Free Math Worksheets

- 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 \(1\).

- Find the limiting value of the function. - Integrate the difference between the graph and the asymptote. - Evaluate the improper limit.

1. Since \(e^{3-x}\to0\) as \(x\to\infty\), the horizontal asymptote is \(y=5\). 2. The vertical difference is \(\frac{1}{2}e^{3-x}\), so \(A=\int_0^\infty\frac{1}{2}e^{3-x}\,dx\). 3. Therefore, \(A=\lim_{b\to\infty}\left[-\frac{1}{2}e^{3-x}\right]_0^b =\frac{e^3}{2}\).

a) \(y=5\) b) The area is finite and equals \(\frac{e^3}{2}\) 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 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 limit with a variable. - Find an antiderivative of the exponential function. - Use the fact that \(e^{-cx}\to0\) for \(c>0\).

1. Write the area as \(A=\lim_{b\to\infty}\int_0^b1.5e^{-0.25x}\,dx\). 2. An antiderivative is \(-6e^{-0.25x}\). 3. Therefore, \(A=\lim_{b\to\infty}\left[-6e^{-0.25x}\right]_0^b =\lim_{b\to\infty}\left(6-6e^{-0.25b}\right)=6\). The finite limit shows that the integral converges.

The improper integral converges, and the area is \(6\) square units.

- Rewrite the function with a negative exponent. - Use a variable upper limit. - Evaluate the limit as the upper boundary increases.

1. The area is \(A=\lim_{b\to\infty}\int_2^b8x^{-2}\,dx\). 2. An antiderivative is \(-\frac{8}{x}\). 3. Thus, \(A=\lim_{b\to\infty}\left[-\frac{8}{x}\right]_2^b =\lim_{b\to\infty}\left(4-\frac{8}{b}\right)=4\). The region has finite area.

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

- Examine the difference between the function and the proposed asymptote. - Express the unbounded area as an improper integral. - Use the limiting behavior of an exponential with a negative exponent.

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 lies above the asymptote, so \(A=\int_{-\infty}^{2}2e^{0.5x-1}\,dx\). 3. Thus, \(A=\lim_{a\to-\infty}\left[4e^{0.5x-1}\right]_a^2 =4\).

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. - Check convergence from both sides separately.

1. The vertical difference is \(f(x)-g(x)=\frac{1}{\sqrt[3]{x^2}}\). 2. Split the improper integral at \(x=0\): \(A=\int_{-1}^{0}\frac{1}{\sqrt[3]{x^2}}\,dx+\int_0^8\frac{1}{\sqrt[3]{x^2}}\,dx\). 3. Using the real cube root, \(\int x^{-2/3}\,dx=3\sqrt[3]{x}\). The left integral converges to \(3\), and the right integral converges to \(6\). 4. Both parts converge, so \(A=3+6=9\).

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

- Form the difference between the functions. - Identify the interior vertical asymptote. - It is enough to show that one of the one-sided integrals diverges.

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

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, \(k=3\).

\(k=3\)

- Substitute using the denominator. - Replace the infinite limit with a variable. - Evaluate the rational expression as the variable grows.

1. Write \(A=\int_0^\infty\frac{10x}{(x^2+1)^2}\,dx\). 2. With \(u=x^2+1\), an antiderivative is \(-\frac{5}{x^2+1}\). 3. Therefore, \(A=\lim_{b\to\infty}\left[-\frac{5}{x^2+1}\right]_0^b=5\). The area is finite.

The area is finite and equals \(5\) square units.

- Form the vertical difference. - Use integration by parts. - Compare exponential decay with linear growth.

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

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

- Convert all distances to meters. - Integrate the inverse-square function. - Interpret an infinite upper limit as moving arbitrarily far away.

1. Convert distances: \(R=4\times10^6\,\text{m}\) and \(r_2=12\times10^6\,\text{m}\). 2. Since an antiderivative is \(-\frac{k}{r}\), \(W=k\left(\frac{1}{R}-\frac{1}{r_2}\right) =4\times10^6\,\text{J}=4\,\text{MJ}\). 3. For an infinite final distance, \(\int_R^\infty\frac{k}{r^2}\,dr =\frac{k}{R} =6\times10^6\,\text{J} =6\,\text{MJ}\). 4. This is the idealized work required to move the satellite infinitely far from the planet.

a) \(W=4\,\text{MJ}\) b) \(6\,\text{MJ}\); this is the idealized escape work to move the satellite infinitely far away.

