Questions: ∫ from 0 to 1 (u+2)(u-3) du

Solution Steps

To evaluate the integral, we first expand the integrand \((u+2)(u-3)\) and then integrate the resulting polynomial term by term over the interval from 0 to 1.

1. Expand the integrand \((u+2)(u-3)\).
2. Integrate the resulting polynomial term by term.
3. Evaluate the definite integral from 0 to 1.

We start with the integrand \((u + 2)(u - 3)\). Expanding this expression gives: \[ (u + 2)(u - 3) = u^2 - 3u + 2u - 6 = u^2 - u - 6 \]

Next, we integrate the polynomial \(u^2 - u - 6\) over the interval from 0 to 1: \[ \int_{0}^{1} (u^2 - u - 6) \, du \] Calculating the integral term by term: \[ \int u^2 \, du = \frac{u^3}{3}, \quad \int u \, du = \frac{u^2}{2}, \quad \int 6 \, du = 6u \] Thus, the integral becomes: \[ \left[ \frac{u^3}{3} - \frac{u^2}{2} - 6u \right]_{0}^{1} \]

Now we evaluate the expression at the bounds: \[ \left( \frac{1^3}{3} - \frac{1^2}{2} - 6 \cdot 1 \right) - \left( \frac{0^3}{3} - \frac{0^2}{2} - 6 \cdot 0 \right) = \left( \frac{1}{3} - \frac{1}{2} - 6 \right) \] Calculating this gives: \[ \frac{1}{3} - \frac{3}{6} - \frac{36}{6} = \frac{1 - 3 - 36}{6} = \frac{-38}{6} = -\frac{19}{3} \]

Final Answer