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

2024
Views:

250

Pages:

7
Author:

Cristian Santos

Question 1 Calculate the double integral: ∬ 𝑅 6 𝑥 1 + 𝑥 𝑦 𝑑 𝐴 , 𝑅 = [ 0 , 6 ] × [ 0 , 1 ] ∬ R 1+xy 6x dA,R=[0,6]×[0,1] Answer ∬ 𝑅 6 𝑥 1 + 𝑥 𝑦 𝑑 𝐴 = ∫ 0 6 [ 6 𝑥 ln ⁡ ( 1 + 𝑥 𝑦 ) ] 0 1 𝑑 𝑥 ∬ R 1+xy 6x dA=∫ 0 6 [6xln(1+xy)] 0 1 dx = ∫ 0 6 6 𝑥 ln ⁡ ( 1 + 𝑥 ) 𝑑 𝑥 =∫ 0 6 6xln(1+x)dx = 6 [ 𝑥 ln ⁡ ( 1 + 𝑥 ) − ∫ 𝑥 1 + 𝑥 𝑑 𝑥 ] 0 6 =6[xln(1+x)−∫ 1+x x dx] 0 6 = 6 [ 𝑥 ln ⁡ ( 1 + 𝑥 ) − ( 𝑥 − ln ⁡ ( 1 + 𝑥 ) ) ] 0 6 =6[xln(1+x)−(x−ln(1+x))] 0 6 = 6 [ 7 ln ⁡ ( 7 ) − 6 + 1 ] =6[7ln(7)−6+1] ≈ 45.72 ≈45.72 Question 2 Integrate tan ⁡ 2 𝑥 tan 2 x. Answer ∫ tan ⁡ 2 𝑥 𝑑 𝑥 = ∫ sin ⁡ 2 𝑥 cos ⁡ 2 𝑥 𝑑 𝑥 = ∫ ( 1 − 1 cos ⁡ 2 𝑥 ) 𝑑 𝑥 ∫tan 2 xdx=∫ cos 2 x sin 2 x dx=∫(1− cos 2 x 1 )dx = ∫ 1 𝑑 𝑥 − ∫ sec ⁡ 2 𝑥 𝑑 𝑥 =∫1dx−∫sec 2 xdx = 𝑥 − tan ⁡ 𝑥 + 𝐶 =x−tanx+C Question 3 Evaluate the indefinite integral as an infinite series: ∫ cos ⁡ 𝑥 − 1 𝑥 𝑑 𝑥 ∫ x cosx−1 dx Answer Using the Maclaurin series for cos ⁡ 𝑥 cosx: cos ⁡ 𝑥 = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 𝑥 2 𝑛 ( 2 𝑛 ) ! cosx= n=0 ∑ ∞ (2n)! (−1) n x 2n cos ⁡ 𝑥 − 1 = ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑥 2 𝑛 ( 2 𝑛 ) ! cosx−1= n=1 ∑ ∞ (2n)! (−1) n x 2n Thus: cos ⁡ 𝑥 − 1 𝑥 = ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑥 2 𝑛 − 1 ( 2 𝑛 ) ! x cosx−1 = n=1 ∑ ∞ (2n)! (−1) n x 2n−1 Integrating term by term: ∫ cos ⁡ 𝑥 − 1 𝑥 𝑑 𝑥 = ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑥 2 𝑛 ( 2 𝑛 ) ( 2 𝑛 ) ! + 𝐶 ∫ x cosx−1 dx= n=1 ∑ ∞ (2n)(2n)! (−1) n x 2n +C Question 4 The velocity function is 𝑣 ( 𝑡 ) = 3 𝑡 − 8 v(t)=3t−8, 0 ≤ 𝑡 ≤ 3 0≤t≤3. (a) Find the displacement: Displacement = ∫ 0 3 𝑣 ( 𝑡 ) 𝑑 𝑡 = ∫ 0 3 ( 3 𝑡 − 8 ) 𝑑 𝑡 Displacement=∫ 0 3 v(t)dt=∫ 0 3 (3t−8)dt = [ 3 𝑡 2 2 − 8 𝑡 ] 0 3 =[ 2 3t 2 −8t] 0 3 = ( 3 ( 9 ) 2 − 24 ) = − 10.5 =( 2 3(9) −24)=−10.5 (b) Find the total distance: Break into intervals where 𝑣 ( 𝑡 ) v(t) changes sign at 𝑡 = 8 3 t= 3 8 . Total Distance = ∫ 0 8 / 3 ∣ 3 𝑡 − 8 ∣ 𝑑 𝑡 + ∫ 8 / 3 3 ∣ 3 𝑡 − 8 ∣ 𝑑 𝑡 Total Distance=∫ 0 8/3 ∣3t−8∣dt+∫ 8/3 3 ∣3t−8∣dt Simplify and compute: Total Distance = 65 6 ≈ 10.833 Total Distance= 6 65 ≈10.833 Question 5 Calculate the double integral: ∬ 𝑅 4 𝑥 1 + 𝑥 𝑦 𝑑 𝐴 , 𝑅 = [ 0 , 4 ] × [ 0 , 1 ] ∬ R 1+xy 4x dA,R=[0,4]×[0,1] Answer ∬ 𝑅 4 𝑥 1 + 𝑥 𝑦 𝑑 𝐴 = ∫ 0 4 [ 4 𝑥 ln ⁡ ( 1 + 𝑥 𝑦 ) ] 0 1 𝑑 𝑥 ∬ R 1+xy 4x dA=∫ 0 4 [4xln(1+xy)] 0 1 dx = ∫ 0 4 4 𝑥 ln ⁡ ( 1 + 𝑥 ) 𝑑 𝑥 =∫ 0 4 4xln(1+x)dx Using integration by parts: ∫ 0 4 𝑥 ln ⁡ ( 1 + 𝑥 ) 𝑑 𝑥 = [ 𝑥 ln ⁡ ( 1 + 𝑥 ) − ∫ 𝑥 1 + 𝑥 𝑑 𝑥 ] 0 4 ∫ 0 4 xln(1+x)dx=[xln(1+x)−∫ 1+x x dx] 0 4 = 4 [ 5 ln ⁡ 5 − 4 ] =4[5ln5−4] = 20 ln ⁡ 5 − 16 =20ln5−16