- 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.

- Express the area as an improper integral. - Find the logarithmic antiderivative. - Examine the logarithm as the upper limit grows.

1. The area would be \(A=\int_0^\infty\frac{4}{x+2}\,dx\). 2. With upper limit \(b\), \(\int_0^b\frac{4}{x+2}\,dx =4\ln(b+2)-4\ln2\). 3. This expression approaches infinity as \(b\to\infty\), so the area is not finite.

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 \(2\).

- Rewrite the function as a power. - Evaluate the definite integral in terms of \(u\). - Take the one-sided limit.

1. The area is \(A(u)=\int_u^4x^{-1/2}\,dx\). 2. Therefore, \(A(u)=\left[2\sqrt{x}\right]_u^4 =4-2\sqrt{u}\). 3. As \(u\to0^+\), \(A(u)\to4\).

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

- Integrate the exponential function. - Remember that \(e^0=1\). - Use exponential decay to evaluate the limit.

1. The area is \(I(k)=\int_0^k4e^{-0.5x}\,dx\). 2. An antiderivative is \(-8e^{-0.5x}\), so \(I(k)=8-8e^{-0.5k}\). 3. As \(k\to\infty\), the exponential term approaches \(0\). Therefore, \(\lim_{k\to\infty}I(k)=8\).

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

- Use exponential decay for the first limit. - Factor out \(e^{-x}\) to find the zero. - Evaluate an antiderivative at the moving upper limit. - Let the exponential terms approach zero.

1. Both exponential terms approach \(0\), so \(\lim_{x\to\infty}f(x)=0\). 2. Factor: \(e^{-x}(3-e^{-x})=0\). Thus, \(e^{-x}=3\) and \(x_0=-\ln3\). 3. An antiderivative is \(F(x)=-3e^{-x}+\frac{1}{2}e^{-2x}\). 4. Therefore, \(A(k)=F(k)-F(-\ln3) =\frac{9}{2}-3e^{-k}+\frac{1}{2}e^{-2k}\). 5. Hence, \(\lim_{k\to\infty}A(k)=\frac{9}{2}\).

a) \(0\) b) \(x_0=-\ln3\) c) \(A(k)=\frac{9}{2}-3e^{-k}+\frac{1}{2}e^{-2k}\), with limit \(\frac{9}{2}\)

- Analyze the signs of numerator and denominator. - Substitute using \(x^2+1\). - Take the limit of the area formula.

1. For \(x>0\), both \(4x\) and \((x^2+1)^2\) are positive, so \(g(x)>0\). 2. With \(u=x^2+1\), an antiderivative is \(-\frac{2}{x^2+1}\). 3. Therefore, \(A(z)=\int_0^zg(x)\,dx =2-\frac{2}{z^2+1}\). 4. Taking the limit, \(\lim_{z\to\infty}A(z)=2\). Thus, the unbounded region has finite area.

a) \(g(x)>0\) for \(x>0\). b) \(A(z)=2-\frac{2}{z^2+1}\) c) The total area is \(2\) square units.

- Find the tangent and its \(x\)-intercept. - Subtract the area under the tangent from the area under the curve. - Use exponential decay for the limit.

1. Since \(f'(0)=-1\), the tangent is \(t(x)=-x+2\), which crosses the \(x\)-axis at \(x=2\). 2. For \(k>2\), \(A(k)=\int_0^kf(x)\,dx-\int_0^2t(x)\,dx\). 3. Thus, \(A(k)=\left[-4e^{-0.5x}\right]_0^k-2 =2-4e^{-0.5k}\). 4. For \(k=6\), \(A=2-4e^{-3}\approx1.801\). As \(k\to\infty\), \(A(k)\to2\).

a) \(t(x)=-x+2\) b) \(A=2-4e^{-3}\approx1.801\) square units c) \(A(k)=2-4e^{-0.5k}\), with limit \(2\)

- Find the tangent and where it meets the \(x\)-axis. - Split the lower boundary between the tangent and the axis. - Evaluate the moving-boundary area formula.

