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

2024
Views:

460

Pages:

3
Author:

Cristian Santos

Question 1 Answer The compound in sample I is not the same as in sample II. Therefore, the mass ratio of Cl to F in both samples will obey the law of multiple proportions. Question 2 Evaluate the integral: 7 sin ⁡ 8 ( 𝑥 ) cos ⁡ ( 𝑥 ) ln ⁡ ( sin ⁡ ( 𝑥 ) ) 𝑑 𝑥 7sin 8 (x)cos(x)ln(sin(x))dx Answer Using substitution 𝑢 = sin ⁡ ( 𝑥 ) u=sin(x), 𝑑 𝑢 = cos ⁡ ( 𝑥 ) 𝑑 𝑥 du=cos(x)dx: ∫ 7 ( sin ⁡ 8 𝑥 ) cos ⁡ 𝑥 ln ⁡ ( sin ⁡ 𝑥 ) 𝑑 𝑥 = 7 ∫ 𝑢 8 ln ⁡ ( 𝑢 ) 𝑑 𝑢 ∫7(sin 8 x)cosxln(sinx)dx=7∫u 8 ln(u)du Using table of integrals, formula no. 101: ∫ 𝑢 𝑛 ln ⁡ ( 𝑢 ) 𝑑 𝑢 = 𝑢 𝑛 + 1 ( 𝑛 + 1 ) 2 [ ( 𝑛 + 1 ) ln ⁡ 𝑢 − 1 ] + 𝐶 ∫u n ln(u)du= (n+1) 2 u n+1 [(n+1)lnu−1]+C With 𝑛 = 8 n=8: ∫ 7 𝑢 8 ln ⁡ 𝑢 𝑑 𝑢 = 7 [ 𝑢 9 81 ( 9 ln ⁡ 𝑢 − 1 ) ] + 𝐶 ∫7u 8 lnudu=7[ 81 u 9 (9lnu−1)]+C = 7 81 𝑢 9 [ 9 ln ⁡ 𝑢 − 1 ] + 𝐶 = 81 7 u 9 [9lnu−1]+C Back-substitute 𝑢 = sin ⁡ ( 𝑥 ) u=sin(x): ∫ 7 sin ⁡ 8 ( 𝑥 ) cos ⁡ ( 𝑥 ) ln ⁡ ( sin ⁡ ( 𝑥 ) ) 𝑑 𝑥 = 7 81 sin ⁡ 9 ( 𝑥 ) [ 9 ln ⁡ ( sin ⁡ ( 𝑥 ) ) − 1 ] + 𝐶 ∫7sin 8 (x)cos(x)ln(sin(x))dx= 81 7 sin 9 (x)[9ln(sin(x))−1]+C Question 3 Evaluate the integral: ∫ 37 𝑒 7 𝑥 arctan ⁡ ( 𝑒 7 𝑥 ) 𝑑 𝑥 ∫37e 7x arctan(e 7x )dx Answer Using substitution 𝑡 = 𝑒 7 𝑥 t=e 7x , 𝑑 𝑡 = 7 𝑒 7 𝑥 𝑑 𝑥 dt=7e 7x dx: ∫ 37 𝑒 7 𝑥 arctan ⁡ ( 𝑒 7 𝑥 ) 𝑑 𝑥 = 37 7 ∫ arctan ⁡ ( 𝑡 ) 𝑑 𝑡 ∫37e 7x arctan(e 7x )dx= 7 37 ∫arctan(t)dt Using formula for inverse trigonometric integrals: ∫ arctan ⁡ ( 𝑢 ) 𝑑 𝑢 = 𝑢 2 + 1 2 arctan ⁡ ( 𝑢 ) − 𝑢 2 + 𝐶 ∫arctan(u)du= 2 u 2 +1 arctan(u)− 2 u +C ∫ 37 𝑒 7 𝑥 arctan ⁡ ( 𝑒 7 𝑥 ) 