Lista 5 Mat1 Resolvida-2022 2

1) A = \int_{a}^{b} [t(x) - q(x)] dx A = \int_{0}^{2} (3x - x^2) dx = \int_{0}^{2} 3x - x^2 dx = \left[\frac{2x^2}{2} - \frac{x^3}{3}\right]_{0}^{2} = \frac{x^2}{2} - \frac{x^3}{3}\Bigg|_{0}^{2} = \left(4 - \frac{8}{3}\right) - 0 = \left(\frac{12 - 8}{3}\right) = \boxed{\frac{4}{3} u.a.} 2) A = \int_{a}^{b} [t(x) - q(x)] dx A = \int_{0}^{1} e^x - x^2 dx = \int_{0}^{1} e^x dx - \int_{0}^{1} x^2 dx = \left[e^x - \frac{x^3}{3}\right]_{0}^{1} = \left(e^1 - e^0\right) - \left(\frac{1}{3} - 0\right) = e - 1 - \frac{1}{3} = e - \frac{4}{3} u.a. 3) v(t,k) = t^2 e^{-t} Como v(t,k) \geq 0, \forall x \in \mathbb{R}, basta calculamoss \int_{0}^{10} t^2 e^{-t} dt \ (Integracao\ por\ partes), u = t^2 \hspace{5mm} dv = e^{-t} \\ du = 2t \cdot dt \hspace{5mm} v = -e^{-t} - t^2 e^{-t} + 2 \left[\int t e^{-t} dt\right] = -t^2 e^{-t} + 2 \left[\int u e^{-t} du\right] = - t^2 e^{-t} + 2 \left( -t \cdot e^{-t} + \int e^{-t} dt \right) = - t^2 e^{-t} - 2 t e^{-t} - 2 e^{-t} = -e^{-t} (t^2 + 2t + 2) \left[-e^{-t} (t^2 + 2t + 2)\right]_{0}^{10} = -e^{-10} (122) + 2 = \frac{2 - 122}{e^{10}} = \boxed{1,99}