a) \( \int_{-2}^{0} x^{2} e^{x^{3} / 2} d x \)

To evaluate the integral \[ I = \int_{-2}^{0} x^{2} e^{x^{3} / 2} \, dx, \] we can use integration by parts. We will let: - \( u = e^{x^{3}/2} \) (which we will differentiate) - \( dv = x^2 \, dx \) (which we will integrate) First, we need to compute \( du \) and \( v \): 1. Differentiate \( u \): \[ du = \frac{d}{dx} e^{x^{3}/2} \, dx = e^{x^{3}/2} \cdot \frac{3}{2} x^2 \, dx. \] 2. Integrate \( dv \): \[ v = \int x^2 \, dx = \frac{x^3}{3}. \] Now we apply integration by parts, which states: \[ \int u \, dv = uv - \int v \, du. \] Substituting our expressions for \( u \), \( v \), \( du \), and \( dv \): \[ I = \left[ e^{x^{3}/2} \cdot \frac{x^3}{3} \right]_{-2}^{0} - \int_{-2}^{0} \frac{x^3}{3} \cdot \left( e^{x^{3}/2} \cdot \frac{3}{2} x^2 \right) \, dx. \] Evaluating the boundary term: At \( x = 0 \): \[ e^{0} \cdot \frac{0^3}{3} = 0. \] At \( x = -2 \): \[ e^{(-2)^{3}/2} \cdot \frac{(-2)^3}{3} = e^{-4} \cdot \frac{-8}{3} = -\frac{8}{3} e^{-4}. \] Thus, the boundary term evaluates to: \[ 0 - \left(-\frac{8}{3} e^{-4}\right) = \frac{8}{3} e^{-4}. \] Now we need to evaluate the integral: \[ \int_{-2}^{0} \frac{x^3}{3} \cdot \left( e^{x^{3}/2} \cdot \frac{3}{2} x^2 \right) \, dx = \frac{1}{2} \int_{-2}^{0} x^5 e^{x^{3}/2} \, dx. \] Thus, we have: \[ I = \frac{8}{3} e^{-4} - \frac{1}{2} \int_{-2}^{0} x^5 e^{x^{3}/2} \, dx. \] Let’s denote \[ J = \int_{-2}^{0} x^5 e^{x^{3}/2} \, dx. \] Now we can express \( I \) in terms of \( J \): \[ I = \frac{8}{3} e^{-4} - \frac{1}{2} J. \] To evaluate \( J \), we can again use integration by parts. Let: - \( u = e^{x^{3}/2} \) - \( dv = x^5 \, dx \) Following similar steps as before, we find: 1. \( du = e^{x^{3}/2} \cdot \frac{3}{2} x^2 \, dx \) 2. \( v = \frac{x^6}{6} \) Thus, \[ J = \left[ e^{x^{3}/2} \cdot \frac{x^6}{6} \right]_{-2}^{0} - \int_{-2}^{0} \frac{x^6}{6} \cdot \left( e^{x^{3}/2} \cdot \frac{3}{2} x^2 \right) \, dx. \] Evaluating the boundary term: At \( x = 0 \): \[ e^{0} \cdot \frac{0^6}{6} = 0. \] At \( x = -2 \): \[ e^{-4} \cdot \frac{(-2)^6}{6} = e^{-4} \cdot \frac{64}{6} = \frac{32}{3} e^{-4}. \] Thus, the boundary term evaluates to: \[ 0 - \frac{32}{3}