Let's go through each part step by step. --- ### (a) What is \( g(6) + 2g(3) \)? Given: \[ g(t) = \int_0^1 \frac{tx^t}{1+x^3} dx \] So: \[ g(6) = \int_0^1 \frac{6x^6}{1+x^3} dx \] \[ g(3) = \int_0^1 \frac{3x^3}{1+x^3} dx \] \[ 2g(3) = 2 \int_0^1 \frac{3x^3}{1+x^3} dx = \int_0^1 \frac{6x^3}{1+x^3} dx \] Adding: \[ g(6) + 2g(3) = \int_0^1 \frac{6x^6}{1+x^3} dx + \int_0^1 \frac{6x^3}{1+x^3} dx = \int_0^1 \frac{6x^6 + 6x^3}{1+x^3} dx \] \[ = \int_0^1 \frac{6x^3(x^3 + 1)}{1 + x^3} dx = \int_0^1 6x^3 dx = 6 \int_0^1 x^3 dx = 6 \left[\frac{x^4}{4}\right]_0^1 = 6 \cdot \frac{1}{4} = \frac{3}{2} \] **Final answer:** \[ g(6) + 2g(3) = \boxed{\frac{3}{2}} \] --- ### (b) What method(s) do you use to compute \( g(3) = \int_0^1 \frac{3x^3}{1+x^3} dx \)? We notice that \(\frac{3x^3}{1+x^3}\) can be simplified using **basic substitution** (since numerator is derivative of denominator up to a factor): \[ \frac{3x^3}{1+x^3} = 3 \cdot \frac{x^3}{1+x^3} \] Let \(u = 1 + x^3\), \(du = 3x^2 dx\). But actually, it's more direct to use **basic substitution**. **Final answer:** \[ \boxed{\text{D. Basic substitution}} \] --- ### (c) What method(s) do you use to compute \( g(5) = \int_0^1 \frac{5x^5}{1+x^3} dx \)? Here, the degree of numerator is higher than denominator, so we use **partial fractions**. Expand \(x^5\) in terms of \(x^3\), and then decompose resulting expression. **Final answer:** \[ \boxed{\text{A. Partial fractions}} \] --- ### (d) Compute \( g(7) = \int_0^1 \frac{7x^7}{1+x^3} dx \) using partial fractions. Let’s write \(x^7 = x^7\), but notice \(x^7 = x^4(x^3) = (x^7) = (x^3)(x^4)\), but more useful is to divide \(x^7\) by \(1 + x^3\): Do long division: \[ x^7 = (x^7 + x^6 - x^6) = (x^7 + x^6) - x^6 = x^6(x + 1) - x^6 \] But more generally, perform the division: \(x^7 \div (x^3 + 1)\): \(x^7 \div x^3 = x^4\), so \(x^7 - x^4(x^3 + 1) = x^7 - x^7 - x^4 = -x^4\) \(-x^4 \div x^3 = -x\), so \(-x^4 - (-x)(x^3 + 1) = -x^4 + x^4 + x = x\) So, \(x^7 = (x^3 + 1)(x^4 - x) + x\) So, \[ \frac{7x^7}{1 + x^3} = 7(x^4 - x) + \frac{7x}{1 + x^3} \] Now, integrate term by term: \[ g(7) = \int_0^1 7(x^4 - x) dx + \int_0^1 \frac{7x}{1 + x^3} dx \] Now, compare to the options. The correct answer will have this structure. Looking at the options, the partial fraction decomposition for \(\frac{x}{1 + x^3}\) is: \[ x^3 + 1 = (x + 1)(x^2 - x + 1) \] So, \[ \frac{x}{x^3 + 1} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1} \] Multiply through by \(x^3 + 1\): \[ x = A(x^2 - x + 1) + (Bx + C)(x + 1) \] Expand: \[ x = A(x^2 - x + 1) + Bx(x + 1) + C(x + 1) \\ x = A x^2 - A x + A + B x^2 + B x + C x + C \] \[ x = (A + B)x^2 + (-A + B + C)x + (A + C) \] Set equal to \(x\): - Coefficient of \(x^2\): \(A + B = 0\) - Coefficient of \(x\): \(-A + B + C = 1\) - Constant: \(A + C = 0\) Now, solve the system: - \(A + B = 0 \implies B = -A\) - \(A + C = 0 \implies C = -A\) - \(-A + B + C = 1\) Plug \(B = -A\), \(C = -A\) in: \[ -A + (-A) + (-A) = 1 \implies -A - A - A = 1 \implies -3A = 1 \implies A = -\frac{1}{3} \] So, - \(B = -A = \frac{1}{3}\) - \(C = -A = \frac{1}{3}\) So, \[ \frac{x}{x^3 + 1} = \frac{-1/3}{x+1} + \frac{(1/3)x + 1/3}{x^2 - x + 1} \] \[ = -\frac{1}{3(x+1)} + \frac{1}{3} \frac{x+1}{x^2 - x + 1} \] So, \[ g(7) = 7\int_0^1 (x^4 - x) dx + 7\int_0^1 \left(-\frac{1}{3(x+1)} + \frac{1}{3}\frac{x+1}{x^2 - x + 1}\right)dx \] Now, match to the options: Option A fits this structure, with the correct signs and terms. **Final answer:** \[ \boxed{\text{A}} \] --- ### (e) What methods/theorems do you use to compute \( \lim_{t \to \infty} g(t) \)? Here, you would use **comparison theorem** (to compare \( \frac{tx^t}{1 + x^3} \) to a function that goes to 0 as \( t \to \infty \)), and possibly **basic substitution** to evaluate the limit inside the integral. **Final answer:** \[ \boxed{\text{C. Basic substitution,\ E. Comparison theorem}} \] --- ### (f) What is \( \lim_{t \to \infty} g(t) \)? Recall: \[ g(t) = \int_0^1 \frac{t x^t}{1 + x^3} dx \] As \( t \to \infty \), \( x^t \) goes to 0 for all \( x \in [0,1) \), except at \( x = 1 \), where it is 1. However, the "mass" of the function becomes more and more concentrated near \( x = 1 \), but the length of the interval where it's significant shrinks to 0. So, as \( t \to \infty \), the integral goes to 0. **Final answer:** \[ \lim_{t \to \infty} g(t) = \boxed{0} \] --- ## **Summary of Answers** (a) \( \boxed{\frac{3}{2}} \) (b) \( \boxed{\text{D. Basic substitution}} \) (c) \( \boxed{\text{A. Partial fractions}} \) (d) \( \boxed{\text{A}} \) (e) \( \boxed{\text{C, E}} \) (f) \( \boxed{0} \)

Let's analyze each part carefully, providing detailed calculations, reasoning, and explanations for the options where applicable. --- ### **Given:** \[ g(t) = \int_0^1 \frac{t x^t}{1 + x^3} dx \quad \text{for } t > 3 \] --- ### **(a) Find \( g(6) + 2g(3) \)** **Step 1: Write expressions for \(g(6)\) and \(g(3)\)** \[ g(6) = \int_0^1 \frac{6 x^6}{1 + x^3} dx \] \[ g(3) = \int_0^1 \frac{3 x^3}{1 + x^3} dx \] **Step 2: Express the sum** \[ g(6) + 2g(3) = \int_0^1 \frac{6 x^6}{1 + x^3} dx + 2 \int_0^1 \frac{3 x^3}{1 + x^3} dx \] \[ = \int_0^1 \frac{6 x^6}{1 + x^3} dx + \int_0^1 \frac{6 x^3}{1 + x^3} dx \] **Step 3: Combine the integrals** \[ = \int_0^1 \frac{6 x^6 + 6 x^3}{1 + x^3} dx \] **Step 4: Factor numerator** \[ = \int_0^1 \frac{6 x^3 (x^3 + 1)}{1 + x^3} dx \] Since numerator and denominator share the factor \(1 + x^3\), they cancel: \[ = \int_0^1 6 x^3 dx \] **Step 5: Compute the integral** \[ = 6 \int_0^1 x^3 dx = 6 \left[ \frac{x^4}{4} \right]_0^1 = 6 \times \frac{1}{4} = \frac{6}{4} = \frac{3}{2} \] **Final answer:** \[ \boxed{\frac{3}{2}} \] --- ### **(b) Which method to compute \( g(3) = \int_0^1 \frac{3 x^3}{1 + x^3} dx \)?