Let's solve each part step by step: ### Given: For **d)** and **e)**, set \( k = 1 \), so: \[ f(x) = \sqrt{x} \cdot e^{-x} \] --- ## **d) Antiderivative of \( g(x) = (f(x))^3 \)** So, \[ g(x) = (f(x))^3 = (\sqrt{x} \cdot e^{-x})^3 = x^{3/2} \cdot e^{-3x} \] We want to find \[ \int g(x)\, dx = \int x^{3/2} e^{-3x} \, dx \] Use **integration by parts** or substitution. Let's use substitution: Let \( u = x \), \( dv = x^{3/2} e^{-3x} dx \). Alternatively, use the gamma function: \[ \int_0^\infty x^a e^{-bx} dx = \frac{\Gamma(a+1)}{b^{a+1}} \] But for indefinite integral, let's use substitution: Let \( t = 3x \Rightarrow x = \frac{t}{3}, dx = \frac{dt}{3} \): \[ \int x^{3/2} e^{-3x} dx = \int \left(\frac{t}{3}\right)^{3/2} e^{-t} \cdot \frac{dt}{3} = \frac{1}{3^{5/2}} \int t^{3/2} e^{-t} dt \] \[ = \frac{1}{3^{5/2}} \cdot \gamma\left(\frac{5}{2}, t\right) + C \] where \(\gamma(s, t)\) is the lower incomplete gamma function. But for the indefinite integral: \[ \boxed{ \int x^{3/2} e^{-3x} dx = \frac{1}{3^{5/2}} \Gamma\left(\frac{5}{2}, 3x\right) + C } \] Or, in terms of the incomplete gamma function if considering limits. --- ## **e) Volume of Solid of Revolution** The volume \( V \) generated by rotating \( f(x) \) around the x-axis (Ox axis) from \( x = 0 \) to \( x = \infty \) is: \[ V = \pi \int_{0}^{\infty} [f(x)]^2 dx \] \[ f(x)^2 = (\sqrt{x} e^{-x})^2 = x e^{-2x} \] So, \[ V = \pi \int_{0}^{\infty} x e^{-2x} dx \] This is a standard integral: Let \( a = 2 \): \[ \int_0^\infty x e^{-ax} dx = \frac{1}{a^2} \] So, \[ \int_0^\infty x e^{-2x} dx = \frac{1}{2^2} = \frac{1}{4} \] Therefore, \[ \boxed{ V = \frac{\pi}{4} } \] --- ## **f) Solve the Differential Equation** Given: \[ (1 - 2x)y = 2xy' \] Write as: \[ (1 - 2x)y = 2x \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{(1-2x)}{2x} y \] This is a **first-order linear ODE**; separate variables: \[ \frac{1}{y} dy = \frac{1-2x}{2x} dx \] \[ \int \frac{1}{y} dy = \int \left(\frac{1}{2x} - 1\right) dx \] \[ \ln|y| = \frac{1}{2} \ln|x| - x + C \] \[ |y| = e^C x^{1/2} e^{-x} \] Let \( A = \pm e^C \): \[ \boxed{ y(x) = A \sqrt{x} e^{-x} } \] --- ### **Summary of Answers** **d)** \(\int x^{3/2} e^{-3x} dx = \frac{1}{3^{5/2}} \Gamma\left(\frac{5}{2}, 3x\right) + C\) **e)** Volume \( V = \frac{\pi}{4} \) **f)** General solution: \( y(x) = A \sqrt{x} e^{-x} \), where \(A\) is a constant.