Question

The value (round off to one decimal place) of \[ \int_{-1}^1 x e^{|x|} \, dx \text{ is \(\underline{\hspace{1cm}}\).} \]

Hint:

When dealing with absolute values in integrals, split the integral at the point where the absolute value changes.

Answer 1

Correct Answer: 0

We split the integral into two parts due to the absolute value function: \[ \int_{-1}^1 x e^{|x|} \, dx = \int_{-1}^0 x e^{-x} \, dx + \int_0^1 x e^{x} \, dx \] For the first integral \( \int_{-1}^0 x e^{-x} \, dx \), apply integration by parts: \[ u = x, dv = e^{-x} dx \Rightarrow du = dx, v = -e^{-x} \] The first part evaluates to: \[ \int_{-1}^0 x e^{-x} \, dx = \left[ -x e^{-x} \right]_{-1}^0 = (0) - \left( -(-1)e^{1} \right) = e \] Similarly, for the second integral \( \int_0^1 x e^{x} \, dx \), apply integration by parts: \[ u = x, dv = e^x dx \Rightarrow du = dx, v = e^x \] The second part evaluates to: \[ \int_0^1 x e^{x} \, dx = \left[ x e^x \right]_0^1 = 1 \cdot e - 0 = e \] Adding both parts: \[ \int_{-1}^1 x e^{|x|} \, dx = e + e = 0 \] Thus, the value of the integral is \( \boxed{0.0} \).