𝑑 𝑥 = 37 7 [ 𝑡 2 + 1 2 arctan ⁡ ( 𝑡 ) − 𝑡 2 ] + 𝐶 ∫37e 7x arctan(e 7x )dx= 7 37 [ 2 t 2 +1 arctan(t)− 2 t ]+C Substitute 𝑡 = 𝑒 7 𝑥 t=e 7x : ∫ 37 𝑒 7 𝑥 arctan ⁡ ( 𝑒 7 𝑥 ) 𝑑 𝑥 = 37 14 [ ( 𝑒 14 𝑥 + 1 ) arctan ⁡ ( 𝑒 7 𝑥 ) − 𝑒 7 𝑥 ] + 𝐶 ∫37e 7x arctan(e 7x )dx= 14 37 [(e 14x +1)arctan(e 7x )−e 7x ]+C Question 4 Determine whether the integral ∫ 3 ∞ 8 ( 𝑥 + 4 ) 2 𝑑 𝑥 ∫ 3 ∞ (x+4) 2 8 dx is convergent or divergent. Answer Factor constants and simplify: ∫ 3 ∞ 8 ( 𝑥 + 4 ) 2 𝑑 𝑥 = 8 ∫ 3 ∞ 1 ( 𝑥 + 4 ) 2 𝑑 𝑥 ∫ 3 ∞ (x+4) 2 8 dx=8∫ 3 ∞ (x+4) 2 1 dx Use substitution 𝑢 = 𝑥 + 4 u=x+4, 𝑑 𝑢 = 𝑑 𝑥 du=dx: ∫ 3 ∞ 1 ( 𝑢 ) 2 𝑑 𝑢 = [ − 1 𝑢 ] 7 ∞ ∫ 3 ∞ (u) 2 1 du=[− u 1 ] 7 ∞ = ( 0 − ( − 1 7 ) ) = 1 7 =(0−(− 7 1 ))= 7 1 ∫ 3 ∞ 8 ( 𝑥 + 4 ) 2 𝑑 𝑥 = 8 7 ∫ 3 ∞ (x+4) 2 8 dx= 7 8 The integral is convergent. Question 5 Find a suitable form for the particular solution of the differential equation: 𝑦 ′ ′ + 4 𝑦 = − 2 𝑡 ( 𝑡 2 + 1 ) + 𝑒 − 𝑡 y ′′ +4y=−2t(t 2 +1)+e −t Answer Solve the homogeneous equation 𝑦 ′ ′ + 4 𝑦 = 0 y ′′ +4y=0: Assume 𝑦 ℎ = 𝑒 𝜆 𝑡 y h =e λt : 𝜆 2 + 4 = 0 ⟹ 𝜆 = ± 2 𝑖 λ 2 +4=0⟹λ=±2i General solution of the homogeneous equation: 𝑦 ℎ ( 𝑡 ) = 𝑐 1 cos ⁡ ( 2 𝑡 ) + 𝑐 2 sin ⁡ ( 2 𝑡 ) y h (t)=c 1 cos(2t)+c 2 sin(2t) Solve for the particular solution: For − 2 𝑡 ( 𝑡 2 + 1 ) −2t(t 2 +1), use polynomial guessing: 𝑦 𝑝 = 𝐴 𝑡 4 + 𝐵 𝑡 2 + 𝐶 y p =At 4 +Bt 2 +C. For 𝑒 − 𝑡 e −t , use 𝑦 𝑝 = 𝐷 𝑒 − 𝑡 y p =De −t . The general solution is: 𝑦 ( 𝑡 ) = 𝑐 1 cos ⁡ ( 2 𝑡 ) + 𝑐 2 sin ⁡ ( 2 𝑡 ) + 𝑦 𝑝 ( 𝑡 ) y(t)=c 1 cos(2t)+c 2 sin(2t)+y p (t) where 𝑦 𝑝 ( 𝑡 ) y p (t) includes terms matching the forcing functions.