What is the value of \( \int_{0}^{3} x^{2} e^{x} d x \) ? Enter your answer as an expression in terms of "e" below. To symbolize exponents, use "^".

To evaluate the integral \( \int_{0}^{3} x^{2} e^{x} \, dx \), we can use integration by parts. We will apply integration by parts twice. First, let's set: - \( u = x^2 \) and \( dv = e^x \, dx \). Then we have: - \( du = 2x \, dx \) and \( v = e^x \). Using integration by parts: \[ \int u \, dv = uv - \int v \, du \] This gives us: \[ \int x^2 e^x \, dx = x^2 e^x - \int e^x (2x) \, dx \] Now we need to evaluate \( \int 2x e^x \, dx \) using integration by parts again. Set: - \( u = 2x \) and \( dv = e^x \, dx \). Then: - \( du = 2 \, dx \) and \( v = e^x \). Applying integration by parts again: \[ \int 2x e^x \, dx = 2x e^x - \int e^x (2) \, dx = 2x e^x - 2e^x \] Now we can substitute this back into our first integration result: \[ \int x^2 e^x \, dx = x^2 e^x - (2x e^x - 2e^x) = x^2 e^x - 2x e^x + 2e^x = (x^2 - 2x + 2)e^x \] Now, we evaluate this from \( 0 \) to \( 3 \): \[ \int_{0}^{3} x^{2} e^{x} \, dx = \left[ (x^2 - 2x + 2)e^x \right]_{0}^{3} \] Calculating at the upper limit \( x = 3 \): \[ (3^2 - 2(3) + 2)e^3 = (9 - 6 + 2)e^3 = 5e^3 \] Calculating at the lower limit \( x = 0 \): \[ (0^2 - 2(0) + 2)e^0 = 2 \cdot 1 = 2 \] Thus, the definite integral evaluates to: \[ \int_{0}^{3} x^{2} e^{x} \, dx = 5e^3 - 2 \] So the final answer is: \[ 5e^3 - 2 \]