frac{c}{x+1}\right]_0^b=c\). 2. b) To make the total area \(6\), choose \(c=6\). 3. c) The accumulated area through \(k\) is \(A(k)=c-\frac{c}{k+1}=\frac{ck}{k+1}\). The remaining tail is \(T(k)=\frac{c}{k+1}\). Set \(A(k)=T(k)\): \(\frac{ck}{k+1}=\frac{c}{k+1}\). Since \(c>0\), \(k=1\).

a) The total area is \(c\) square units. b) \(c=6\) c) \(k=1\)

- Check the signs of the two factors on the stated domain. - The derivative of the exponent is a constant multiple of the factor \(x\). - Obtain the total area by taking the limit of the finite accumulation formula. - Translate \(90\%\) of the total into an equation before taking logarithms.

1. For \(x\ge0\), both factors \(2x\) and \(e^{-0.5x^2}\) are nonnegative, so \(f(x)\ge0\). 2. An antiderivative is \(-2e^{-0.5x^2}\), so \(A(k)=2-2e^{-0.5k^2}\). 3. Letting \(k\to\infty\) gives the total area \(2\). 4. Set \(2-2e^{-0.5k^2}=1.8\). Then \(e^{-0.5k^2}=0.1\), so \(k=\sqrt{2\ln10}\approx2.146\).

a) \(f(x)\ge0\) for \(x\ge0\) b) \(A(k)=2-2e^{-0.5k^2}\) c) \(2\) square units d) \(k=\sqrt{2\ln10}\approx2.146\)

- Differentiate the polynomial-exponential product carefully. - For the tail, evaluate the antiderivative from a finite lower limit to a variable upper limit before taking the limit. - The area before \(x=2\) and the tail after it must add to the total area.

1. By the product rule, \(G'(x)=-\frac12(2x+2)e^{-x}+\frac12(x^2+2x+2)e^{-x}=\frac12x^2e^{-x}=g(x)\). 2. Since a polynomial times \(e^{-x}\) approaches \(0\), \(T(k)=\lim_{b\to\infty}[G(x)]_k^b=-G(k)=\frac12(k^2+2k+2)e^{-k}\). 3. Setting \(k=0\) gives the total area \(T(0)=1\). 4. The tail after \(x=2\) is \(T(2)=5e^{-2}\), so the area on \([0,2]\) is \(1-5e^{-2}\).

a) \(G'(x)=g(x)\) b) \(T(k)=\frac12(k^2+2k+2)e^{-k}\) c) \(1\) square unit d) On \([0,2]\): \(1-5e^{-2}\approx0.323\); remaining tail: \(5e^{-2}\approx0.677\)

- Use the product rule on the proposed linear-exponential form. - Match the coefficient of \(x\) and the constant term separately. - Check the sign of the function before interpreting the improper integral as area.

1. Differentiate the proposed form: \(F'(x)=(-ax+a-b)e^{-x}\). 2. Matching coefficients with \((2x+3)e^{-x}\) gives \(-a=2\) and \(a-b=3\), so \(a=-2\) and \(b=-5\). Thus \(F(x)=(-2x-5)e^{-x}\). 3. Since \(f(x)>0\) for \(x\ge0\), the improper integral is also the total area. Therefore \(\int_0^\infty f(x)\,\text{d}x=\lim_{k\to\infty}[F(x)]_0^k=5\).

a) \(F(x)=(-2x-5)e^{-x}\) b) \(5\)

- Compare the polynomial factor with exponential decay. - Differentiate the full polynomial-exponential product before matching coefficients. - Match quadratic, linear, and constant coefficients separately.

1. Exponential decay dominates polynomial growth, so \(x^2e^{-0.5x}\to0\). 2. Differentiating the proposed form gives \(F'(x)=\left(2ax+b-\frac12(ax^2+bx+c)\right)e^{-0.5x}\). 3. Matching coefficients gives \(a=-2\), \(b=-8\), and \(c=-16\), so \(F(x)=(-2x^2-8x-16)e^{-0.5x}\). 4. Since \(F(x)\to0\), \(\int_0^\infty f(x)\,\text{d}x=0-F(0)=16\).

a) \(0\) b) \(F(x)=(-2x^2-8x-16)e^{-0.5x}\) c) \(16\)

- Apply the chain rule to the logarithm in the proposed antiderivative. - Evaluate the antiderivative at \(b\) and \(0\). - Combine the linear and logarithmic terms before taking the limit.