1. Since \(f(1)=1\) and \(f'(1)=-1\), the tangent is \(t(x)=-x+2\). It crosses the \(x\)-axis at \(x=2\). 2. For \(k\ge2\), \(A(k)=\int_1^k\frac{4}{(x+1)^2}\,dx-\int_1^2(-x+2)\,dx\). 3. Therefore, \(A(k)=\frac{3}{2}-\frac{4}{k+1}\). 4. For \(k=3\), \(A=\frac{1}{2}\). As \(k\to\infty\), \(A(k)\to\frac{3}{2}\).

a) \(t(x)=-x+2\) b) \(A=\frac{1}{2}\) square unit c) The limiting area is \(\frac{3}{2}\) square units.

- Differentiate with the product rule. - Match the coefficient of \(x\) and the constant term. - Use exponential decay at infinity.

1. Differentiate the proposed form: \(F'(x)=(-ax+a-b)e^{-x}\). 2. Match coefficients with \((2x+3)e^{-x}\): \(-a=2\) and \(a-b=3\). Thus, \(a=-2\) and \(b=-5\). 3. Therefore, \(F(x)=(-2x-5)e^{-x}\). 4. The area is \(\int_0^\infty f(x)\,dx =\lim_{k\to\infty}[F(x)]_0^k =0-(-5)=5\).

a) \(F(x)=(-2x-5)e^{-x}\) b) The area is \(5\) square units.

- Apply the chain rule to the logarithm. - Combine logarithms before taking the limit. - Divide numerator and denominator by the dominant exponential.

1. Differentiate: \(F'(x)=4-\frac{4e^{2x}}{e^{2x}+1} =\frac{4}{e^{2x}+1}=f(x)\). 2. Therefore, \(A(b)=F(b)-F(0) =4b-2\ln(e^{2b}+1)+2\ln2\). 3. Rewrite: \(A(b)=2\ln\left(\frac{e^{2b}}{e^{2b}+1}\right)+\ln4\). 4. The fraction approaches \(1\), so \(\lim_{b\to\infty}A(b)=\ln4\).

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

- Write a separate improper integral for each region. - Rewrite the integrand using a rational exponent. - Replace each improper endpoint with a variable and examine the corresponding limit.

1. Rewrite the function as \(f(x)=3x^{-3/2}\), with antiderivative \(-6x^{-1/2}=-\frac{6}{\sqrt{x}}\). 2. For \(A_1\), \(\lim_{a\to0^+}\int_a^1 3x^{-3/2}\,\mathrm{d}x =\lim_{a\to0^+}\left(-6+\frac{6}{\sqrt{a}}\right) =\infty\). Thus \(A_1\) does not have finite area. 3. For \(A_2\), \(\lim_{b\to\infty}\int_1^b 3x^{-3/2}\,\mathrm{d}x =\lim_{b\to\infty}\left(6-\frac{6}{\sqrt{b}}\right) =6\). Thus \(A_2\) has finite area \(6\).

\(A_1\) does not have finite area. \(A_2\) has finite area \(6\) square units.

- Examine the function value as \(x\) increases without bound. - Replace the infinite upper limit with a variable. - Integrate \((x+1)^{-2}\), then evaluate the limit.

1. Since \(\lim_{x\to\infty}\frac{6}{(x+1)^2}=0\), the horizontal asymptote is \(y=0\). 2. The area is \(A=\lim_{b\to\infty}\int_0^b\frac{6}{(x+1)^2}\,\mathrm{d}x\). 3. An antiderivative is \(-\frac{6}{x+1}\). Therefore, \(A=\lim_{b\to\infty}\left[-\frac{6}{x+1}\right]_0^b =\lim_{b\to\infty}\left(6-\frac{6}{b+1}\right) =6\).

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

- Replace the infinite upper limit with a variable. - Use the power rule after rewriting the denominator with an exponent. - Evaluate the resulting limit.

1. Write \(A=\lim_{b\to\infty}\int_2^b\frac{4}{(x+2)^{3/2}}\,\mathrm{d}x\). 2. An antiderivative is \(-\frac{8}{\sqrt{x+2}}\). 3. Therefore, \(A=\lim_{b\to\infty}\left[-\frac{8}{\sqrt{x+2}}\right]_2^b =\lim_{b\to\infty}\left(4-\frac{8}{\sqrt{b+2}}\right) =4\). The region has finite area.

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

- Subtract the lower function from the upper function. - Replace the infinite upper limit with a variable. - Examine the logarithmic expression as the variable approaches infinity.

