Question4. $[2.945 .88$ Points]
DETAILS MY NOTES SCALCET9 6.XP.1.UOZ. Sketch the region endosed by the given curves. Decide whether to integrate with respect to $x$ or $y$. Draw a typical approximating rectangle. $$y=x^{2}-3 x, y=2 x+6$$ fod the wee ef the regian. $\square$ Aned Heip? thanes

Studdy Solution
Evaluate the integral: $$A = \int_{-1}^{6} (-x^2 + 5x + 6) \, dx$$
Find the antiderivative: $$\int (-x^2 + 5x + 6) \, dx = -\frac{x^3}{3} + \frac{5x^2}{2} + 6x$$
Evaluate the antiderivative at the limits $x = 6$ and $x = -1$: $$A = \left[ -\frac{6^3}{3} + \frac{5 \cdot 6^2}{2} + 6 \cdot 6 \right] - \left[ -\frac{(-1)^3}{3} + \frac{5 \cdot (-1)^2}{2} + 6 \cdot (-1) \right]$$ $$A = \left[ -\frac{216}{3} + \frac{5 \cdot 36}{2} + 36 \right] - \left[ \frac{1}{3} + \frac{5}{2} - 6 \right]$$ $$A = \left[ -72 + 90 + 36 \right] - \left[ \frac{1}{3} + \frac{5}{2} - 6 \right]$$ $$A = 54 - \left[ \frac{1}{3} + \frac{15}{6} - 6 \right]$$ $$A = 54 - \left[ \frac{1 + 15 - 36}{6} \right]$$ $$A = 54 - \left[ \frac{-20}{6} \right]$$ $$A = 54 + \frac{10}{3}$$ $$A = 54 + 3.33$$ $$A = 57.33$$
Solution: $$A = 57.33$$