Lista - Integrais Impróprias - Cálculo 2 2021-2

Integrais Impr´oprias FAMAT-UFU P´agina 1 Este trabalho deve ser resolvido e entregue dia 05/07/22 INTEGRAIS IMPR´OPRIAS Determine se cada integral ´e convergente ou divergente. Calcule aquelas que s˜ao convergentes. a) ∞∫ 3 1 (x − 2) 3 2 dx b) 0∫ −∞ 1 3 − 4xdx c) ∞∫ −∞ xe−x2dx d) ∞∫ e 1 x (ln x)3dx e) ∞∫ 1 ln x x dx f) 1∫ 0 3 x5dx g) 0∫ −∞ xe2xdx h) 14∫ −2 1 4√x + 2dx i) 0∫ −∞ 1 1 + x2dx j) +∞ ∫ 2 1 x2 − 1dx k) 2∫ 0 1 √2 − xdx l) 1∫ 0 x √ 1 − x2dx m) π 2∫ − π 2 tg (x) dx n) 1∫ −1 1 |x|dx lais@ufu.br sites.google.com/site/laisufu La´ıs Rodrigues a) ∫[3,∞] 1/(x-2)^(3/2) dx -> ∫[1,∞] 1/(x-2)^(3/2) dx du = x-2 e du = dx ∫ [1,u^(3/2)] du/u^(3/2) = -2/sqrt(u) voltando = -2/(x-2) => aplicando os limites: -2/sqrt(x-2) [3,∞) (lim (x->∞) -2/sqrt(x-2)) - (-2) = 0 - (-2) = 2 b) ∫[3-4x,∞] 1/(3-4x) dx = -1/4 ln|3-4x| + C aplicando os limites => ∫[3-4x,∞] 0 dx DIVERGE c) ∫[-∞,∞] x * e^(-x^2) dx vs integral indefinida ∫x * e^(-x^2) dx substituição: u = -x^2 du = -2x dx => x dx = -du/2 ∫ -e^u/2 du => -1/2 ∫ e^u du = -1/2 e^u + C voltando u = -x^2 => -1/2 e^(-x^2) + C aplicando os limites: ∫[-∞,∞] x * e^(-x^2) dx = -1/2 e^(x^2) [∞,-∞] = -1/2 (e^(∞) - e^(-∞)) = 0 d) ∫[e,∞] dx/(x(ln x)^3) => ∫ dx/(x(ln x)^3) substituição u = ln x du = 1/x dx ∫ du/u^3 = -1/2u^2 (...) = -1/(2(ln x)^2) + C aplicando os limites: ∫[e,∞] x(ln x)^3 dx = lim (x->∞) (-1/(ln x)^2) - (-1) = 0 + 1 = 1/2 e) ∫[1,∞] ln x/x dx => ∫ ln x/x dx (u = ln x) du = 1/x dx ∫ du = u^2/2 + C = (ln x)^2/2 + C lim (x->∞) (ln x)^2/2 - ln⁡(x)^2/2 => DIVERGE f) ∫[0,∞] 3/(x^5) dx = 3∫[0,∞] dx/x^5 = 3 [-1/4x^4] [0,∞] = -3/4 + ∞ => DIVERGE g) ∫[-∞,∞] x*e^(2x) dx => ∫ x*e^(2x) dx (u = x) du = dx e^(u) = du e^(u) du = 1/4 [u e^(2x) - ∫ e^(2x) du] (INTEGRAÇÃO POR PARTES) 1/4 (e^(u) * u - ∫ e^(u) du) = 1/4 (e^(u)u - e^(u)) = 1/4 [e^(2x) * x - e^(2x)] + C aplicando os limites: ∫[-∞,∞] x*e^(2x) dx = -1/4 h) ∫[-2,∞] 14/(sqrt(x+2)) dx u = x + 2 du = 1*dx ∫[0,∞] 14/u^(1/2) du = (1/3) u^(3/2) => (4/3) (x+2)^(3/2) |_2 = ((32/3) - 0) = 32/3 i) ∫[-∞,∞] dx/(1+x^2) n-> ∫ dx = arctg(x)+C Aplicando os limites: ∫[-∞,∞] dx arctg(0) - lim arctg(x) (x->-∞) = 0 - (-π/2) = π/2 j) ∫[∞,∞] dx/(x^2-x) = ∫ dx/(x(x-1)) = (1/x(x-1)) dx a integração -(ln|x+1| - ln|x-1|)/2 + C ➯ aplicando os limites ∫[∞,∞] = -(ln|x+1|-ln|x-1|)/2 => 0 - (-ln(3)/2) k) ∫[0,2] 1/sqrt(2x-x) dx u substituição: u = 2-x du = -dx -∫[0,2] du/sqrt(u) -(-∫[0,2] du/sqrt(u)) = -(2u^(1/2)/1/2) |⟩0 = 2(sqrt(2)) e) \int_0^1 \frac{x}{\sqrt{1-x^2}} dx \text{ substituicao: } u = 1-x^2 \ xdx = -\frac{du}{2} \ \rightarrow -du = 2xdx -\int \frac{du}{2\cdot \sqrt{u}} = -\frac{1}{2} \int \frac{du}{\sqrt{u}} = -\frac{1}{2} (2\sqrt{u}) \text{dado que } u = 1-x^2: - \frac{1}{2} \left. (x\cdot\sqrt{1-x^2}) \right|_0^1 = -\left. \sqrt{1-x^2} \right|_0^1 = \sqrt{0} - (-\sqrt{1}) = 1 m) \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \text{tg}(x)dx \rightarrow \text{funcao impar: } \ f(x) = \text{tg}(x) \ f(-x) = \text{tg}(-x) = -\text{tg}(x) \text{logo: } \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \text{tg}(x)dx = \boxed{0} n) \int_{-1}^1 \frac{1}{|x|} dx \Rightarrow \int_{-1}^0 \frac{1}{-x} dx + \int_0^1 \frac{1}{x} dx \left(\text{\small DIVERGE} \right) \approx \frac{1}{0} \int_{-1}^1 \frac{1}{|x|} dx \rightarrow \boxed{\text{DIVERGE}}