1. The vertical distance between the graphs is \(f(x)-g(x)=\frac{2}{x}\). 2. Therefore, \(A=\lim_{b\to\infty}\int_2^b\frac{2}{x}\,\mathrm{d}x\). 3. Since an antiderivative is \(2\ln x\), \(A=\lim_{b\to\infty}\left(2\ln b-2\ln 2\right)\). Because \(\ln b\to\infty\), the integral diverges. The region does not have finite area.

The region does not have finite area.

- Compare the growth of the linear factor with the decay of the exponential factor. - Replace the infinite upper limit with a variable. - Use integration by parts for the product of a linear function and an exponential function.

1. The exponential factor \(e^{-0.5x}\) approaches \(0\) faster than the linear factor \(3x+2\) grows. Therefore, \(\lim_{x\to\infty}(3x+2)e^{-0.5x}=0\). 2. The area is \(A=\lim_{b\to\infty}\int_0^b(3x+2)e^{-0.5x}\,\mathrm{d}x\). 3. Integration by parts gives the antiderivative \(-(6x+16)e^{-0.5x}\). 4. Thus, \(A=\lim_{b\to\infty}\left[-(6x+16)e^{-0.5x}\right]_0^b =0-(-16)=16\). The region has finite area.

a) Exponential decay dominates linear growth, so \(f(x)\to0\). b) The region has finite area \(16\) square units.

- 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 is never zero. - Compare linear growth with exponential decay. - Use the product rule to verify the antiderivative. - Use an improper integral for the unbounded region.

1. Since the exponential factor is positive, the only zero is \(x=2\). 2. Exponential decay dominates the linear factor, so \(\lim_{x\to-\infty}f(x)=0\). 3. By the product rule, \(F'(x)=(-4)e^{0.5x}+(16-4x)(0.5)e^{0.5x} =(4-2x)e^{0.5x}=f(x)\). 4. The Quadrant I area is \(\int_0^2f(x)\,dx=F(2)-F(0)=8e-16\). 5. Since \(f(x)\ge0\) for \(x\le2\), \(\int_{-\infty}^{2}f(x)\,dx =F(2)-\lim_{a\to-\infty}F(a)=8e\).

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

- Set the first derivative equal to zero. - Use the second derivative to classify the critical points. - Apply the product rule to verify the antiderivative. - Compare polynomial growth with exponential decay.

1. The derivative is \(g'(x)=(x^2+2x-3)e^x=(x+3)(x-1)e^x\). The critical points are \(x=-3\) and \(x=1\). 2. Using \(g''(x)=(x^2+4x-1)e^x\), there is a local maximum at \((-3, 6e^{-3})\) and a local minimum at \((1, -2e)\). 3. By the product rule, \(G'(x)=g(x)\). 4. Therefore, \(I(k)=G(-1)-G(k) =\frac{2}{e}-(k^2-2k-1)e^k\). 5. Since a polynomial times \(e^k\) approaches \(0\) as \(k\to-\infty\), \(\lim_{k\to-\infty}I(k)=\frac{2}{e}\).

a) Local maximum: \((-3, 6e^{-3})\); local minimum: \((1, -2e)\) b) \(G'(x)=g(x)\) c) \(I(k)=\frac{2}{e}-(k^2-2k-1)e^k\) d) The limit is \(\frac{2}{e}\approx0.736\).

- Write the rectangle's area as width times height. - What derivative condition is needed at an interior maximum? - How do you evaluate an integral with an upper limit of \(\infty\)? - How is one area expressed as a percentage of another?

1. The rectangle's area is \(A(x)=x f(x)=24xe^{-0.2x}\). 2. Differentiate: \(A'(x)=24e^{-0.2x}-4.8xe^{-0.2x}=(24-4.8x)e^{-0.2x}\). 3. Since the exponential factor is positive, \(A'(x)=0\) when \(24-4.8x=0\), so \(x=5\). The derivative changes from positive to negative, so this gives the maximum. 4. The maximum rectangle area is \(A(5)=120e^{-1}\approx 44.15\). 5. The total area is the improper integral \(A_{\text{total}}=\int_0^{\infty}24e^{-0.2x}\,dx\). 6. Evaluate the limit: \(A_{\text{total}}=\lim_{b\to\infty}\left[-120e^{-0.2x}\right]_0^b=120\). 7. The percentage is \(\frac{120e^{-1}}{120}\cdot 100\%=\frac{100}{e}\%\approx 36.79\%\).

a) \(x=5\), and the maximum area is \(120e^{-1}\approx 44.15\). b) \(A_{\text{total}}=120\) c) \(\frac{100}{e}\%\approx 36.79\%\)

- The exponential factor is always positive. - Use the discriminant of the critical-point equation in part b. - Differentiate the proposed antiderivative using the product rule. - Treat the unbounded lower limit as a limit, and remember that area is nonnegative.

