Integrals #2 - Questions and Answers

Question 1 Which of the following integrals are improper? ∫ 4 5 1 5 𝑥 − 1   𝑑 𝑥 ∫ 4 5 ​ 5x−1 1 ​ dx ∫ 0 1 1 2 𝑥 − 1   𝑑 𝑥 ∫ 0 1 ​ 2x−1 1 ​ dx ∫ 0 ∞ sin ⁡ ( 𝑥 ) 1 + 3 𝑥 2   𝑑 𝑥 ∫ 0 ∞ ​ 1+3x 2 sin(x) ​ dx ∫ 1 3 ln ⁡ ( 𝑥 − 1 )   𝑑 𝑥 ∫ 1 3 ​ ln(x−1)dx Answer 1 Analysis: ∫ 4 5 1 5 𝑥 − 1   𝑑 𝑥 ∫ 4 5 ​ 5x−1 1 ​ dx This integral is not improper because the integrand is continuous and defined within the limits 𝑥 = 4 x=4 to 𝑥 = 5 x=5. ∫ 0 1 1 2 𝑥 − 1   𝑑 𝑥 ∫ 0 1 ​ 2x−1 1 ​ dx The integrand becomes undefined at 𝑥 = 1 2 x= 2 1 ​ , which is within the range [ 0 , 1 ] [0,1]. This is an improper integral. ∫ 0 ∞ sin ⁡ ( 𝑥 ) 1 + 3 𝑥 2   𝑑 𝑥 ∫ 0 ∞ ​ 1+3x 2 sin(x) ​ dx The integration range includes infinity, which makes it improper. ∫ 1 3 ln ⁡ ( 𝑥 − 1 )   𝑑 𝑥 ∫ 1 3 ​ ln(x−1)dx The integrand becomes undefined at 𝑥 = 1 x=1, which is within the integration range. This is an improper integral. Conclusion: The improper integrals are: ( 2 )   ∫ 0 1 1 2 𝑥 − 1   𝑑 𝑥 (2)∫ 0 1 ​ 2x−1 1 ​ dx ( 3 )   ∫ 0 ∞ sin ⁡ ( 𝑥 ) 1 + 3 𝑥 2   𝑑 𝑥 (3)∫ 0 ∞ ​ 1+3x 2 sin(x) ​ dx ( 4 )   ∫ 1 3 ln ⁡ ( 𝑥 − 1 )   𝑑 𝑥 (4)∫ 1 3 ​ ln(x−1)dx. Question 2 Use a double integral to find the volume of the indicated solid. 𝑧 = 6 − 2 𝑦 z=6−2y, 0 ≤ 𝑥 ≤ 4 ,   0 ≤ 𝑦 ≤ 2 0≤x≤4,0≤y≤2. Answer: Set up the double integral for the volume: 𝑉 = ∫ 0 4 ∫ 0 2 ( 6 − 2 𝑦 )   𝑑 𝑦   𝑑 𝑥 . V=∫ 0 4 ​ ∫ 0 2 ​ (6−2y)dydx. Integrate with respect to 𝑦 y: ∫ 0 2 ( 6 − 2 𝑦 )   𝑑 𝑦 = [ 6 𝑦 − 𝑦 2 ] 0 2 = ( 6 ( 2 ) − 2 2 ) − ( 0 − 0 ) = 12 − 4 = 8. ∫ 0 2 ​ (6−2y)dy=[6y−y 2 ] 0 2 ​ =(6(2)−2 2 )−(0−0)=12−4=8. Integrate with respect to 𝑥 x: 𝑉 = ∫ 0 4 8   𝑑 𝑥 = 8 𝑥 ∣ 0 4 = 8 ( 4 ) − 8 ( 0 ) = 32. V=∫ 0 4 ​ 8dx=8x ​ 0 4 ​ =8(4)−8(0)=32. Final Volume: 32 32. Question 3 Evaluate the double integral ∬ 𝐷 ( 𝑥 2 + 2 𝑦 )   𝑑 𝐴 ∬ D ​ (x 2 +2y)dA, where 𝐷 D is bounded by 𝑦 = 𝑥 ,   𝑦 = 𝑥 3 ,   𝑥 ≥ 0 y=x,y=x 3 ,x≥0. Answer: Region 𝐷 D: The curves 𝑦 = 𝑥 y=x and 𝑦 = 𝑥 3 y=x 3 intersect at 𝑥 = 0 x=0 and 𝑥 = 1 x=1. 