1. \(F'(x)=4-\frac{4e^{2x}}{e^{2x}+1}=\frac4{e^{2x}+1}=f(x)\). 2. By the Fundamental Theorem, \(A(b)=4b-2\ln(e^{2b}+1)+2\ln2\). 3. Rewrite \(A(b)=2\ln\left(\frac{e^{2b}}{e^{2b}+1}\right)+\ln4\). The fraction approaches \(1\), so \(A(b)\to\ln4\).

a) \(F'(x)=f(x)\) b) \(A(b)=4b-2\ln(e^{2b}+1)+2\ln2\) c) \(\ln4=2\ln2\approx1.386\)

- Start with the universal bound on \(\sin^2(x)\). - The comparison function is a standard convergent power tail. - Distinguish an inequality between integrals from an equation between integrals.

1. Since \(0\le\sin^2(x)\le1\) and \(x^2>0\), dividing by \(x^2\) gives \(0\le f(x)\le b(x)\). 2. \(\int_1^\infty\frac1{x^2}\,\text{d}x=1\). 3. By comparison, \(\int_1^\infty f(x)\,\text{d}x\) converges and satisfies \(0\le\int_1^\infty f(x)\,\text{d}x\le1\). 4. The inequality gives bounds only; it does not show the two integrals are equal.

a) \(0\le f(x)\le b(x)\) b) \(1\) c) The integral of \(f\) converges and lies between \(0\) and \(1\). d) Comparison gives bounds, not an exact equality.

- Compare the positive denominators before taking reciprocals. - The antiderivative of \(1/x\) is logarithmic. - For divergence by comparison, the target function must be at least as large as a divergent positive comparison function.

1. For \(x\ge1\), \(x+1\le2x\). Taking reciprocals of positive quantities gives \(\frac1{x+1}\ge\frac1{2x}>0\). 2. \(\int_1^t\frac1{2x}\,\text{d}x=\frac12\ln t\), which grows without bound. Thus the improper integral of \(h\) diverges. 3. Because \(g\ge h>0\) and the smaller comparison integral diverges, \(\int_1^\infty g(x)\,\text{d}x\) diverges. 4. Approaching zero is not sufficient; a positive function can decrease too slowly for its accumulated area to be finite.

a) \(g(x)\ge h(x)>0\) b) \(\int_1^\infty h(x)\,\text{d}x\) diverges. c) \(\int_1^\infty g(x)\,\text{d}x\) diverges by lower comparison. d) A function can approach \(0\) too slowly for its improper integral to converge.

- Write a separate one-sided improper integral for each unbounded endpoint. - Rewrite the integrand with a rational exponent before applying the power rule. - Do not assume the same function must behave the same way near zero and at infinity.

1. Rewrite \(f(x)=3x^{-3/2}\). An antiderivative is \(-6x^{-1/2}\). 2. For \(A_1\), \(\lim_{a\to0^+}\int_a^1 3x^{-3/2}\,\text{d}x=\infty\), so \(A_1\) has infinite area. 3. For \(A_2\), \(\lim_{b\to\infty}\int_1^b3x^{-3/2}\,\text{d}x=6\), so \(A_2\) has finite area \(6\).

\(A_1\) has infinite area. \(A_2\) has finite area \(6\) square units.

- Use upper function minus lower function to get the vertical distance. - Keep the finite left endpoint symbolic in part b). - The final limit extends the region toward the shared singularity at \(x=0\).

1. The vertical distance is \(f(x)-g(x)=\frac6{x^2}\). Thus the area for \(x\ge1\) is \(\int_1^\infty\frac6{x^2}\,\text{d}x=6\). 2. With left boundary \(a\), \(A(a)=\int_a^\infty\frac6{x^2}\,\text{d}x=\frac6a\). 3. As \(a\to0^+\), \(A(a)\to\infty\). Therefore the region between the graphs over all \(x>0\) has infinite area.

a) \(6\) square units b) \(A(a)=\frac6a\) c) \(A(a)\to\infty\); the entire region on \(x>0\) has infinite area.

- Treat \(a\) as a positive constant when differentiating. - Evaluate the antiderivative at \(0\) and use exponential decay at infinity. - Apply the area formula to \(2a\) before comparing.

1. Treating \(a\) as constant, \(F_a'(x)=-a^2e^{-x/a}+a(x+a)e^{-x/a}=axe^{-x/a}=f_a(x)\). 2. Since \(f_a\ge0\) and \(F_a(x)\to0\), \(\int_0^\infty f_a(x)\,\text{d}x=0-F_a(0)=a^3\). 3. Replacing \(a\) by \(2a\) changes the area to \((2a)^3=8a^3\), so the factor is \(8\). 4. From \(a^3=64\) and \(a>0\), \(a=4\).

a) \(F_a'(x)=f_a(x)\) b) \(a^3\) square units c) A factor of \(8\) d) \(a=4\)

- Read the finite x-boundary from the dashed vertical segment in the graph. - Use the given vertical separation as the integrand for area between the curves. - Examine the logarithmic antiderivative as the upper endpoint grows.