1. Since \(e^x>0\), the x-intercepts satisfy \(a-x^2=0\), or \(x^2=a\). Therefore, there are no x-intercepts when \(a<0\), one x-intercept when \(a=0\), and two x-intercepts when \(a>0\). 2. Differentiate: \(g_a'(x)=(-x^2-2x+a)e^x\). The critical numbers satisfy \(x^2+2x-a=0\). This quadratic has two distinct real roots exactly when \(\Delta=4+4a>0\), which is equivalent to \(a>-1\). Because the quadratic factor in \(g_a'\) is negative outside the two roots and positive between them, the derivative changes sign at both roots. Thus, both critical points are local extrema. 3. Differentiate \(G_a\): \(G_a'(x)=(-2x+2)e^x+(-x^2+2x+a-2)e^x=(a-x^2)e^x=g_a(x)\). 4. For \(a=0\), \(g_0(x)=-x^2e^x\le0\). Therefore, the area is \(-\int_{-\infty}^{0}g_0(x)\,\text{d}x\). An antiderivative is \(G_0(x)=(-x^2+2x-2)e^x\). Since \(G_0(x)\to0\) as \(x\to-\infty\) and \(G_0(0)=-2\), \(\int_{-\infty}^{0}g_0(x)\,\text{d}x=-2\). Hence, the area is \(2\) square units.

a) No x-intercepts for \(a<0\), one for \(a=0\), and two for \(a>0\). b) Exactly two local extrema if and only if \(a>-1\). c) \(G_a'(x)=g_a(x)\), so \(G_a\) is an antiderivative. d) \(2\) square units

- Convert all quantities to SI units. - Use infinity as the final separation. - Integrate the inverse-square force.

1. Convert units: \(r_0=0.1\,\text{m}\), \(Q_1=5\times10^{-5}\,\text{C}\), and \(Q_2=2\times10^{-5}\,\text{C}\). 2. The work is \(W=\int_{r_0}^{\infty}kQ_1Q_2r^{-2}\,dr =\frac{kQ_1Q_2}{r_0}\). 3. Substituting, \(W=\frac{(8.99\times10^9)(5\times10^{-5})(2\times10^{-5})}{0.1} =89.9\,\text{J}\).

The electric field does \(89.9\,\text{J}\) of work.

- Find where the curve reaches height \(k\). - Split the region into a rectangle and an area under the curve. - Evaluate the limit of the area formula.

1. The graph meets \(y=k\) where \(x^{-2/3}=k\), so \(x=k^{-3/2}\). 2. The rectangular part has area \(k\cdot k^{-3/2}=k^{-1/2}\). 3. The remaining part has area \(\int_{k^{-3/2}}^1x^{-2/3}\,dx =3-\frac{3}{\sqrt{k}}\). 4. Therefore, \(A(k)=3-\frac{2}{\sqrt{k}}\), and \(\lim_{k\to\infty}A(k)=3\).

\(A(k)=3-\frac{2}{\sqrt{k}}\), and the limit is \(3\).

- Find the tangent equation and its \(x\)-intercept. - Split the area where the lower boundary changes. - Take the limit of the area formula.

1. Since \(f(3)=2\) and \(f'(3)=-\frac{4}{3}\), \(t(x)=-\frac{4}{3}x+6\). 2. The tangent crosses the \(x\)-axis at \(x=\frac{9}{2}\). Thus, \(A_b=\int_3^{9/2}(f-t)\,dx+\int_{9/2}^{b}f(x)\,dx\). 3. The first area is \(\frac{1}{2}\), and the second is \(4-\frac{18}{b}\). Hence, \(A_b=\frac{9}{2}-\frac{18}{b}\). 4. Therefore, \(\lim_{b\to\infty}A_b=\frac{9}{2}\). The region extending indefinitely to the right has finite area.

a) \(t(x)=-\frac{4}{3}x+6\) b) \(A_b=\frac{9}{2}-\frac{18}{b}\) c) The limit is \(\frac{9}{2}\), so the unbounded region has finite area.

