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

2024
Views:

364

Pages:

4
Author:

Cristian Santos

Question 1 Evaluate the integral: ∫ 1 𝑒 𝑑 𝑥 𝑥 ( 1 + ln ⁡ 𝑥 ) ∫ 1 e x(1+lnx) dx Answer Using substitution 𝑣 = 1 + ln ⁡ 𝑥 v=1+lnx, 𝑑 𝑣 = 𝑑 𝑥 𝑥 dv= x dx : Change the limits: When 𝑥 = 1 x=1, 𝑣 = 1 v=1 When 𝑥 = 𝑒 x=e, 𝑣 = 2 v=2 Rewrite the integral: ∫ 1 𝑒 𝑑 𝑥 𝑥 ( 1 + ln ⁡ 𝑥 ) = ∫ 1 2 1 𝑣 𝑑 𝑣 ∫ 1 e x(1+lnx) dx =∫ 1 2 v 1 dv Integrate: ∫ 1 2 1 𝑣 𝑑 𝑣 = ln ⁡ 𝑣 ∣ 1 2 = ln ⁡ 2 − ln ⁡ 1 ∫ 1 2 v 1 dv=lnv 1 2 =ln2−ln1 Result: ln ⁡ 2 Result: ln2 Question 2 Evaluate the integral: ∫ 𝑥 2 49 − 𝑥 2 𝑑 𝑥 ∫x 2 49−x 2 dx Answer Use substitution 𝑢 = 49 − 𝑥 2 u= 49−x 2 , 𝑥 = 49 − 𝑢 2 x= 49−u 2 , and simplify. After simplification, the integral becomes: ∫ 𝑥 2 49 − 𝑥 2 𝑑 𝑥 = − 49 ( 𝑢 3 3 ) + 𝑢 5 5 + 𝐶 ∫x 2 49−x 2 dx=−49( 3 u 3 )+ 5 u 5 +C Back-substitute 𝑢 = 49 − 𝑥 2 u= 49−x 2 : Result: − 49 3 ( 49 − 𝑥 2 ) 3 / 2 + 1 5 ( 49 − 𝑥 2 ) 5 / 2 + 𝐶 Result: − 3 49 (49−x 2 ) 3/2 + 5 1 (49−x 2 ) 5/2 +C Question 3 Find ∫ 0 3 𝑥 𝑓 ( 𝑥 2 ) 𝑑 𝑥 ∫ 0 3 xf(x 2 )dx if ∫ 0 9 𝑓 ( 𝑥 ) 𝑑 𝑥 = 4 ∫ 0 9 f(x)dx=4. Answer Use substitution 𝑥 2 = 𝑢 x 2 =u, 2 𝑥 𝑑 𝑥 = 𝑑 𝑢 2xdx=du: ∫ 0 3 𝑥 𝑓 ( 𝑥 2 ) 𝑑 𝑥 = 1 2 ∫ 0 9 𝑓 ( 𝑢 ) 𝑑 𝑢 ∫ 0 3 xf(x 2 )dx= 2 1 ∫ 0 9 f(u)du Given ∫ 0 9 𝑓 ( 𝑢 ) 𝑑 𝑢 = 4 ∫ 0 9 f(u)du=4, substitute: ∫ 0 3 𝑥 𝑓 ( 𝑥 2 ) 𝑑 𝑥 = 1 2 ⋅ 4 = 2 ∫ 0 3 xf(x 2 )dx= 2 1 ⋅4=2 Result: 2 Question 4 Find the volume of the solid bounded by 𝑥 = − 1 x=−1, 𝑥 = 1 x=1, and cross-sections perpendicular to the x-axis as squares whose base runs from 1 − 𝑥 2 1−x 2 to 𝑦 = 1 − 𝑥 2 y= 1−x 2 . Answer Cross-sectional area 𝐴 ( 𝑥 ) = 4 ( 1 − 𝑥 2 ) A(x)=4(1−x 2 ). Integrate the area from 𝑥 = − 1 x=−1 to 𝑥 = 1 x=1: 𝑉 = ∫ − 1 1 4 ( 1 − 𝑥 2 ) 𝑑 𝑥 V=∫ −1 1 4(1−x 2 )dx 𝑉 = 4 [ 𝑥 − 𝑥 3 3 ] − 1 1 V=4[x− 3 x 3 ] −1 1 Simplify: 𝑉 = 4 ( 2 3 ) = 16 3 V=4( 3 2 )= 3 16 Result: 16 3 3 16 Question 5 Put spherical coordinates to use for the triple integral ∫ 𝐸 ( 𝑥 2 + 𝑦 2 ) 𝑑 𝑉 ∫ E (x 2 +y 2 )dV, where 𝐸 E lies between the spheres 𝑥 2 + 𝑦 2 + 𝑧 2 = 4 x 2 +y 2 +z 2 =4 and 𝑥 2 + 𝑦 2 + 𝑧 2 = 9 x 2 +y 2 +z 2 =9. Answer Spherical coordinates: 𝑥 2 + 𝑦 2 = 𝜌 2 sin ⁡ 2 𝜙 x 2 +y 2 =ρ 2 sin 2 ϕ, 𝑑 𝑉 = 𝜌 2 sin ⁡ 𝜙 𝑑 𝜌 𝑑 𝜙 𝑑 𝜃 dV=ρ 2 sinϕdρdϕdθ. Bounds: 𝜌 ∈ [ 2 , 3 ] ρ∈[2,3], 𝜙 ∈ [ 0 , 𝜋 ] ϕ∈[0,π], 𝜃 ∈ [ 0 , 2 𝜋 ] θ∈[0,2π]. Integral: ∫ 0 2 𝜋 ∫ 0 𝜋 ∫ 2 3 𝜌 2 sin ⁡ 2 𝜙 ⋅ 𝜌 2 sin ⁡ 𝜙 𝑑 𝜌 𝑑 𝜙 𝑑 𝜃 ∫ 0 2π ∫ 0 π ∫ 2 3 ρ 2 sin 2 ϕ⋅ρ 2 sinϕdρdϕdθ Simplify: ∫ 0 2 𝜋 𝑑 𝜃 ⋅ ∫ 0 𝜋 sin ⁡ 3 𝜙 𝑑 𝜙 ⋅ ∫ 2 3 𝜌 4 𝑑 𝜌 ∫ 0 2π dθ⋅∫ 0 π sin 3 ϕdϕ⋅∫ 2 3 ρ 4 dρ Evaluate step by step to get: 𝑉 = 168 𝜋 15 V= 15 168π Result: 168 𝜋 15 15 168π