𝑥 x runs from 0 0 to 1 1. For each 𝑥 x, 𝑦 y runs from 𝑥 3 x 3 (lower curve) to 𝑥 x (upper curve). Set up the integral: ∬ 𝐷 ( 𝑥 2 + 2 𝑦 )   𝑑 𝐴 = ∫ 0 1 ∫ 𝑥 3 𝑥 ( 𝑥 2 + 2 𝑦 )   𝑑 𝑦   𝑑 𝑥 . ∬ D ​ (x 2 +2y)dA=∫ 0 1 ​ ∫ x 3 x ​ (x 2 +2y)dydx. Integrate with respect to 𝑦 y: ∫ 𝑥 3 𝑥 ( 𝑥 2 + 2 𝑦 )   𝑑 𝑦 = [ 𝑥 2 𝑦 + 𝑦 2 ] 𝑥 3 𝑥 . ∫ x 3 x ​ (x 2 +2y)dy=[x 2 y+y 2 ] x 3 x ​ . Substitute the limits 𝑦 = 𝑥 y=x and 𝑦 = 𝑥 3 y=x 3 : 𝑥 2 ( 𝑥 ) + ( 𝑥 ) 2 − [ 𝑥 2 ( 𝑥 3 ) + ( 𝑥 3 ) 2 ] = 𝑥 3 + 𝑥 2 − ( 𝑥 5 + 𝑥 6 ) . x 2 (x)+(x) 2 −[x 2 (x 3 )+(x 3 ) 2 ]=x 3 +x 2 −(x 5 +x 6 ). Integrate with respect to 𝑥 x: ∫ 0 1 ( 𝑥 3 + 𝑥 2 − 𝑥 5 − 𝑥 6 )   𝑑 𝑥 = [ 𝑥 4 4 + 𝑥 3 3 − 𝑥 6 6 − 𝑥 7 7 ] 0 1 . ∫ 0 1 ​ (x 3 +x 2 −x 5 −x 6 )dx=[ 4 x 4 ​ + 3 x 3 ​ − 6 x 6 ​ − 7 x 7 ​ ] 0 1 ​ . Evaluate at 𝑥 = 1 x=1: 1 4 + 1 3 − 1 6 − 1 7 . 4 1 ​ + 3 1 ​ − 6 1 ​ − 7 1 ​ . Combine terms: 1 4 + 1 3 = 7 12 , 1 6 + 1 7 = 13 42 . 4 1 ​ + 3 1 ​ = 12 7 ​ , 6 1 ​ + 7 1 ​ = 42 13 ​ . Subtract: 7 12 − 13 42 = 23 84 . 12 7 ​ − 42 13 ​ = 84 23 ​ . Final Answer: 23 84 84 23 ​ . Question 4 Find the area of the part of the plane 5 𝑥 + 4 𝑦 + 𝑧 = 20 5x+4y+z=20 that lies in the first octant. Answer: Equation of the plane: 𝑧 = 20 − 5 𝑥 − 4 𝑦 z=20−5x−4y. Boundaries: 𝑥 ≥ 0 ,   𝑦 ≥ 0 ,   𝑧 ≥ 0 x≥0,y≥0,z≥0. Intercepts are: 𝑥 = 4 x=4 (when 𝑦 = 0 ,   𝑧 = 0 y=0,z=0), 𝑦 = 5 y=5 (when 𝑥 = 0 ,   𝑧 = 0 x=0,z=0), 𝑧 = 20 z=20 (when 𝑥 = 0 ,   𝑦 = 0 x=0,y=0). Area of the surface: Use the formula for surface area: 𝐴 = ∬ 𝐷 1 + ( ∂ 𝑧 ∂ 𝑥 ) 2 + ( ∂ 𝑧 ∂ 𝑦 ) 2   𝑑 𝐴 . A=∬ D ​ 1+( ∂x ∂z ​ ) 2 +( ∂y ∂z ​ ) 2 ​ dA. Partial derivatives: ∂ 𝑧 ∂ 𝑥 = − 5 , ∂ 𝑧 ∂ 𝑦 = − 4. ∂x ∂z ​ =−5, ∂y ∂z ​ =−4. Substitute: 𝐴 = ∬ 𝐷 1 + ( − 5 ) 2 + ( − 4 ) 2   𝑑 𝐴 = ∬ 𝐷 1 + 25 + 16   𝑑 𝐴 = ∬ 𝐷 42   𝑑 𝐴 . A=∬ D ​ 1+(−5) 2 +(−4) 2 ​ dA=∬ D ​ 1+25+16 ​ dA=∬ D ​ 42 ​ dA. Region 𝐷 D: 𝐷 D is a triangle with vertices ( 0 , 0 ) , ( 4 , 0 ) , ( 0 , 5 ) (0,0),(4,0),(0,5). The area of 𝐷 D is: Area of  𝐷 = 1 2 ⋅ 4 ⋅ 5 = 10. Area of D= 2 1 ​ ⋅4⋅5=10. Compute the surface area: 𝐴 = 42 ⋅ 10 = 10 42 . A= 42 ​ ⋅10=10 42 ​ . Final Area: 10 42 10 42 ​ .