- Locate the vertical asymptote. - A proper integral requires bounded behavior on the interval. - Test the two one-sided improper integrals separately.

1. An antiderivative is \(H(x)=\ln|x-3|\). Therefore, \(\int_4^6\frac{1}{x-3}\,dx =\ln3\). 2. The interval \([2, 4]\) contains the vertical asymptote \(x=3\). The function is unbounded there, so the proper integral is not defined. 3. As an improper integral, it must be split at \(x=3\). The left one-sided integral tends to \(-\infty\), and the right one-sided integral tends to \(+\infty\). Since the two integrals do not both converge, the improper integral diverges.

a) \(\ln3\) b) The interval contains the singularity \(x=3\); the proper integral is undefined and the corresponding improper integral diverges.

- Find the positive critical point. - The tangent at a local maximum is horizontal. - Use substitution based on the exponent. - Apply exponential decay to the limit.

1. The derivative is \(f'(x)=2(1-x^2)e^{-0.5x^2}\). The positive local maximum occurs at \(x=1\), with value \(\frac{2}{\sqrt e}\). Its tangent is horizontal: \(y=\frac{2}{\sqrt e}\). 2. The bounded area is \(A=\int_0^1\left(\frac{2}{\sqrt e}-2xe^{-0.5x^2}\right)\,dx =\frac{4}{\sqrt e}-2\). 3. The accumulation area is \(A(k)=\int_0^k2xe^{-0.5x^2}\,dx =2-2e^{-0.5k^2}\). 4. As \(k\to\infty\), \(A(k)\to2\).

a) \(y=\frac{2}{\sqrt e}\) b) \(A=\frac{4}{\sqrt e}-2\approx0.426\) square unit c) \(A(k)=2-2e^{-0.5k^2}\), with limit \(2\)

- Differentiate to verify the maximum. - Use the horizontal tangent as the upper boundary. - Integrate the polynomial-exponential product by parts. - Exponential decay dominates polynomial growth.

1. The derivative is \(g'(x)=\left(x-\frac{x^2}{2}\right)e^{-x}\), so \(g'(2)=0\). The derivative changes from positive to negative, so \(x=2\) is a local maximum. 2. Since \(g(2)=2e^{-2}\), the tangent is \(y=2e^{-2}\). 3. An antiderivative is \(-\frac{1}{2}(x^2+2x+2)e^{-x}\). 4. The area between the tangent and graph on \([0, 2]\) is \(9e^{-2}-1\). 5. For the improper integral, \(\int_0^\infty g(x)\,dx =\lim_{b\to\infty}\left[-\frac{1}{2}(x^2+2x+2)e^{-x}\right]_0^b =1\).

a) The tangent is \(y=2e^{-2}\). b) \(A=9e^{-2}-1\approx0.218\) square unit c) \(\int_0^\infty g(x)\,dx=1\)

- Compare exponential decay with polynomial growth. - Differentiate the proposed form and match coefficients. - Use the limit of the antiderivative at infinity.

1. Exponential decay dominates polynomial growth, so \(\lim_{x\to\infty}x^2e^{-0.5x}=0\). 2. Differentiating the proposed form and matching coefficients gives \(a=-2\), \(b=-8\), and \(c=-16\). 3. Thus, \(F(x)=(-2x^2-8x-16)e^{-0.5x}\). 4. Therefore, \(\int_0^\infty f(x)\,dx =\lim_{k\to\infty}[F(x)]_0^k =0-(-16)=16\).

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

- Differentiate and find the sign change. - Eliminate the parameter between the coordinates. - Integrate the power terms separately. - Let the reciprocal terms approach zero.

1. The derivative is \(f_a'(x)=\frac{3a-2x}{x^4}\). It changes from positive to negative at \(x=\frac{3a}{2}\). 2. Thus, \(H_a=\left(\frac{3a}{2}, \frac{4}{27a^2}\right)\). 3. Since \(a=\frac{2x}{3}\) at a maximum, substitution gives \(y=\frac{1}{3x^2}\). 4. An antiderivative is \(-\frac{1}{x}+\frac{a}{2x^2}\). Therefore, \(A(u)=\frac{1}{2a}-\frac{1}{u}+\frac{a}{2u^2}\). 5. Hence, \(\lim_{u\to\infty}A(u)=\frac{1}{2a}\).

a) \(H_a=\left(\frac{3a}{2}, \frac{4}{27a^2}\right)\) b) The locus is \(y=\frac{1}{3x^2}\). c) \(A(u)=\frac{1}{2a}-\frac{1}{u}+\frac{a}{2u^2}\), with limit \(\frac{1}{2a}\)

- Differentiate after rewriting with negative powers. - Eliminate \(k\) from the maximum coordinates. - Evaluate the improper integral from its antiderivative.

