University:
Johns Hopkins University
Course:
AS.110.601 | Algebra I
Academic year:

2024
Views:

206

Pages:

5
Author:

Cristian Santos

Question 1 Evaluate the integral: ∫ 0 ∞ ln ⁡ ( 𝑥 2 + 1 ) 𝑥 2 + 1 𝑑 𝑥 . ∫ 0 ∞ x 2 +1 ln(x 2 +1) dx. Answer 1 Use substitution 𝑥 = tan ⁡ ( 𝜃 ) x=tan(θ), so 𝑑 𝑥 = sec ⁡ 2 ( 𝜃 ) 𝑑 𝜃 dx=sec 2 (θ)dθ and 𝑥 2 + 1 = sec ⁡ 2 ( 𝜃 ) x 2 +1=sec 2 (θ). Rewrite the integral: ∫ 0 𝜋 / 2 ln ⁡ ( sec ⁡ 2 ( 𝜃 ) ) 𝑑 𝜃 . ∫ 0 π/2 ln(sec 2 (θ))dθ. Simplify: ∫ 0 𝜋 / 2 2 ln ⁡ ( cos ⁡ ( 𝜃 ) ) 𝑑 𝜃 . ∫ 0 π/2 2ln(cos(θ))dθ. Let 𝐼 = ∫ 0 𝜋 / 2 ln ⁡ ( cos ⁡ ( 𝜃 ) ) 𝑑 𝜃 I=∫ 0 π/2 ln(cos(θ))dθ. Similarly: 𝐼 = ∫ 0 𝜋 / 2 ln ⁡ ( sin ⁡ ( 𝜃 ) ) 𝑑 𝜃 . I=∫ 0 π/2 ln(sin(θ))dθ. Add both equations: 2 𝐼 = ∫ 0 𝜋 / 2 ln ⁡ ( sin ⁡ ( 2 𝜃 ) ) 𝑑 𝜃 − ∫ 0 𝜋 / 2 ln ⁡ ( 2 ) 𝑑 𝜃 . 2I=∫ 0 π/2 ln(sin(2θ))dθ−∫ 0 π/2 ln(2)dθ. Simplify: 2 𝐼 = ∫ 0 𝜋 / 2 ln ⁡ ( sin ⁡ ( 2 𝜃 ) ) 𝑑 𝜃 − 𝜋 2 ln ⁡ ( 2 ) . 2I=∫ 0 π/2 ln(sin(2θ))dθ− 2 π ln(2). Evaluate: 𝐼 = − 𝜋 2 ln ⁡ ( 2 ) . I=− 2 π ln(2). Thus: ∫ 0 ∞ ln ⁡ ( 𝑥 2 + 1 ) 𝑥 2 + 1 𝑑 𝑥 = 𝜋 ln ⁡ ( 2 ) . ∫ 0 ∞ x 2 +1 ln(x 2 +1) dx=πln(2). Question 2 Solving limits: 𝑓 ( 𝑥 ) = lim ⁡ 𝑛 → ∞ ( 𝑛 ∫ 0 𝜋 / 4 ( tan ⁡ 𝑥 ) 𝑛 𝑑 𝑥 ) . f(x)= n→∞ lim (n∫ 0 π/4 (tanx) n dx). Answer 1 Use the substitution 𝑢 = tan ⁡ ( 𝑥 ) u=tan(x), so 𝑑 𝑢 = sec ⁡ 2 ( 𝑥 ) 𝑑 𝑥 du=sec 2 (x)dx. The integral becomes: 𝐼 𝑛 = ∫ 0 1 𝑢 𝑛 1 + 𝑢 2 𝑑 𝑢 . I n =∫ 0 1 1+u 2 u n du. Integrate by parts and analyze the bounds as 𝑛 → ∞ n→∞. Final result: lim ⁡ 𝑛 → ∞ 𝑛 ∫ 0 𝜋 / 4 ( tan ⁡ 𝑥 ) 𝑛 𝑑 𝑥 = 1 2 . n→∞ lim n∫ 0 π/4 (tanx) n dx= 2 1 . Question 3 Calculate: ∫ 0 𝜋 / 2 arccos ⁡ ( sin ⁡ 𝑥 ) 𝑑 𝑥 . ∫ 0 π/2 arccos(sinx)dx. Answer 1 Use symmetry properties: arccos ⁡ ( sin ⁡ 𝑥 ) = 𝜋 2 − arcsin ⁡ ( 𝑥 ) . arccos(sinx)= 2 π −arcsin(x). Combine integrals and simplify: ∫ 0 𝜋 / 2 arccos ⁡ ( sin ⁡ 𝑥 ) 𝑑 𝑥 = 𝜋 2 8 . ∫ 0 π/2 arccos(sinx)dx= 8 π 2 . Question 4 Evaluate: ∫ ln ⁡ ( 𝑥 ) 𝑑 𝑥 . ∫ln( x )dx. Answer 1 Simplify ln ⁡ ( 𝑥 ) = 1 2 ln ⁡ ( 𝑥 ) ln( x )= 2 1 ln(x). Use integration by parts: ∫ ln ⁡ ( 𝑥 ) 𝑑 𝑥 = 𝑥 ln ⁡ ( 𝑥 ) − 𝑥 + 𝐶 . ∫ln(x)dx=xln(x)−x+C. Thus: ∫ ln ⁡ ( 𝑥 ) 𝑑 𝑥 = 1 2 ( 𝑥 ln ⁡ ( 𝑥 ) − 𝑥 ) + 𝐶 . ∫ln( x )dx= 2 1 (xln(x)−x)+C. Result: ∫ ln ⁡ ( 𝑥 ) 𝑑 𝑥 = 𝑥 ln ⁡ ( 𝑥 ) 2 − 𝑥 2 + 𝐶 . ∫ln( x )dx= 2 xln(x) − 2 x +C. Question 5 Evaluate the definite integral: ∫ 0 1 1 + 7 𝑥 𝑑 𝑥 . ∫ 0 1 1+7x dx. Answer Use substitution 𝑢 = 1 + 7 𝑥 u=1+7x, so 𝑑 𝑢 = 7 𝑑 𝑥 du=7dx. Change limits: When 𝑥 = 0 , 𝑢 = 1 x=0,u=1. When 𝑥 = 1 , 𝑢 = 8 x=1,u=8. Rewrite the integral: ∫ 1 8 𝑢 ⋅ 1 7 𝑑 𝑢 . ∫ 1 8 u ⋅ 7 1 du. Simplify: 1 7 ∫ 1 8 𝑢 1 / 2 𝑑 𝑢 = 1 7 [ 2 3 𝑢 3 / 2 ] 1 8 . 7 1 ∫ 1 8 u 1/2 du= 7 1 [ 3 2 u 3/2 ] 1 8 . Evaluate: 1 7 ⋅ 2 3 ( 8 3 / 2 − 1 3 / 2 ) . 7 1 ⋅ 3 2 (8 3/2 −1 3/2 ). Simplify: 8 3 / 2 = 8 3 = 512 = 8 8 . 8 3/2 = 8 3 = 512 =8 8 . Thus: ∫ 0 1 1 + 7 𝑥 𝑑 𝑥 = 3 20 ( 8 8 − 1 ) = 45 28 . ∫ 0 1 1+7x dx= 20 3 (8 8 −1)= 28 45 . Result: ∫ 0 1 1 + 7 𝑥 𝑑 𝑥 = 45 28 . ∫ 0 1 1+7x dx= 28 45 .