Exercícios Resolvidos Integrais-2021 2

02)\nf(x) = x^2 + 2\ng(x) = -x^2 + 10\nA = \int_{-2}^{2}[(x^2+2) - (-x^2+10)]dx\n\nA = \int_{-2}^{2}(x^2+2)dx - \int_{-2}^{2}(-x^2+10)dx\n\nA = \left[\frac{x^3}{3} + 2x\right]_{-2}^{2} - \left[\frac{-x^3}{3} +10x\right]_{-2}^{2}\n\nA = \left[\frac{2^3}{3}+2\cdot2\right] - \left[\left(-\frac{2^3}{3}\right)+10\cdot2\right] - \left[(-\frac{-2^3}{3})+10\cdot(-2)\right]\n\nA = \left[\frac{8}{3} + 4\right] - \left[-\frac{8}{3} - 9\right] - \left[-\frac{8}{3} + 20\right] - \left[+\frac{8}{3} - 20\right]\n\nA = \left[\frac{20}{3} + \frac{20}{3}\right] - \left[\frac{50}{3} - \left(-\frac{50}{3}\right)\right]\n\nA = \frac{40}{3} - \left[\frac{10}{3}\right] = \frac{64}{3}\n\nV = \pi\int_{-2}^{2}[(x^2+2)^2 - (-x^2+10)^2]dx\n\nV = \pi\int_{-2}^{2}[x^4 + 4x^2 + 4 - (x^4 - 20x^2 + 100)]dx\n\nV = \pi\int_{-2}^{2}[x^4 + 4x^2 + 4 - x^4 + 20x^2 -100]dx = \pi\int_{-2}^{2}[24x^2 - 96]dx\n\nV = \pi\left[\frac{8x^3}{3} - 96x\right]_{-2}^{2} = \pi\left[(8\cdot(2)^3 - 96\cdot2) - (8\cdot(-2)^3 - 96\cdot(-2))\right]\n\nV = \pi\left[(64 - 192) - (-64 + 192)\right]\n\nV = \pi(256) = 256\pi 03)\na)\n\int_{0}^{2}2x-1 \, dx = \left[x^2-x\right]_0^2 = \left[2^2-2\cdot2-(0^2-0)\right] = 2J\\nb)\nf(x) = 2\sqrt[3]{x}\nh) \text{at}\ x = 1 \text{ até}\ x = 3\n\int_{1}^{3}2\sqrt[3]{x} \, dx = F\x^{3/2}, dx = 2\left[\frac{2}{5} x^{5/2}\right]_1^3 = \frac{4}{5}\left[(3^{5/2}) - (1^{5/2})\right]\n= \frac{4}{5}\left(\sqrt{243}-1\right) = \frac{4\sqrt{243}}{5} - \frac{4}{5}\nConceito = passo4) \nAntiderivada de\ f(x) = x^2-2x\, tal\ que\,\ F(1)= -1\n\int x^2-2x \, dx = \frac{x^3}{3} - x^2 + C\n\frac{1^3}{3} - 1^2 + C = -1\n\frac{1}{3} - 1 + C = -1\nC = -\frac{1}{3}\n\n05)\n\int_{0}^{3} (x^2+1)dx\n\int_{0}^{3} (x^2+1) \, dx = \left[\frac{x^3}{3} + x\right]_0^{3} = \left[\left(\frac{3^3}{3} + 3\right) - \left(0^3/3 + 0\right)\right] = 12\n\n06)\nf(x) = 2x [\text{Inclinação da reta tangente}]\nF(x) = \int 2x dx = \frac{2x^2}{2} + C = x^2 + C\nComo a reta tangente \;e \; no \; ponto (1,4), \text{ então} \;F(1) = 4.\nF(1) = (1)^2 + C = 4\nC = 4 - 1\nC = 3.\nLogo, \;F(x) = x^2 + 3. 07)\n\nA = \lim_{\Delta x \to 0}\sum_{n=a}^{m}f(\xi_{c}) \Delta x^{*}\n\n\Delta x = \frac{5-2}{3} = \Delta x = 1\n\nS_3 = f(2),1 + f(3),1 + f(4),1\nS_3 = 9\n\n\int_{3}^{5} x \, dx =\n\n\frac{x^2}{2} = \frac{25}{2} + 2 = \frac{29}{2} = 10.5\n\n08)\na(t) = 3 \,m/s^2\nv(t) = 3t+C\nv(t) = 3t - 2\n\np(t) = \frac{3}{2}t^2 - 2t + C\nv(1) = 3\cdot1 + C = 1\nC = -2\np(1) = \frac{3}{2} - 2 + C = 2\nC = -\frac{3}{2} f(x) = x^2 + C = x^2 + 1 \;\text{então}\, \int_{3}^{5}x\, dx = \frac{29}{2} - 2 = 10.5 \,\text{em}, \n\n v(t) = 3t - 2 \text{(m/s)}