1. The derivative is \(g_k'(x)=\frac{3k^2-2kx}{x^4}\). It changes from positive to negative at \(x=\frac{3k}{2}\). 2. Thus, \(H_k=\left(\frac{3k}{2}, \frac{4}{27k}\right)\). Since \(k=\frac{2x}{3}\), the locus is \(y=\frac{2}{9x}\), so \(c=\frac{2}{9}\). 3. An antiderivative is \(-\frac{k}{x}+\frac{k^2}{2x^2}\). 4. Therefore, \(\int_k^\infty g_k(x)\,dx =0-\left(-1+\frac{1}{2}\right) =\frac{1}{2}\). 5. At the lower limit, the factors of \(k\) cancel; at infinity, the remaining terms vanish.

a) \(H_k=\left(\frac{3k}{2}, \frac{4}{27k}\right)\), and \(c=\frac{2}{9}\) b) \(\frac{1}{2}\) c) The parameter cancels at the lower limit, and all upper-limit terms approach zero.

- Find the upper function minus the lower function. - Express the right-unbounded area as a limit of definite integrals. - For the last part, examine what happens to \(\frac{6}{a}\) as \(a\) approaches \(0\) from the right.

1. The vertical distance between the graphs is \(f(x)-g(x)=\frac{6}{x^2}\). 2. For part a), \(A=\lim_{b\to\infty}\int_1^b\frac{6}{x^2}\,\mathrm{d}x =\lim_{b\to\infty}\left[-\frac{6}{x}\right]_1^b =6\). 3. For part b), \(A(a)=\lim_{b\to\infty}\int_a^b\frac{6}{x^2}\,\mathrm{d}x =\frac{6}{a}\). 4. For part c), \(\lim_{a\to0^+}A(a)=\infty\). Therefore, the entire region between the graphs on \(x>0\) does not have finite area.

a) \(A=6\) square units b) \(A(a)=\frac{6}{a}\) c) \(A(a)\to\infty\), so the entire region between the graphs on \(x>0\) does not have finite area.

- Use the product rule and chain rule. - The exponential factor is never zero. - Differentiate the proposed antiderivative. - Evaluate the improper integral with a limit.

1. The first derivative is \(f_a'(x)=(a-x)e^{-x/a}\). It is zero only at \(x=a\). The second derivative is \(f_a''(x)=\left(\frac{x}{a}-2\right)e^{-x/a}\), and \(f_a''(a)=-\frac{1}{e}<0\). Thus, the local maximum is \(\left(a, \frac{a^2}{e}\right)\). 2. The second derivative is zero only at \(x=2a\), and its sign changes there. The inflection point is \(\left(2a, \frac{2a^2}{e^2}\right)\). 3. Differentiating \(F_a\) gives \(F_a'(x)=axe^{-x/a}=f_a(x)\). 4. Since \(f_a(x)\ge0\) for \(x\ge0\), the area is \(\int_0^\infty f_a(x)\,dx\). Using the antiderivative, \(\lim_{b\to\infty}F_a(b)=0\) and \(F_a(0)=-a^3\), so the area is \(a^3\).

a) Local maximum: \(\left(a, \frac{a^2}{e}\right)\) b) Inflection point: \(\left(2a, \frac{2a^2}{e^2}\right)\) c) \(A=a^3\)

- Use the product rule, then analyze the sign of the derivative. - Compare exponential decay with linear growth. - Use integration by parts or verify a proposed antiderivative by differentiation. - Compare the locations and heights of the maxima.

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

a) The maximum occurs at \(x=\frac{1}{k}\). b) \(\lim_{x\to\infty}h_k(x)=0\), so the \(x\)-axis is a horizontal asymptote. c) \(\int_0^\infty h_k(x)\,\mathrm{d}x=1\); the area is independent of \(k\). d) Curve \(a\): \(k=1\); curve \(b\): \(k=2\)