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

2024
Views:

244

Pages:

6
Author:

Cristian Santos

Question 1 Evaluate ∬ 𝐷 𝑥 𝑦 2 𝑑 𝐴 ∬ D xy 2 dA, where 𝐷 D is bounded by 𝑥 = 0 x=0 and 𝑥 = 1 − 𝑦 2 x= 1−y 2 , 𝑦 ∈ [ − 1 , 1 ] y∈[−1,1]: Setup the integral: ∫ − 1 1 ∫ 0 1 − 𝑦 2 𝑥 𝑦 2 𝑑 𝑥 𝑑 𝑦 . ∫ −1 1 ∫ 0 1−y 2 xy 2 dxdy. Solve the inner integral with respect to 𝑥 x: ∫ 0 1 − 𝑦 2 𝑥 𝑦 2 𝑑 𝑥 = 𝑦 2 2 𝑥 2 ∣ 0 1 − 𝑦 2 = 𝑦 2 2 ( 1 − 𝑦 2 ) . ∫ 0 1−y 2 xy 2 dx= 2 y 2 x 2 0 1−y 2 = 2 y 2 (1−y 2 ). Solve the outer integral with respect to 𝑦 y: ∫ − 1 1 𝑦 2 2 ( 1 − 𝑦 2 ) 𝑑 𝑦 = 2 15 . ∫ −1 1 2 y 2 (1−y 2 )dy= 15 2 . Result: 2 15 15 2 . Question 2 Find the volume of the solid bounded by the coordinate planes and 3 𝑥 + 2 𝑦 + 𝑧 = 6 3x+2y+z=6: Set the bounds: 𝑥 x varies from 0 0 to 2 2, 𝑦 y varies from 0 0 to 3 − 3 2 𝑥 3− 2 3 x, and 𝑧 = 6 − 3 𝑥 − 2 𝑦 z=6−3x−2y. Setup and evaluate: ∫ 0 2 ∫ 0 3 − 3 2 𝑥 ∫ 0 6 − 3 𝑥 − 2 𝑦 𝑑 𝑧 𝑑 𝑦 𝑑 𝑥 . ∫ 0 2 ∫ 0 3− 2 3 x ∫ 0 6−3x−2y dzdydx. Solve step by step to get the volume: 𝑉 = 6. V=6. Question 3 Antiderivative of ln ⁡ ( 𝑥 ) ln(x): Using integration by parts ∫ 𝑢 𝑑 𝑣 = 𝑢 𝑣 − ∫ 𝑣 𝑑 𝑢 ∫udv=uv−∫vdu, where: 𝑢 = ln ⁡ ( 𝑥 ) , 𝑑 𝑣 = 𝑑 𝑥 u=ln(x),dv=dx, so 𝑑 𝑢 = 1 𝑥 𝑑 𝑥 , 𝑣 = 𝑥 du= x 1 dx,v=x. Apply the formula: ∫ ln ⁡ ( 𝑥 ) 𝑑 𝑥 = 𝑥 ln ⁡ ( 𝑥 ) − ∫ 𝑥 ⋅ 1 𝑥 𝑑 𝑥 = 𝑥 ln ⁡ ( 𝑥 ) − 𝑥 + 𝐶 . ∫ln(x)dx=xln(x)−∫x⋅ x 1 dx=xln(x)−x+C. Result: 𝑥 ln ⁡ ( 𝑥 ) − 𝑥 + 𝐶 xln(x)−x+C. Question 4 Find the region corresponding to lim ⁡ 𝑛 → ∞ ∑ 𝑘 = 1 𝑛 𝜋 4 𝑛 tan ⁡ ( 𝑘 𝜋 4 𝑛 ) lim n→∞ ∑ k=1 n 4n π tan( 4n kπ ): The sum represents the Riemann sum of ∫ 0 𝜋 4 tan ⁡ ( 𝑥 ) 𝑑 𝑥 ∫ 0 4 π tan(x)dx. This corresponds to the area under 𝑦 = tan ⁡ ( 𝑥 ) y=tan(x) from 𝑥 = 0 x=0 to 𝑥 = 𝜋 4 x= 4 π . Question 5 Evaluate ∫ − 3 0 ( 1 + 9 − 𝑥 2 ) 𝑑 𝑥 ∫ −3 0 (1+ 9−x 2 )dx: Split the integral: ∫ − 3 0 ( 1 + 9 − 𝑥 2 ) 𝑑 𝑥 = ∫ − 3 0 1 𝑑 𝑥 + ∫ − 3 0 9 − 𝑥 2 𝑑 𝑥 . ∫ −3 0 (1+ 9−x 2 )dx=∫ −3 0 1dx+∫ −3 0 9−x 2 dx. Solve each part: ∫ − 3 0 1 𝑑 𝑥 = 3 ∫ −3 0 1dx=3. ∫ − 3 0 9 − 𝑥 2 𝑑 𝑥 ∫ −3 0 9−x 2 dx represents a quarter circle with radius 3 3: Area = 1 4 𝜋 ( 3 ) 2 = 9 𝜋 4 . Area= 4 1 π(3) 2 = 4 9π . Combine results: 𝐼 = 3 + 9 𝜋 4 . I=3+ 4 9π . Result: 3 + 9 𝜋 4 3+ 4 9π .

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