P2 - Cálculo 1 2021 2

Universidade Federal do Mato Grosso Faculdade de engenharia de Várzea Grande Disciplina: Cálculo I Professor Dr. Fábio Santiago Data:14/07/2022 Nome: RGA: Nome: RGA: Nome: RGA: Nome: RGA: Nome: RGA: Exercício 1) (5 pontos). Resolva as integrais a seguir, apresente todos os passos utilizados em sua resolução. a) ∫ 𝑠𝑒𝑛 (𝑥 2) 𝑑𝑥 f) 𝑏) ∫(𝑥 + 𝑒𝑥)𝑑𝑥 b) ∫ 1 3 cos(3𝑥) − 1 7 𝑠𝑒𝑛(7𝑥)𝑑𝑥 g) 𝑑) ∫ cos(3𝑥) 𝑐𝑜𝑠(7𝑥)𝑑𝑥 c) ∫ 𝑥𝑒−𝑥2 1 0 𝑑𝑥 h) 𝑓) ∫ 𝑒−𝑥 1 0 cos (2𝑥)𝑑𝑥 d) ∫ 𝑥 + 2 𝑥3 + 2𝑥2 + 5𝑥 i) ℎ) ∫ 𝑥2√𝑥 − 1 𝑑𝑥 e) ∫ cos4(𝑥) 𝑠𝑒𝑛4(𝑥) 𝑑𝑥 j) ∫ 𝑥2 + 5𝑥 + 4 𝑑𝑥 Exercício2) (1,50 pontos) Seja 𝑛 um número natural, com 𝑛 ≥ 2. Mostre que: ∫ cos𝑛(𝑥) 𝑑𝑥 = 1 𝑛 cosn−1(𝑥) 𝑠𝑒𝑛(𝑥) + 𝑛 − 1 𝑛 ∫ cos𝑛−2 (𝑥) 𝑑𝑥 Exercício 3) (1,50 pontos) Verifique que para todo natural 𝑛 ≥ 1 e todo real 𝑠 > 0 ∫ 𝑡𝑛𝑒−𝑠𝑡 𝑑𝑡 = − 1 𝑠 𝑡𝑛𝑒−𝑠𝑡 + 𝑛 𝑠 ∫ 𝑡𝑛−1𝑒−𝑠𝑡 𝑑𝑡 Exercício 4 ) (2,0 pontos) Seja 𝑓: ℝ → ℝ derivável até segunda ordem, e tal que, para todo 𝑥, 𝑓′′(𝑥) − 𝑓(𝑥) = 0. Prove que 𝑔(𝑥) = 𝑒𝑥[𝑓′(𝑥) − 𝑓(𝑥)], 𝑥 ∈ ℝ, é constante. 1. \( \int \sin(\frac{x}{2})dx \) => \frac{x}{2}=u => dx=2du => \int 2 \sin(u)du = -2 \cos(u) + c = -2 \cos(\frac{x}{2}) + c 2) \left[ \frac{1}{3} \int \cos(3x) - \frac{1}{7} \sin(7x) \right] dx = \frac{1}{3} \int \cos(3x) dx - \frac{1}{7} \int \sin(7x) dx \int \cos(3x) dx => 3x = u => 3dx=du => dx = \frac{du}{3} => \frac{1}{3} \int \cos(u) du = \frac{1}{3} \sin(u) + C_1 = \frac{1}{3} \sin(3x) + C_1 \int \sin(7x)dx => 7x = u => 7dx=du => dx = \frac{du}{7} => \frac{1}{7} \int \sin(u) du = -\frac{1}{7} \cos(u) + C_2 = -\frac{1}{7} \cos(7x) + C_2 => \int \left[ \frac{1}{3} \cos(3x) - \frac{1}{7} \sin(7x) \right] dx = \frac{1}{9} \sin(3x) + \frac{1}{49} \cos(7x) + c 3) \int_{0}^{1} x e^{-x^2} dx => -x^2 = u => -2xdx = du => xdx = -\frac{du}{2} x=0 => u=0 \quad x=1 => u=-1 => \int_{0}^{-1} e^u \left(-\frac{du}{2}\right) = -\int_{0}^{-1} e^u \left(-\frac{du}{2}\right) = \frac{1}{2} \int_{-1}^{0} e^u du => \frac{1}{2} e^u \big|_{-1}^{0} = \frac{1}{2} [e^0 - e^{-1}] = \frac{1}{2}[1 - \frac{1}{e}] 4) \int \frac{x+2}{x^3+2x^2+5x} dx \quad x^3 + 2x^2 + 5x = x(x^2 + 2x + 5) => x((x+1)^2 + 4) => \int \frac{x+2}{x((x+1)^2 + 4)} dx = \int \frac{A}{x} + \frac{Bx + C}{(x+1)^2 + 4} dx => A((x+1)^2 +4) + x(Bx + C) = x+2 So, x=0 => A(1^2+4) + 0 = 2 => A = \frac{2}{5} => \frac{2}{5}((x+1)^2 + 4) + x(Bx + C) = x+2 => \frac{2}{5}(x^2 + 2x + 5) + Bx^2 + Cx = x+2 => \frac{2}{5} + B = 0 => B = -\frac{2}{5} => 2 \cdot 2 + C = 1 => C = \frac{1}{5} \int \frac{x+2}{x^3 + 2x^2 + 5x} dx = \int \left[\frac{1/5}{x} + \frac{-2x/5 + 1/5}{(x+1)^2 + 4}\right] dx = \frac{1}{5} \int \frac{dx}{x} + \int \frac{-2x+1}{(x+1)^2+4} dx \int \frac{dx}{x} = \ln |x| + C_1 -\frac{\int -2x + 3 \ dx}{x^2 + 2x + 5} = \int \frac{-2x - 2 + 3 \ dx}{x^2 + 2x+5} = \int \frac{-(2x+2) \ dx}{x^2 + 2x+5} + \int \frac{3 \ dx}{(x+1)^2+4} \int \frac{-(2x+2) \ dx}{x^2 + 2x+5} = \rightarrow x^2 + 2x + 5 = u = \rightarrow (2x+2) dx = du = -\int \frac{du}{u} = -\ln |u| + C_2 = -\ln(x^2 + 2x + 5) + C_2 \int \frac{3 \ dx}{(x+1)^2+4} = \int \frac{3 dx}{4((x+1)^2 / 4 + 1)} = \frac{3}{4} \int \frac{dx}{(\frac{x+1}{2})^2+1} = \rightarrow \frac{x+1}{2} = u \Rightarrow \frac{dx}{2} = du \Rightarrow \frac{3}{2} \int \frac{du}{u^2 + 1} = \frac{3}{2} \arctan(u) + C_3 = \frac{3}{2} \arctan (\frac{x+1}{2}) + C_3 => \frac{1}{10} \int \cos(u)\ du = \frac{1}{10} \sen(u) + C_1 = \frac{1}{10} \sen(10x) + C_1 \int \cos(4x)\ dx => 4x = u => 4\ dx = du => \frac{1}{4} \int \cos(u)\ du = \frac{1}{4} \sen(u) + C_2 = \frac{1}{4} \sen(4x) + C_2 \int \cos(x) \cos(3x)\ dx = \frac{1}{10} \sen(10x) + \frac{1}{4} \sen(4x) + C \int e^{-x} \cos(2x)\ dx \int e^{-x} \cos(2x)\ dx => \cos(2x) = u => -2 \sen(2x)\ dx = du dx.e^{-x} = dv => v = -e^{-x} \int e^{-x} \cos(2x)\ dx = -e^{-x} \cos(2x) - 2 \int e^{-x} \sen(2x)\ dx \int e^{-x} \sen(2x)\ dx => \sen(2x) = u => 2 \cos(2x)\ dx = du e^{-x} \ dx = dv => v = -e^{-x} \int e^{-x} \sen(2x)\ dx = -e^{-x} \sen(2x) + 2 \int e^{-x} \cos(2x)\ dx => \int e^{-x} \cos(2x)\ dx = -e^{-x} \cos(2x) + 2e^{-x} \sen(2x) - 4 \int e^{-x} \cos(2x)\ dx + C_3 => 5 \int e^{-x} \cos(2x)\ dx = e^{-x} (2 \sen(2x) - \cos(2x)) + C_3 => \int e^{-x} \cos(2x)\ dx = \frac{e^{-x} (2 \sen(2x) - \cos(2x))}{5} + C \int x^2 \sqrt{x-1}\ dx => u = \sqrt{x-1} => du = \frac{dx}{2 \sqrt{x-1}} => dx = 2u\ du u^2 = x - 1 => x = u^2 + 1 => \int (u^2+1)^2 \cdot u = 2u\ du = \int (u^4 + 2u^2 + 1) \cdot 2u^2\ du = \int (2u^6 + 4u^4 + 2u^2)\ du = \frac{2u^7}{7} + \frac{4u^5}{5} + \frac{2u^3}{3} + C = 2u^3 \left| \frac{u^4}{7} + \frac{2u^2}{5} + \frac{1}{3} \right| + C = 2(x-1)^{\frac{3}{2}} \left( \frac{(x-1)^2}{7} + \frac{2(x-1)}{5} + \frac{1}{3} \right) + C * \int (x^2+5x+9)\ dx = \frac{x^3}{3} + \frac{5x^2}{2} + 4x + c 2. \int cos^m(x) dx = \int cos^{m-1}(x) cos(x) dx cos^{m-1}(x) = u => (m-1) cos^{m-2} \cdot (-sin(x)) dx = du cos(x) dx = dv => v = sin(x) => \int cos^m(x) dx = cos^{m-1}(x) sin(x) + (m-1) \int cos^{m-2}(x) sin^2(x) dx = cos^{m-1}(x) sin(x) + (m-1) \int cos^{m-2}(x) (1 - cos^2(x)) dx = cos^{m-1}(x) sin(x) + (m-1) [ \int cos^{m-2}(x) - cos^m(x) ] dx = cos^{m-1}(x) sin(x) + (m-1) \int cos^{m-2}(x) dx - (m-1) \int cos^m(x) dx => m \int cos^m(x) dx = cos^{m-1}(x) sin(x) + (m-1) \int cos^{m-2}(x) dx => \int cos^m(x) dx = \frac{1}{m} cos^{m-1}(x) sin(x) + \frac{m-1}{m} \int cos^{m-2}(x) dx 3. \int t^m e^{-st} dt => fm = u => mt^{m-1} dt = du e^{-st} dt = dv => v = -\frac{e^{-st}}{s} \int t^m e^{-st} dt = -\frac{t^m e^{-st}}{s} + \frac{m}{s} \int t^{m-1} e^{-st} dt 4. f''(x) - f^0(x) = 0 g(x) = e^x(f'(x)-f(x)) f''(x) - f(x) = 0 => f''(x) = f(x) g'(x) = e^x(f'(x)-f(x)) + e^x(f''(x)-f'(x)) = e^x(f'(x)-f(x) + f''(x)-f'(x)) = 0 g''(x) = 0 => g(x) = cte