1. The dashed vertical boundary is at \(x=2\). 2. Since the vertical separation is \(\frac{2}{x}\), the shaded area is \(A=\lim_{b\to\infty}\int_2^b\frac{2}{x}\,\text{d}x\). 3. Thus \(A=\lim_{b\to\infty}(2\ln b-2\ln2)\). 4. Because \(\ln b\to\infty\), the improper integral diverges, so the region has infinite area.

\(A=\lim_{b\to\infty}\int_2^b\frac{2}{x}\,\text{d}x\), and the integral diverges. The shaded region has infinite area.

- Compare the long-term behavior of the linear and exponential factors separately. - Replace the infinite upper endpoint with a variable before evaluating. - Use integration by parts or a linear-exponential antiderivative form.

1. The exponential factor \(e^{-0.5x}\) approaches \(0\) faster than the linear factor \(3x+2\) grows, so \(f(x)\to0\). 2. Write \(A=\lim_{b\to\infty}\int_0^b(3x+2)e^{-0.5x}\,\text{d}x\). 3. An antiderivative is \(-(6x+16)e^{-0.5x}\). Therefore \(A=\lim_{b\to\infty}[-(6x+16)e^{-0.5x}]_0^b=16\). 4. The finite limit shows that the improper integral converges.

a) Exponential decay dominates linear growth, so \(f(x)\to0\). b) The improper integral converges to \(16\).

- Inspect the integrand for points inside the interval where it is undefined. - An interior singularity divides one written integral into separate one-sided limit problems. - Convergence requires each required improper piece to have a finite limit on its own.

1. The integrand is undefined at the interior point \(x=2\), so the integral must be split there before any antiderivative values are compared. 2. The correct setup is \(\lim_{a\to2^-}\int_0^a\frac{1}{x-2}\,\text{d}x+\lim_{b\to2^+}\int_b^4\frac{1}{x-2}\,\text{d}x\). 3. For the left part, \([\ln|x-2|]_0^a=\ln|a-2|-\ln2\to-\infty\). 4. For the right part, \([\ln|x-2|]_b^4=\ln2-\ln|b-2|\to+\infty\). 5. Because the required one-sided improper integrals do not both converge to finite values, the original improper integral diverges. Opposite infinities cannot be canceled in the ordinary improper-integral definition.

The calculation crosses the singularity at \(x=2\) without splitting the integral. The correct setup is \(\lim_{a\to2^-}\int_0^a\frac{1}{x-2}\,\text{d}x+\lim_{b\to2^+}\int_b^4\frac{1}{x-2}\,\text{d}x\). The left part tends to \(-\infty\) and the right part to \(+\infty\), so the improper integral diverges.

- Substitute \(u=\ln x\). - Compare the result with a \(p\)-integral. - Treat \(a=1\) separately.

1. Substitute \(u=\ln x\): \(\int_e^\infty\frac{1}{x(\ln x)^a}\,dx =\int_1^\infty u^{-a}\,du\). 2. This \(p\)-integral converges exactly when \(a>1\). 3. For \(a>1\), \(A(a)=\lim_{b\to\infty}\left[\frac{u^{1-a}}{1-a}\right]_1^b =\frac{1}{a-1}\). 4. For \(a=3\), \(A=\frac{1}{2}\).

The area is finite exactly when \(a>1\), with \(A(a)=\frac{1}{a-1}\). For \(a=3\), \(A=\frac{1}{2}\) square unit.

- Apply the power rule and note the exceptional exponent. - Examine \(x^{1-n}\) as \(x\to\infty\). - Recall the antiderivative of \(\frac{1}{x}\).

1. For \(C=150\) and \(n=4\), \(\int_1^\infty150x^{-4}\,dx =\lim_{b\to\infty}\left[-\frac{50}{x^3}\right]_1^b=50\). 2. For \(n>1\), \(\int_1^\infty Cx^{-n}\,dx =\lim_{b\to\infty}\left[\frac{Cx^{1-n}}{1-n}\right]_1^b =\frac{C}{n-1}\). 3. When \(n=1\), an antiderivative is \(C\ln x\), which grows without bound as \(x\to\infty\).

a) \(50\) b) \(\frac{C}{n-1}\) c) It diverges because \(C\ln x\to\infty\).

- The exponential factor cannot create an x-intercept. - Compare exponential decay with linear growth for the limit at \(-\infty\). - Use the product rule to verify the proposed antiderivative. - Determine the sign of \(f\) from its factors before interpreting each integral as area.