** **Analysis:** The integrand: \[ \frac{3 x^3}{1 + x^3} \] Note that numerator \(3x^3\) looks like the derivative of the denominator \(1 + x^3\) (since \(d/dx [1 + x^3] = 3x^2\)). But here, numerator is \(3x^3\), not \(3x^2\). So, direct substitution based on derivative of denominator doesn't work straightforwardly. Alternatively, consider substitution: - Let \( u = 1 + x^3 \Rightarrow du = 3 x^2 dx \). But the numerator has \(3 x^3\), so: \[ x^3 = x \times x^2 \] Express \(x^3\) as \(x \times x^2\), then: \[ \frac{3 x^3}{1 + x^3} = 3 x \cdot \frac{x^2}{1 + x^3} \] Now, using \(u = 1 + x^3\), \(du = 3 x^2 dx\), or: \[ x^2 dx = \frac{du}{3} \] Thus, \[ g(3) = \int_0^1 3x \cdot \frac{x^2}{1 + x^3} dx = \int_0^1 3x \cdot \frac{1}{u} \cdot x^2 dx \] But \(x^2 dx = du/3\), so \[ g(3) = \int_{x=0}^{x=1} 3x \cdot \frac{1}{u} \cdot x^2 dx = \int_{u=1}^{u=2} 3x \cdot \frac{1}{u} \cdot \frac{du}{3} = \int_{u=1}^{u=2} x \cdot \frac{1}{u} du \] But \(x\) is expressed in terms of \(u\): \[ u = 1 + x^3 \Rightarrow x = (u - 1)^{1/3} \] So, \[ g(3) = \int_{u=1}^{2} (u - 1)^{1/3} \cdot \frac{1}{u} du \] This integral looks complicated, but it is a substitution method involving transforming the integrand into a simpler form. **Conclusion:** The key method is **basic substitution**, specifically \( u = 1 + x^3 \). **Answer:** \[ \boxed{\text{D. Basic substitution}} \] --- ### **(c) Which method to compute \( g(5) = \int_0^1 \frac{5 x^5}{1 + x^3} dx \)?** **Step 1: Recognize the numerator** - \(x^5\) over \(1 + x^3\). - Since numerator degree \(5\) > denominator degree \(3\), polynomial division or partial fractions is appropriate. **Step 2: Polynomial division** Divide \(x^5\) by \(1 + x^3\): - \(x^5 \div x^3 = x^2\) Multiply: \[ (1 + x^3) \times x^2 = x^2 + x^5 \] Subtract: \[ x^5 - (x^5 + x^2) = - x^2 \] Now, express: \[ x^5 = (1 + x^3) x^2 - x^2 \] Thus, \[ \frac{5 x^5}{1 + x^3} = 5 [(x^2) - \frac{x^2}{1 + x^3}] = 5 x^2 - 5 \frac{x^2}{1 + x^3} \] The integral becomes: \[ g(5) = \int_0^1 5 x^2 dx - 5 \int_0^1 \frac{x^2}{1 + x^3} dx \] **Step 3: To evaluate the second integral**, use partial fractions on \(\frac{x^2}{1 + x^3}\). Factor denominator: \[ 1 + x^3 = (x + 1)(x^2 - x + 1) \] Decompose: \[ \frac{x^2}{(x + 1)(x^2 - x + 1)} = \frac{A}{x + 1} + \frac{B x + C}{x^2 - x + 1} \] Multiply through: \[ x^2 = A (x^2 - x + 1) + (B x + C)(x + 1) \] Expand: \[ x^2 = A x^2 - A x + A + B x (x + 1) + C (x + 1) \] \[ = A x^2 - A x + A + B x^2 + B x + C x + C \] Combine like terms: \[ x^2 = (A + B) x^2 + (-A + B + C) x + (A + C) \] Set coefficients equal: - \(x^2\): \(A + B = 1\) - \(x\): \(-A + B + C = 0\) - Constant: \(A + C = 0\) Solve: From \(A + C = 0 \Rightarrow C = -A\). Plug into \(-A + B + C = 0\): \[ -A + B - A = 0 \Rightarrow B = 2A \] From \(A + B = 1\): \[ A + 2A = 1 \Rightarrow 3A = 1 \Rightarrow A = \frac{1}{3} \] Then: \[ B = 2A = \frac{2}{3} \] \[ C = -A = -\frac{1}{3} \] **Step 4: Write the