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Engenharia de Computação ·
Cálculo 1
· 2021/2
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Resolva as integrais a seguir, apresente todos os passos utilizados em sua resolução. a) ∫ \sin(\frac {x}{2}) dx b) ∫_{3}^{5} [ \cos(3x) - \frac {1}{7} \sin(7x)] dx c) ∫ e^x x^2 dx d) ∫ \frac {x^2 + 2}{x^2 + 5} dx e) ∫ \frac {\cos^4(x)}{\sin^4(x)} dx f) ∫ (x + e^x)^2 dx g) ∫ \cos(3x) \cos(7x) dx h) ∫_{0}^{\infty} e^{-x^2} \cos(2x) dx i) ∫ \sqrt{x^2 - 1} dx j) ∫ x^3 + 5x + 4 dx Exercício 2 (1,50 pontos) Seja n um número natural, com n >= 2. Mostre que: ∫ \cos^n(x) dx = \frac{1}{n} \cos^{n-1}(x) \sin(x) + \frac{n-1}{n} ∫ \cos^{n-2}(x) dx Exercicio 3 (1,50 pontos) Verifique que para todo natural n >= 1 e todo real s > 0 ∫ [e^{s-nt} dt = \frac{1}{s} [ e^s + e^{-nt}] ∫ [e^{s-nt} dt Exercicio 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) = e^x [f'(x) - f(x)], x ∈ R, é constante. 1. a) ∫ \sin(\frac{x}{2}) dx => \frac{x}{2} = u => dx = du => dx = 2 du => ∫ 2 \sin(u) du = -2 \cos(u) + c = -2 \cos(\frac{x}{2}) + c b) ∫_{3}^{5} [\frac{1}{3} \cos(3x) - \frac{1}{7} \sin(7x)] dx = \frac {1}{3} ∫ \cos(3x) dx - \frac {1}{7} ∫ \sin(7x) dx ∫ \cos(3x) dx => 3x = u => 3dx = du => dx = du/3 ∫ \cos(u) du = \frac{1}{3} \sin(u) + C = \frac{1}{3} \sin(3x) + C_1 ∫ \sin(7x) dx => 7x = u => 7dx = du => dx = du/7 => ∫ \sin(u) du = -\frac{1}{7} \cos(u) + C_2 = -\frac{1}{7} \cos(7x) + C_2 => ∫ [\frac{1}{3} \cos(3x) - \frac{1}{7} \sin(7x)] dx = \frac{1}{9} \sin(3x) + \frac{1}{49} \cos(7x) + c Digitalizado com CamScanner (a) ∫₀¹ x e^{-x²} dx = => -x² = u => -2x dx = du => x dx = -du/2 x=0 => u=0 x=1 => u=-1 => ∫₀⁻¹ e^{u} [-du/2] = -∫⁻¹⁰ e^{u} [-du/2] = 1/2 ∫⁻¹⁰ e^{u} du => 1/2 e^{u} |⁰⁻¹ = 1/2 [e⁰ - e⁻¹] = 1/2 [-1 - 1/e] @ ∫ x+2 dx / x³+2x²+5x x³+2x ²+5x = x(x²+2x+5) => x((x+1)² + 4) => ∫ x+2 dx / x³+2x²+5x = ∫[ x+2 / x((x+1)²+4)] dx = ∫ [A/x + Bx+C/(x+1)²+4] dx => A((x+1)²+4) + (Bx+C)x = x+2 So x=0 => A(1²+4) + 0 = 2 => A=2/5 => 2/5 [(x+1)²+4] + x(Bx+C) = x+2 => 2/5 (x²+2x+5) + Bx²+Cx = x+2 => 2/5 + B=0 => B=-2/5 => 2.2 + C = 1 => C=1/5 ∫ x+2 dx / x³+2x²+5x = ∫ [1/5/x + -2x/5+1/5 / (x+1)²+4] dx = 1/5 ∫ dx/x + ∫ -2x+1 dx / (x+1)²+4 ∫ dx/x = ln |x| + C₁ ∫ -2x+1 dx / x²+2x+5 = ∫ -2x-2+3 dx / x²+2x+5 = ∫ [-2x+2/ x²+2x+5 + 3 dx / (x+1)²+4] ∫ -(2x+2) dx / x²+2x+5 => x²+2x+5 = u => (2x+2)dx=du = -∫ du/u = -ln |u| + C₂ = -ln(x²+2x+5) + C₂ ∫ 3 dx / (x+1)²+4 = ∫ 3 dx/4((x+1)²/4+1) = 3/4 ∫ dx/(x+1)²/4+1 => dx/2 =du => 3/2 ∫ du/u²+1 = 3/2 arctig(u) + C₃ = 3/2 arctg(x+1/2) + C₃ 2) \int \frac{cos^4(x) \ dx}{sin^4(x)} = \int cotg^4(x) \ dx = \int cotg^2(x) \cdot cotg^2(x) \ dx =\int (1+cosec^2(x)) \ cdot cotg^2(x) \ dx = \int -cotg^2(x) + cotg^2(x) \cdot cosec^2(x) \ dx = -\int cotg^2(x) \ dx + \int cotg^2(x) \cdot cosec^2(x) \ dx = x + cotg(x) + C_1 \int cotg^2(x) \cdot cosec^2(x) \ dx \Rightarrow cotg(x) = u \Rightarrow -cosec^2(x) dx = du =-\int u^2du = -\frac{u^3}{3} + C_2 = -\frac{cotg^3(x)}{3} + C_2 \Rightarrow\int \frac{cos^4(x) \ dx}{sin^4(x)} = x + cotg(x) - \frac{cotg^3(x)}{3} + C 3) \int (x+e^x) \ dx = \int x \,dx + \int e^x \,dx = \frac{x^2}{2} + e^x + c @\int cos(3x) cos(4x) \ dx \Rightarrow A = 9x, \ B = 4x \Rightarrow\int cos(3x) cos(4x) \ dx = \int \frac{cos(8x) + cos(4x)}{2} \ dx = \frac{1}{2} \int cos(10x) \ dx + \int cos(4x) \, dx \Rightarrow \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 \Rightarrow 4x = u \Rightarrow 4 dx = du \Rightarrow \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 1. \int e^x \cos(2x) dx \int e^x \cos(2x) dx => \cos(2x) = u => -2\sin(2x)dx = du dx \cdot e^{-x} = dv => v = -e^{-x} \int e^x \cos(2x) dx = -e^{-x} \cos(2x) - 2 \int e^{-x} \sin(2x) dx \int e^{-x} \sin(2x) dx => \sin(2x) = u => 2\cos(2x)dx = du e^{-x} dx = dv => v = -e^{-x} \int e^{-x} \sin(2x) dx = -e^{-x} \sin(2x) + 2 \int e^{-x} \cos(2x) dx => \int e^{-x} \cos(2x) dx = -e^{-x} \cos(2x) + 2 e^{-x} \sin(2x) - 4 \int e^x \cos(2x) dx + C_3 => 5 \int e^{-x} \cos(2x) dx = e^{-x} (\sin(2x) - \cos(2x)) + C_3 => \int e^{-x} \cos(2x) dx = \frac{e^x (\sin(2x) - \cos(2x))}{5} + C 1) \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 = 2 \frac{u^7}{7} + 4 \frac{u^5}{5} + 2 \frac{u^3}{3} + C = 2u^3 (\frac{u^4}{7} + \frac{2u^2}{5} + \frac{1}{3}) + C = 2(x-1)^{\frac{3}{2}} (\frac{(x-1)^2}{7} + \frac{2(x-1)}{5} + \frac{1}{3}) + C e 2) \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 => 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 fme^{-st}dt => fm = u => mfm^{-s}dtdt = du e^{-st}dt = dv => v = -\frac{e^{-st}}{s} \int fme^{-st}dt = \frac{fme^{-st}}{s} + \frac{m}{s} \int fm^{-s} e^{-st} dt 4. f''(x) - f^\theta(x) = 0 \quad 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 \quad g''(x) = 0 g'(x) = 0 => g(x) = cte