1. The exponential factor is always positive, so the zero comes from \(4-2x=0\). Thus \(x=2\). 2. As \(x\to-\infty\), exponential decay dominates the linear factor, so \(f(x)\to0\). 3. By the product rule, \(F'(x)=-4e^{0.5x}+(16-4x)(0.5)e^{0.5x}=(4-2x)e^{0.5x}=f(x)\). 4. On \([0,2]\), \(f\ge0\), so the Quadrant I area is \(\int_0^2f(x)\,\text{d}x=F(2)-F(0)=8e-16\). 5. Since \(f(x)\ge0\) for \(x\le2\), the total area is \(\int_{-\infty}^2f(x)\,\text{d}x=F(2)-\lim_{a\to-\infty}F(a)=8e\).

a) \(x=2\) b) \(0\) c) \(F'(x)=f(x)\) d) \(8e-16\approx5.746\) square units e) \(8e\approx21.746\) square units

- Treat the case \(p=1\) separately because the power-rule antiderivative changes to a logarithm. - For \(p\ne1\), keep the exponent \(1-p\) symbolic after applying the power rule. - Decide whether \(b^{1-p}\) approaches \(0\) or grows without bound. - State both the convergence condition and the value in the convergent case.

1. First consider \(p\ne1\). For an upper limit \(b\), \(\int_1^b x^{-p}\,dx=\left[\frac{x^{1-p}}{1-p}\right]_1^b=\frac{b^{1-p}-1}{1-p}\). 2. If \(p>1\), then \(1-p<0\), so \(b^{1-p}\to0\). Therefore, \(I(p)=\frac{-1}{1-p}=\frac{1}{p-1}\). 3. If \(0<p<1\), then \(1-p>0\), so \(b^{1-p}\to\infty\). The integral diverges. 4. If \(p=1\), then \(\int_1^b\frac{1}{x}\,dx=\ln b\to\infty\). Thus, the integral converges exactly when \(p>1\).

The integral converges exactly for \(p>1\). In that case, \(I(p)=\frac{1}{p-1}\).

- Solve the curve-line intersection before splitting the region. - For general \(k\), express the intersection x-coordinate as a power of \(k\). - Separate the region into a rectangle and an area under the curve. - Relate the limiting region to the improper integral at \(x=0\).

1. For \(k=4\), solve \(x^{-2/3}=4\), giving \(x=\frac18\). The area is \(4\left(\frac18\right)+\int_{1/8}^1x^{-2/3}\,\text{d}x=\frac12+[3x^{1/3}]_{1/8}^1=2\). 2. In general, the intersection is \(x=k^{-3/2}\). The rectangle has area \(k^{-1/2}\), while \(\int_{k^{-3/2}}^1x^{-2/3}\,\text{d}x=3-\frac3{\sqrt{k}}\). 3. Hence \(A(k)=3-\frac2{\sqrt{k}}\). 4. As \(k\to\infty\), \(A(k)\to3\), the finite improper area under \(x^{-2/3}\) on \((0,1]\).

a) \(x=\frac18\), and the area is \(2\) square units. b) \(A(k)=3-\frac2{\sqrt{k}}\) c) \(\lim_{k\to\infty}A(k)=3\); this is the improper area under the curve on \((0,1]\).

- Use the product rule, then analyze the sign of the first derivative. - Compare exponential decay with linear growth for the horizontal asymptote. - Differentiate the proposed polynomial–exponential antiderivative to check it before using the improper limit. - Use both the location and height of each maximum to match the labeled curves.

1. a) Differentiate: \(h_k'(x)=k^2e^{-kx}(1-kx)\). Because \(k^2e^{-kx}>0\), the derivative is zero only when \(1-kx=0\), so \(x=1/k\). The derivative changes from positive to negative there, so the point is a maximum. 2. b) As \(x\to\infty\), exponential decay dominates the linear factor \(x\). Thus, \(k^2xe^{-kx}\to0\), and the x-axis is a horizontal asymptote. 3. c) An antiderivative is \(-(kx+1)e^{-kx}\). Therefore, \(\int_0^\infty h_k(x)\,dx=\lim_{b\to\infty}[-(kx+1)e^{-kx}]_0^b=1\). Every graph in the family encloses the same total area, \(1\) square unit, with the x-axis. 4. d) For \(k=1\), the maximum occurs at \(x=1\) and has height \(1/e\). For \(k=2\), it occurs at \(x=1/2\) and has height \(2/e\). Therefore, curve \(a\) corresponds to \(k=1\), and curve \(b\) corresponds to \(k=2\).

a) The maximum occurs at \(x=\frac1k\). b) \(h_k(x)\to0\), so the x-axis is a horizontal asymptote. c) \(\int_0^\infty h_k(x)\,dx=1\); the total area is independent of \(k\). d) Curve \(a\): \(k=1\); curve \(b\): \(k=2\)