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

2024
Views:

274

Pages:

4
Author:

Cristian Santos

Question 1 Suppose that 𝑓 ( 𝑥 ) = 0.125 𝑥 f(x)=0.125x for 0 ≤ 𝑥 ≤ 4 0≤x≤4. Determine the mean and variance of 𝑋 X. Answer The probability density function 𝑓 ( 𝑥 ) = 0.125 𝑥 f(x)=0.125x. Mean 𝐸 ( 𝑋 ) E(X): 𝐸 ( 𝑋 ) = ∫ 0 4 𝑥 𝑓 ( 𝑥 ) 𝑑 𝑥 = ∫ 0 4 0.125 𝑥 2 𝑑 𝑥 = 4 3 E(X)=∫ 0 4 xf(x)dx=∫ 0 4 0.125x 2 dx= 3 4 Variance Var ( 𝑋 ) Var(X): 𝐸 ( 𝑋 2 ) = ∫ 0 4 𝑥 2 𝑓 ( 𝑥 ) 𝑑 𝑥 = ∫ 0 4 0.125 𝑥 3 𝑑 𝑥 = 4 3 E(X 2 )=∫ 0 4 x 2 f(x)dx=∫ 0 4 0.125x 3 dx= 3 4 Var ( 𝑋 ) = 𝐸 ( 𝑋 2 ) − ( 𝐸 ( 𝑋 ) ) 2 = 64 9 Var(X)=E(X 2 )−(E(X)) 2 = 9 64 Mean: 4 3 3 4 Variance: 64 9 9 64 . Question 2 Calculate the double integral: ∬ 𝑅 𝑧 sin ⁡ ( 𝑥 + 𝑦 ) 𝑑 𝐴 , 𝑅 = [ 0 , 𝜋 6 ] × [ 0 , 𝜋 3 ] ∬ R zsin(x+y)dA,R=[0, 6 π ]×[0, 3 π ] Answer Evaluate iteratively: ∫ 0 𝜋 / 6 ∫ 0 𝜋 / 3 𝑧 sin ⁡ ( 𝑥 + 𝑦 ) 𝑑 𝑦 𝑑 𝑥 ∫ 0 π/6 ∫ 0 π/3 zsin(x+y)dydx Simplify: ∫ 0 𝜋 / 6 [ − 𝑧 cos ⁡ ( 𝑥 + 𝑦 ) ] 0 𝜋 / 3 𝑑 𝑥 ∫ 0 π/6 [−zcos(x+y)] 0 π/3 dx = ∫ 0 𝜋 / 6 ( − 𝑧 cos ⁡ ( 𝑥 + 𝜋 3 ) + 𝑧 cos ⁡ ( 𝑥 ) ) 𝑑 𝑥 =∫ 0 π/6 (−zcos(x+ 3 π )+zcos(x))dx After simplification: Result: − 1 + 3 2 − 𝜋 12 Result: 2 −1+ 3 − 12 π Question 3 Evaluate the integral: ∫ sin ⁡ 3 𝜃 cos ⁡ − 2 𝜃 𝑑 𝜃 ∫sin 3 θcos −2 θdθ Answer Using trigonometric identities and substitution: Rewrite using cos ⁡ 2 𝜃 = 1 − sin ⁡ 2 𝜃 cos 2 θ=1−sin 2 θ. Substitution 𝑡 = cos ⁡ 𝜃 t=cosθ, 𝑑 𝑡 = − sin ⁡ 𝜃 𝑑 𝜃 dt=−sinθdθ: ∫ sin ⁡ 3 𝜃 cos ⁡ − 2 𝜃 𝑑 𝜃 = ∫ ( 1 − 𝑡 2 ) 𝑡 − 2 𝑑 𝑡 ∫sin 3 θcos −2 θdθ=∫(1−t 2 )t −2 dt Simplify: ∫ ( 𝑡 − 2 − 1 ) 𝑑 𝑡 = − 1 𝑡 + 𝑡 + 𝐶 ∫(t −2 −1)dt=− t 1 +t+C Substitute back 𝑡 = cos ⁡ 𝜃 t=cosθ: 𝐼 = 1 cos ⁡ 𝜃 + cos ⁡ 𝜃 + 𝐶 I= cosθ 1 +cosθ+C Question 4 Use geometry and properties of integrals to evaluate: ∫ 0 4 8 𝑥 − 𝑥 2 𝑑 𝑥 ∫ 0 4 8x−x 2 dx Answer Complete the square: 8 𝑥 − 𝑥 2 = − ( 𝑥 2 − 8 𝑥 + 16 − 16 ) = − ( 𝑥 − 4 ) 2 + 16 8x−x 2 =−(x 2 −8x+16−16)=−(x−4) 2 +16 Use substitution 𝑢 = 𝑥 − 4 u=x−4: ∫ 0 4 8 𝑥 − 𝑥 2 𝑑 𝑥 = ∫ − 4 0 − 𝑢 2 + 16 𝑑 𝑢 ∫ 0 4 8x−x 2 dx=∫ −4 0 −u 2 +16 du Use trigonometric substitution 𝑢 = 4 sin ⁡ 𝜃 u=4sinθ: ∫ 16 cos ⁡ 2 𝜃 cos ⁡ 𝜃 𝑑 𝜃 = 16 ∫ cos ⁡ 2 𝜃 𝑑 𝜃 ∫ 16cos 2 θ cosθdθ=16∫cos 2 θdθ Simplify and solve: Result: 4 𝜋 Result: 4π Question 5 Evaluate the integral: ∫ sec ⁡ 3 𝑥 tan ⁡ 4 𝑥 𝑑 𝑥 ∫sec 3 xtan 4 xdx Answer Using identities and simplifications: Expand tan ⁡ 4 𝑥 tan 4 x in terms of sec ⁡ 2 𝑥 sec 2 x. ∫ sec ⁡ 3 𝑥 ( sec ⁡ 2 𝑥 − 1 ) 2 𝑑 𝑥 ∫sec 3 x(sec 2 x−1) 2 dx Simplify and separate terms: ∫ sec ⁡ 7 𝑥 𝑑 𝑥 − 2 ∫ sec ⁡ 5 𝑥 𝑑 𝑥 + ∫ sec ⁡ 3 𝑥 𝑑 𝑥 ∫sec 7 xdx−2∫sec 5 xdx+∫sec 3 xdx Solve each term: 𝐼 = ln ⁡ ( tan ⁡ 𝑥 + sec ⁡ 𝑥 ) 16 + sec ⁡ 2 𝑥 tan ⁡ 𝑥 6 − 7 sec ⁡ 4 𝑥 tan ⁡ 𝑥 24 + sec ⁡ 𝑥 arctan ⁡ 𝑥 16 + 𝐶 I= 16 ln(tanx+secx) + 6 sec 2 xtanx − 24 7sec 4 xtanx + 16 secxarctanx +C