partial fractions** \[ \frac{x^2}{1 + x^3} = \frac{\frac{1}{3}}{x+1} + \frac{\frac{2}{3} x - \frac{1}{3}}{x^2 - x + 1} \] **Step 5: Integrate** Now, back to the integral: \[ g(5) = 5 \int_0^1 x^2 dx - 5 \left[ \int_0^1 \frac{\frac{1}{3}}{x+1} dx + \int_0^1 \frac{\frac{2}{3} x - \frac{1}{3}}{x^2 - x + 1} dx \right] \] - The first integral: \[ 5 \int_0^1 x^2 dx = 5 \times \frac{1}{3} = \frac{5}{3} \] - The second integral: \[ \frac{1}{3} \int_0^1 \frac{1}{x+1} dx = \frac{1}{3} [\ln|x+1|]_0^1 = \frac{1}{3} (\ln 2 - \ln 1) = \frac{1}{3} \ln 2 \] - The third integral: \[ \int_0^1 \frac{\frac{2}{3} x - \frac{1}{3}}{x^2 - x + 1} dx \] Note that: \[ \frac{2}{3} x - \frac{1}{3} = \frac{1}{3}(2x - 1) \] and the quadratic in denominator: \[ x^2 - x + 1 \] Complete the square: \[ x^2 - x + 1 = \left( x - \frac{1}{2} \right)^2 + \frac{3}{4} \] This suggests substitution or recognizing a derivative pattern for arctangent. **Step 6: Recognize the form** The numerator \(2x - 1\) is the derivative of \(x^2 - x + 1\): \[ \frac{d}{dx} (x^2 - x + 1) = 2x - 1 \] Thus, \[ \int \frac{2x - 1}{x^2 - x + 1} dx = \ln |x^2 - x + 1| + C \] Therefore, \[ \int_0^1 \frac{\frac{2}{3} x - \frac{1}{3}}{x^2 - x + 1} dx = \frac{1}{3} \left[\ln (x^2 - x + 1) \right]_0^1 \] Calculate at bounds: \[ x=1: 1 - 1 + 1 = 1 \] \[ x=0: 0 - 0 + 1 = 1 \] So, \[ \ln 1 - \ln 1 = 0 \] **Step 7: Final calculation** Putting it all together: \[ g(5) = \frac{5}{3} - 5 \left[ \frac{1}{3} \ln 2 + 0 \right] = \frac{5}{3} - \frac{5}{3} \ln 2 \] **Final answer:** \[ \boxed{\frac{5}{3} (1 - \ln 2)} \] **Which method?** - Polynomial division - Partial fractions - Recognizing derivative patterns for integrals of rational functions **Answer:** \[ \boxed{\text{A. Partial fractions}} \] --- ### **(d) Compute \( g(7) = \int_0^1 \frac{7 x^7}{1 + x^3} dx \) using partial fractions** **Step 1: Polynomial division** Divide \(x^7\) by \(1 + x^3\): - \(x^7 \div x^3 = x^4\) Multiply: \[ (1 + x^3) \times x^4 = x^4 + x^7 \] Subtract: \[ x^7 - (x^7 + x^4) = - x^4 \] Next: \[ - x^4 \div x^3 = - x \] Multiply: \[ (1 + x^3) \times (-x) = -x - x^4 \] Subtract: \[ - x^4 - (-x - x^4) = - x^4 + x + x^4 = x \] Now: \[ x \div x^3 = \text{not divisible}, \text{so polynomial division ends here.} \] Thus, \[ x^7 = (1 + x^3)(x^4 - x) + x \] Express: \[ \frac{7 x^7}{1 + x^3} = 7 (x^

Let's continue from where we left off. We had: \[ x^7 = (1 + x^3)(x^4 - x) + x \] Therefore, \[ g(7) = \int_0^1 \frac{7 x^7}{1 + x^3} dx = 7 \int_0^1 (x^4 - x) dx + 7 \int_0^1 \frac{x}{1 + x^3} dx \] --- ### **Step 2: Compute the first integral** \[ 7 \int_0^1 (x^4 - x) dx = 7 \left[\frac{x^5}{5} - \frac{x^2}{2}\right]_0^1 = 7 \left(\frac{1}{5} - \frac{1}{2}\right) = 7 \left(\frac{2 - 5}{10}\right) = 7 \left(-\frac{3}{10}\right) = -\frac{21}{10} \] --- ### **Step 3: Compute the second integral** \[ I = \int_0^1 \frac{x}{1 + x^3} dx \] **Step 3.1: Partial fraction decomposition** Factor denominator: \[ 1 + x^3 = (x + 1)(x^2 - x + 1) \] Decompose: \[ \frac{x}{(x + 1)(x^2 - x + 1)} = \frac{A}{x + 1} + \frac{B x + C}{x^2 - x + 1} \] Multiply both sides by denominator: \[ x = A(x^2 - x + 1) + (B x + C)(x + 1) \] Expand: \[ x = A x^2 - A x + A + B x (x + 1) + C (x + 1) \] \[ = A x^2 - A x + A + B x^2 + B x + C x + C \] Combine like terms: \[ x = (A + B) x^2 + (-A + B + C) x + (A + C) \] Set coefficients: - For \(x^2\): \[ A + B = 0 \] - For \(x\): \[ -A + B + C = 1 \] - For constant term: \[ A + C = 0 \] Solve: From \(A + C=0 \Rightarrow C = -A\). From \(A + B=0 \Rightarrow B= -A\). Substitute into \(-A + B + C=1\): \[ -A + (-A) + (-A) = 1 \Rightarrow -A - A - A = 1 \Rightarrow -3A=1 \Rightarrow A= -\frac{1}{3} \] Then: \[ B= -A= \frac{1}{3} \] \[ C= -A= \frac{1}{3} \] --- ### **Step 4: Write the partial fractions** \[ \frac{x}{(x + 1)(x^2 - x + 1)} = - \frac{1}{3} \frac{1}{x+1} + \frac{1/3 x + 1/3}{x^2 - x + 1} \] So, \[ I = \int_0^1 \left( - \frac{1}{3} \frac{1}{x+1} + \frac{1/3 x + 1/3}{x^2 - x + 1} \right) dx \] which can be split as: \[ I = - \frac{1}{3} \int_0^1 \frac{1}{x+1} dx + \frac{1}{3} \int_0^1 \frac{x}{x^2 - x + 1} dx + \frac{1}{3} \int_0^1 \frac{1}{x^2 - x + 1} dx \] --- ### **Step 5: Compute each integral** **Integral 1:** \[ - \frac{1}{3} \int_0^1 \frac{1}{x+1} dx = - \frac{1}{3} [\ln|x+1|]_0^1 = - \frac{1}{3} (\ln 2 - \ln 1) = - \frac{1}{3} \ln 2 \] **Integral 2:** \[ \frac{1}{3} \int_0^1 \frac{x}{x^2 - x + 1} dx \] Notice numerator is \(x\). Let's consider substitution: Let: \[ u = x^2 - x + 1 \] \[ du/dx = 2x - 1 \] \[ \Rightarrow du = (2x - 1) dx \] We want to express \(x dx\) in terms of \(du\): \[ x dx = \frac{du + dx}{2} \] But this seems complicated. Alternatively, note that: \[ x = \frac{du + 1 - x^2}{2} \] which may not help straightforwardly. Instead, consider alternative: - Rewrite numerator \(x\) as: \[ x = \frac{1}{2}(2x) = \frac{1}{2} (2x - 1 + 1) \] \[ x = \frac{1}{2} (2x - 1) + \frac{1}{2} \] We know: \[ du/dx = 2x - 1 \] \[ \Rightarrow 2x - 1 = du/dx \] Thus, \[ x dx = \frac{1}{2} (du) + \frac{1}{2} dx \] But integrating this complicates matters. Alternatively, note that: \[ x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4} \] which suggests substitution: \[ x - \frac{1}{2} = t \Rightarrow x = t + \frac{1}{2} \] Then, \[ x^2 - x + 1 = t^2 + \frac{3}{4} \] Express numerator \(x\): \[ x= t + \frac{1}{2} \] Integral becomes: \[ \frac{1}{3} \int_{t=-1/2}^{t=1/2} \frac{t + 1/2}{t^2 + 3/4} dt \] Now, split into two integrals: \[ \frac{1}{3} \left[\int_{-1/2}^{1/2} \frac{t}{t^2 + 3/4} dt + \frac{1}{2} \int_{-1/2}^{1/2} \frac{1}{t^2 + 3/4} dt \right] \] - The first integral (odd function over symmetric limits): \[ \int_{-a}^{a} \frac{t}{t^2 + c} dt = 0 \] since numerator is odd, denominator even. - The second integral: \[ \frac{1}{2} \int_{-1/2}^{1/2} \frac{1}{t^2 + 3/4} dt \] which is an even function, can be computed as: \[ \frac{1}{2} \times 2 \int_0^{1/2} \frac{1}{t^2 + ( \sqrt{3}/2)^2} dt = \int_0^{1/2} \frac{1}{t^2 + (\sqrt{3}/2)^2} dt \] Standard integral: \[ \int \frac{1}{t^2 + a^2} dt = \frac{1}{a} \arctan \left( \frac{t}{a} \right) \] So, \[ a = \frac{\sqrt{3}}{2} \] and \[ \int_0^{1/2} \frac{dt}{t^2 + a^2} = \frac{1}{a} \left[ \arctan \left( \frac{t}{a} \right) \right]_0^{1/2} \] Calculate: \[ \frac{1}{a} = \frac{2}{\sqrt{3}} \] \[ \frac{t}{a} = \frac{1/2}{\sqrt{3}/2} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} \] and at \(t=0\): \[ \arctan 0 = 0 \] so, \[ \int_0^{1/2} \frac{dt}{t^2 + a^2} = \frac{2}{\sqrt{3}} \arctan \left( \frac{1}{\sqrt{3}} \right) = \frac{2}{\sqrt{3}} \times \frac{\pi}{6} = \frac{2}{\sqrt{3}} \times \frac{\pi}{6} = \frac{\pi}{3 \sqrt{3}} \] --- ### **Step 6: Summarize second integral** \[ I = - \frac{1}{3} \ln 2 + \frac{\pi}{3 \sqrt{3}} \] --- ### **Step 7: Final calculation of \(g(7)\)** Recall: \[ g(7) = - \frac{21}{10} + 7 \times I = - \frac{21}{10} + 7 \left( - \frac{1}{3} \ln 2 + \frac{\pi}{3 \sqrt{3}} \right) \] Compute: \[ 7 \times - \frac{1}{3} \ln 2 = - \frac{7}{3} \ln 2 \] \[ 7 \times \frac{\pi}{3 \sqrt{3}} = \frac{7 \pi}{3 \sqrt{3}} \] So, \[ g(7) = - \frac{21}{10} - \frac{7}{3} \ln 2 + \frac{7 \pi}{3 \sqrt{3}} \] --- ### **### **Final Result:** \[ \boxed{ g(7) = - \frac{21}{10} - \frac{7}{3} \ln 2 + \frac{7 \pi}{3 \sqrt{3}} } \] --- ### **Summary:** - We used polynomial division to decompose the integrand. - Integrated each term separately. - Recognized derivatives and standard integrals for rational functions. - Computed definite integrals carefully. --- ### **(e) Limit as \( t \to \infty \) of \( g(t) \):** Since: \[ g(t) = \int_0^1 \frac{t x^t}{1 + x^3} dx \] - For \(0 \leq x < 1\), as \(t \to \infty\), \(x^t \to 0\). - Near \(x=1\), \(x^t \to 1\), but the measure of the point is zero, and the integral concentrates near \(x=1\). By the Dominated Convergence Theorem, the integral tends to 0 because the integrand becomes negligible everywhere except possibly near 1, but the integral over that shrinking neighborhood tends to 0. **Answer:** \[ \boxed{0} \] --- ### **(f) Final answer:** \[ \boxed{0} \] --- **Summary of all parts:** | Part | Final Answer | Explanation Highlights | |---|---|---| | (a) | \(\frac{3}{2}\) | Simplified integrals and cancellation | | (b) | D. Basic substitution | Recognized substitution \( u=1+x^3 \) | | (c) | A. Partial fractions | Polynomial division + partial fractions | | (d) | \( - \frac{21}{10} - \frac{7}{3} \ln 2 + \frac{7 \pi}{3 \sqrt{3}} \) | Decomposed into integrals, used substitution and partial fractions | | (e) | 0 | Limit of the integral as \( t \to \infty \) tends to zero by dominance | | (f) | 0 | Limit value after analysis | Feel free to ask for further clarification or detailed steps on any part!