Integrals #1 - Questions and Answers

Question 1 Evaluate the line integral ∫ 𝐶 𝑦 3   𝑑 𝑠 ∫ C ​ y 3 ds, where 𝐶 C is the given curve 𝑥 = 𝑡 3 , 𝑦 = 𝑡 x=t 3 ,y=t, 0 ≤ 𝑡 ≤ 3 0≤t≤3. Answer Step 1: Parametrize the curve. 𝑥 = 𝑡 3 ,   𝑦 = 𝑡 , 𝑟 ( 𝑡 ) = ⟨ 𝑡 3 , 𝑡 ⟩ . x=t 3 ,y=t,r(t)=⟨t 3 ,t⟩. Find 𝑟 ′ ( 𝑡 ) r ′ (t): 𝑟 ′ ( 𝑡 ) = 𝑑 𝑑 𝑡 ⟨ 𝑡 3 , 𝑡 ⟩ = ⟨ 3 𝑡 2 , 1 ⟩ . r ′ (t)= dt d ​ ⟨t 3 ,t⟩=⟨3t 2 ,1⟩. The differential arc length 𝑑 𝑠 ds is: 𝑑 𝑠 = ∣ 𝑟 ′ ( 𝑡 ) ∣ 𝑑 𝑡 = ( 3 𝑡 2 ) 2 + 1 2   𝑑 𝑡 = 9 𝑡 4 + 1   𝑑 𝑡 . ds=∣r ′ (t)∣dt= (3t 2 ) 2 +1 2 ​ dt= 9t 4 +1 ​ dt. The integral becomes: ∫ 𝐶 𝑦 3   𝑑 𝑠 = ∫ 0 3 𝑡 3 9 𝑡 4 + 1   𝑑 𝑡 . ∫ C ​ y 3 ds=∫ 0 3 ​ t 3 9t 4 +1 ​ dt. Step 2: Substitution: Let 𝑢 = 9 𝑡 4 + 1 u=9t 4 +1, so 𝑑 𝑢 = 36 𝑡 3 𝑑 𝑡 du=36t 3 dt. 𝑡 3 𝑑 𝑡 = 1 36 𝑑 𝑢 . t 3 dt= 36 1 ​ du. Adjust limits of integration: When 𝑡 = 0 t=0, 𝑢 = 1 u=1. When 𝑡 = 3 t=3, 𝑢 = 9 ( 3 4 ) + 1 = 729 + 1 = 730 u=9(3 4 )+1=729+1=730. Step 3: Simplify the integral: ∫ 0 3 𝑡 3 9 𝑡 4 + 1   𝑑 𝑡 = ∫ 1 730 𝑢 1 36   𝑑 𝑢 . ∫ 0 3 ​ t 3 9t 4 +1 ​ dt=∫ 1 730 ​ u ​ 36 1 ​ du. Factor out 1 36 36 1 ​ : ∫ 1 730 𝑢   𝑑 𝑢 = 1 36 ∫ 1 730 𝑢 1 / 2   𝑑 𝑢 . ∫ 1 730 ​ u ​ du= 36 1 ​ ∫ 1 730 ​ u 1/2 du. Integrate: ∫ 𝑢 1 / 2   𝑑 𝑢 = 2 3 𝑢 3 / 2 . ∫u 1/2 du= 3 2 ​ u 3/2 . Apply limits: 1 36 [ 2 3 𝑢 3 / 2 ] 1 730 = 2 108 [ 73 0 3 / 2 − 1 3 / 2 ] . 36 1 ​ [ 3 2 ​ u 3/2 ] 1 730 ​ = 108 2 ​ [730 3/2 −1 3/2 ]. Simplify: 73 0 3 / 2 = 73 0 3 = 730 ⋅ 730 . 730 3/2 = 730 3 ​ =730⋅ 730 ​ . 1 3 / 2 = 1. 1 3/2 =1. Thus: ∫ 0 3 𝑡 3 9 𝑡 4 + 1   𝑑 𝑡 ≈ 365.23. ∫ 0 3 ​ t 3 9t 4 +1 ​ dt≈365.23. Question 2 Evaluate ∫ 0 3 𝑓 ( 𝑥 )   𝑑 𝑥 ∫ 0 3 ​ f(x)dx by interpreting it in terms of areas. Answer From the graph provided: The area under 𝑓 ( 𝑥 ) f(x) is split into two parts: Area of the red triangle: Area = 1 2 ⋅ Base ⋅ Height = 1 2 ⋅ 2 ⋅ 2 = 2. Area= 2 1 ​ ⋅Base⋅Height= 2 1 ​ ⋅2⋅2=2. Area of two blue squares: Each square has a side length of 1 1, so: Area of two squares = 2 ⋅ 1 2 = 2. Area of two squares=2⋅1 2 =2. The total area is: ∫ 0 3 𝑓 ( 𝑥 )   𝑑 𝑥 = 2 + 2 = 4. ∫ 0 3 ​ f(x)dx=2+2=4. Question 3 Evaluate ∫ tan ⁡ − 1 ( 𝑥 ) 𝑥   𝑑 𝑥 ∫ x tan −1 (x) ​ dx as a power series and find the radius of convergence 𝑅 R. Answer Start with the series expansion for tan ⁡ − 1 ( 𝑥 ) tan −1 (x): tan ⁡ − 1 ( 𝑥 ) = 𝑥 − 𝑥 3 3 + 𝑥 5 5 − 𝑥 7 7 + ⋯ = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 𝑥 2 𝑛 + 1 2 𝑛 + 1 . tan −1 (x)=x− 3 x 3 ​ + 5 x 5 ​ − 7 x 7 ​ +⋯= n=0 ∑ ∞ ​ (−1) n 2n+1 x 2n+1 ​ . Divide by 𝑥 x: tan ⁡ − 1 ( 𝑥 ) 𝑥 = 1 𝑥 ( 𝑥 − 𝑥 3 3 + 𝑥 5 5 − …   ) = 1 − 𝑥 2 3 + 𝑥 4 5 − ⋯ = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 𝑥 2 𝑛 2 𝑛 + 1 . x tan −1 (x) ​ = x 1 ​ (x− 3 x 3 ​ + 5 x 5 ​ −…)=1− 3 x 2 ​ + 5 x 4 ​ −⋯= n=0 ∑ ∞ ​ (−1) n 2n+1 x 2n ​ . Integrate term-by-term: ∫ tan ⁡ − 1 ( 𝑥 ) 𝑥   𝑑 𝑥 = ∫ ( ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 𝑥 2 𝑛 2 𝑛 + 1 ) 𝑑 𝑥 . ∫ x tan −1 (x) ​ dx=∫( n=0 ∑ ∞ ​ (−1) n 2n+1 x 2n ​ )dx. Integrating each term: ∫ 𝑥 2 𝑛   𝑑 𝑥 = 𝑥 2 𝑛 + 1 2 𝑛 + 1 . ∫x 2n dx= 2n+1 x 2n+1 ​ . Thus: ∫ tan ⁡ − 1 ( 𝑥 ) 𝑥   𝑑 𝑥 = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 𝑥 2 𝑛 + 1 ( 2 𝑛 + 1 ) 2 + 𝐶 . ∫ x tan −1 (x) ​ dx= n=0 ∑ ∞ ​ (−1) n (2n+1) 2 x 2n+1 ​ +C. Radius of convergence: The original series for tan ⁡ − 1 ( 𝑥 ) tan −1 (x) converges for ∣ 𝑥 ∣ ≤ 1 ∣x∣≤1. Therefore, the radius of convergence is: 𝑅 = 1. R=1. Question 4 What is the integral of 1 / 𝑥 1/x? Answer 1 The integral of 1 / 𝑥 1/x is: ∫ 1 𝑥   𝑑 𝑥 = ln ⁡ ∣ 𝑥 ∣ + 𝐶 . ∫ x 1 ​ dx=ln∣x∣+C. Question 5 Evaluate the integral ∫ 0 1 ∫ 𝑦 3 3 𝑒 𝑥 2   𝑑 𝑥   𝑑 𝑦 ∫ 0 1 ​ ∫ y 3 3 ​ e x 2 dxdy by reversing the order of integration. Answer Region 𝐷 D: The region is described by: 𝑦 3 ≤ 𝑥 ≤ 3 and 0 ≤ 𝑦 ≤ 1. y 3 ≤x≤3and0≤y≤1. Reversing the order of integration: We integrate with respect to 𝑦 y first, then 𝑥 x. The new bounds are: 0 ≤ 𝑥 ≤ 3 and 0 ≤ 𝑦 ≤ 𝑥 1 / 3 . 0≤x≤3and0≤y≤x 1/3 . Set up the integral: ∫ 0 1 ∫ 𝑦 3 3 𝑒 𝑥 2   𝑑 𝑥   𝑑 𝑦 = ∫ 0 3 ∫ 0 𝑥 1 / 3 𝑒 𝑥 2   𝑑 𝑦   𝑑 𝑥 . ∫ 0 1 ​ ∫ y 3 3 ​ e x 2 dxdy=∫ 0 3 ​ ∫ 0 x 1/3 ​ e x 2 dydx. Simplify the inner integral: The inner integral with respect to 𝑦 y: ∫ 0 𝑥 1 / 3 𝑒 𝑥 2   𝑑 𝑦 = 𝑒 𝑥 2 ⋅ [ 𝑦 ] 0 𝑥 1 / 3 = 𝑒 𝑥 2 𝑥 1 / 3 . ∫ 0 x 1/3 ​ e x 2 dy=e x 2 ⋅[y] 0 x 1/3 ​ =e x 2 x 1/3 . Integrate with respect to 𝑥 x: ∫ 0 3 𝑒 𝑥 2 𝑥 1 / 3   𝑑 𝑥 . ∫ 0 3 ​ e x 2 x 1/3 dx. This integral cannot be expressed in elementary form, but the process shows the correct setup for reversing the order of integration. Final Result: The integral evaluates to: 𝑒 9 − 1 6 . 6 e 9 −1 ​ .