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Resolva as integrais a seguir, apresente todos os passos utilizados em sua resolução. a) ∫ \sin(\frac {x}{2}) dx b) ∫_{3}^{5} [ \cos(3x) - \frac {1}{7} \sin(7x)] dx c) ∫ e^x x^2 dx d) ∫ \frac {x^2 + 2}{x^2 + 5} dx e) ∫ \frac {\cos^4(x)}{\sin^4(x)} dx f) ∫ (x + e^x)^2 dx g) ∫ \cos(3x) \cos(7x) dx h) ∫_{0}^{\infty} e^{-x^2} \cos(2x) dx i) ∫ \sqrt{x^2 - 1} dx j) ∫ x^3 + 5x + 4 dx Exercício 2 (1,50 pontos) Seja n um número natural, com n >= 2. Mostre que: ∫ \cos^n(x) dx = \frac{1}{n} \cos^{n-1}(x) \sin(x) + \frac{n-1}{n} ∫ \cos^{n-2}(x) dx Exercicio 3 (1,50 pontos) Verifique que para todo natural n >= 1 e todo real s > 0 ∫ [e^{s-nt} dt = \frac{1}{s} [ e^s + e^{-nt}] ∫ [e^{s-nt} dt Exercicio 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) = e^x [f'(x) - f(x)], x ∈ R, é constante. 1. a) ∫ \sin(\frac{x}{2}) dx => \frac{x}{2} = u => dx = du => dx = 2 du => ∫ 2 \sin(u) du = -2 \cos(u) + c = -2 \cos(\frac{x}{2}) + c b) ∫_{3}^{5} [\frac{1}{3} \cos(3x) - \frac{1}{7} \sin(7x)] dx = \frac {1}{3} ∫ \cos(3x) dx - \frac {1}{7} ∫ \sin(7x) dx ∫ \cos(3x) dx => 3x = u => 3dx = du => dx = du/3 ∫ \cos(u) du = \frac{1}{3} \sin(u) + C = \frac{1}{3} \sin(3x) + C_1 ∫ \sin(7x) dx => 7x = u => 7dx = du => dx = du/7 => ∫ \sin(u) du = -\frac{1}{7} \cos(u) + C_2 = -\frac{1}{7} \cos(7x) + C_2 => ∫ [\frac{1}{3} \cos(3x) - \frac{1}{7} \sin(7x)] dx = \frac{1}{9} \sin(3x) + \frac{1}{49} \cos(7x) + c Digitalizado com CamScanner (a) ∫₀¹ x e^{-x²} dx = => -x² = u => -2x dx = du => x dx = -du/2 x=0 => u=0 x=1 => u=-1 => ∫₀⁻¹ e^{u} [-du/2] = -∫⁻¹⁰ e^{u} [-du/2] = 1/2 ∫⁻¹⁰ e^{u} du => 1/2 e^{u} |⁰⁻¹ = 1/2 [e⁰ - e⁻¹] = 1/2 [-1 - 1/e] @ ∫ x+2 dx / x³+2x²+5x x³+2x ²+5x = x(x²+2x+5) => x((x+1)² + 4) => ∫ x+2 dx / x³+2x²+5x = ∫[ x+2 / x((x+1)²+4)] dx = ∫ [A/x + Bx+C/(x+1)²+4] dx => A((x+1)²+4) + (Bx+C)x = x+2 So x=0 => A(1²+4) + 0 = 2 => A=2/5 => 2/5 [(x+1)²+4] + x(Bx+C) = x+2 => 2/5 (x²+2x+5) + Bx²+Cx = x+2 => 2/5 + B=0 => B=-2/5 => 2.2 + C = 1 => C=1/5 ∫ x+2 dx / x³+2x²+5x = ∫ [1/5/x + -2x/5+1/5 / (x+1)²+4] dx = 1/5 ∫ dx/x + ∫ -2x+1 dx / (x+1)²+4 ∫ dx/x = ln |x| + C₁ ∫ -2x+1 dx / x²+2x+5 = ∫ -2x-2+3 dx / x²+2x+5 = ∫ [-2x+2/ x²+2x+5 + 3 dx / (x+1)²+4] ∫ -(2x+2) dx / x²+2x+5 => x²+2x+5 = u => (2x+2)dx=du = -∫ du/u = -ln |u| + C₂ = -ln(x²+2x+5) + C₂ ∫ 3 dx / (x+1)²+4 = ∫ 3 dx/4((x+1)²/4+1) = 3/4 ∫ dx/(x+1)²/4+1 => dx/2 =du => 3/2 ∫ du/u²+1 = 3/2 arctig(u) + C₃ = 3/2 arctg(x+1/2) + C₃ 2) \int \frac{cos^4(x) \ dx}{sin^4(x)} = \int cotg^4(x) \ dx = \int cotg^2(x) \cdot cotg^2(x) \ dx =\int (1+cosec^2(x)) \ cdot cotg^2(x) \ dx = \int -cotg^2(x) + cotg^2(x) \cdot cosec^2(x) \ dx = -\int cotg^2(x) \ dx + \int cotg^2(x) \cdot cosec^2(x) \ dx = x + cotg(x) + C_1 \int cotg^2(x) \cdot cosec^2(x) \ dx \Rightarrow cotg(x) = u \Rightarrow -cosec^2(x) dx = du =-\int u^2du = -\frac{u^3}{3} + C_2 = -\frac{cotg^3(x)}{3} + C_2 \Rightarrow\int \frac{cos^4(x) \ dx}{sin^4(x)} = x + cotg(x) - \frac{cotg^3(x)}{3} + C 3) \int (x+e^x) \ dx = \int x \,dx + \int e^x \,dx = \frac{x^2}{2} + e^x + c @\int cos(3x) cos(4x) \ dx \Rightarrow A = 9x, \ B = 4x \Rightarrow\int cos(3x) cos(4x) \ dx = \int \frac{cos(8x) + cos(4x)}{2} \ dx = \frac{1}{2} \int cos(10x) \ dx + \int cos(4x) \, dx \Rightarrow \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 \Rightarrow 4x = u \Rightarrow 4 dx = du \Rightarrow \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 1. \int e^x \cos(2x) dx \int e^x \cos(2x) dx => \cos(2x) = u => -2\sin(2x)dx = du dx \cdot e^{-x} = dv => v = -e^{-x} \int e^x \cos(2x) dx = -e^{-x} \cos(2x) - 2 \int e^{-x} \sin(2x) dx \int e^{-x} \sin(2x) dx => \sin(2x) = u => 2\cos(2x)dx = du e^{-x} dx = dv => v = -e^{-x} \int e^{-x} \sin(2x) dx = -e^{-x} \sin(2x) + 2 \int e^{-x} \cos(2x) dx => \int e^{-x} \cos(2x) dx = -e^{-x} \cos(2x) + 2 e^{-x} \sin(2x) - 4 \int e^x \cos(2x) dx + C_3 => 5 \int e^{-x} \cos(2x) dx = e^{-x} (\sin(2x) - \cos(2x)) + C_3 => \int e^{-x} \cos(2x) dx = \frac{e^x (\sin(2x) - \cos(2x))}{5} + C 1) \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 = 2 \frac{u^7}{7} + 4 \frac{u^5}{5} + 2 \frac{u^3}{3} + C = 2u^3 (\frac{u^4}{7} + \frac{2u^2}{5} + \frac{1}{3}) + C = 2(x-1)^{\frac{3}{2}} (\frac{(x-1)^2}{7} + \frac{2(x-1)}{5} + \frac{1}{3}) + C e 2) \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 => 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 fme^{-st}dt => fm = u => mfm^{-s}dtdt = du e^{-st}dt = dv => v = -\frac{e^{-st}}{s} \int fme^{-st}dt = \frac{fme^{-st}}{s} + \frac{m}{s} \int fm^{-s} e^{-st} dt 4. f''(x) - f^\theta(x) = 0 \quad 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 \quad g''(x) = 0 g'(x) = 0 => g(x) = cte