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

2024
Views:

313

Pages:

3
Author:

Cristian Santos

Question 1 Use the Comparison Theorem to determine whether the integral ∫ 0 ∞ 𝑥 𝑥 3 + 1 𝑑 𝑥 ∫ 0 ∞ x 3 +1 x dx is convergent or divergent. Answer We compare: 𝑥 𝑥 3 + 1 ≤ 𝑥 𝑥 3 = 1 𝑥 2 for 𝑥 > 1. x 3 +1 x ≤ x 3 x = x 2 1 forx>1. The integral of 1 𝑥 2 x 2 1 converges: ∫ 1 ∞ 1 𝑥 2 𝑑 𝑥 = [ − 1 𝑥 ] 1 ∞ = 1. ∫ 1 ∞ x 2 1 dx=[− x 1 ] 1 ∞ =1. Thus, ∫ 0 ∞ 𝑥 𝑥 3 + 1 𝑑 𝑥 ∫ 0 ∞ x 3 +1 x dx converges because it is bounded above by a convergent integral. Result: Converges Question 2 Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of: 𝑔 ( 𝑠 ) = ∫ 5 𝑠 𝑥 ( 𝑡 − 𝑡 2 ) 8 𝑑 𝑡 . g(s)=∫ 5 s x(t−t 2 ) 8 dt. Answer The Fundamental Theorem of Calculus states: 𝑑 𝑑 𝑠 ∫ 𝑎 𝑠 𝑓 ( 𝑡 ) 𝑑 𝑡 = 𝑓 ( 𝑠 ) . ds d ∫ a s f(t)dt=f(s). Thus, 𝑔 ′ ( 𝑠 ) = ( 𝑠 − 𝑠 2 ) 8 . g ′ (s)=(s−s 2 ) 8 . Result: 𝑔 ′ ( 𝑠 ) = ( 𝑠 − 𝑠 2 ) 8 g ′ (s)=(s−s 2 ) 8 Question 3 Compute the definite integrals involving absolute value: ∫ − 3 3 ∣ 𝑥 ∣ 𝑑 𝑥 and ∫ − 2 2 ∣ sin ⁡ 𝑥 ∣ 𝑑 𝑥 . ∫ −3 3 ∣x∣dxand∫ −2 2 ∣sinx∣dx. Answer 1 For ∫ − 3 3 ∣ 𝑥 ∣ 𝑑 𝑥 ∫ −3 3 ∣x∣dx: Split at 𝑥 = 0 x=0: ∫ − 3 3 ∣ 𝑥 ∣ 𝑑 𝑥 = ∫ − 3 0 − 𝑥 𝑑 𝑥 + ∫ 0 3 𝑥 𝑑 𝑥 . ∫ −3 3 ∣x∣dx=∫ −3 0 −xdx+∫ 0 3 xdx. Evaluate: = [ − 𝑥 2 2 ] − 3 0 + [ 𝑥 2 2 ] 0 3 = 9 2 + 9 2 = 9. =[− 2 x 2 ] −3 0 +[ 2 x 2 ] 0 3 = 2 9 + 2 9 =9. Answer 2 For ∫ − 2 2 ∣ sin ⁡ 𝑥 ∣ 𝑑 𝑥 ∫ −2 2 ∣sinx∣dx: Since ∣ sin ⁡ 𝑥 ∣ ∣sinx∣ is symmetric, double the integral over [ 0 , 2 ] [0,2]: ∫ − 2 2 ∣ sin ⁡ 𝑥 ∣ 𝑑 𝑥 = 2 ∫ 0 2 sin ⁡ 𝑥 𝑑 𝑥 . ∫ −2 2 ∣sinx∣dx=2∫ 0 2 sinxdx. Evaluate: ∫ sin ⁡ 𝑥 𝑑 𝑥 = − cos ⁡ 𝑥 . ∫sinxdx=−cosx. 2 [ − cos ⁡ 𝑥 ] 0 2 = 2 [ − cos ⁡ ( 2 ) + cos ⁡ ( 0 ) ] . 2[−cosx] 0 2 =2[−cos(2)+cos(0)]. Simplify: = 2 [ − cos ⁡ ( 2 ) + 1 ] . =2[−cos(2)+1]. Question 4 Evaluate the integral: ∫ ln ⁡ 2 ln ⁡ 3 𝑒 𝑥 𝑒 2 𝑥 − 2 𝑒 𝑥 + 1 𝑑 𝑥 . ∫ ln2 ln3 e 2x −2e x +1 e x dx. Answer Substitute 𝑢 = 𝑒 𝑥 u=e x , so 𝑑 𝑢 = 𝑒 𝑥 𝑑 𝑥 du=e x dx. Limits change: 𝑥 = ln ⁡ 2 ⟹ 𝑢 = 2 x=ln2⟹u=2, 𝑥 = ln ⁡ 3 ⟹ 𝑢 = 3 x=ln3⟹u=3. Simplify the denominator: 𝑒 2 𝑥 − 2 𝑒 𝑥 + 1 = ( 𝑒 𝑥 − 1 ) 2 . e 2x −2e x +1=(e x −1) 2 . Thus, the integral becomes: ∫ 2 3 𝑢 ( 𝑢 − 1 ) 2 𝑑 𝑢 . ∫ 2 3 (u−1) 2 u du. Simplify: Split the fraction: 𝑢 ( 𝑢 − 1 ) 2 = 1 𝑢 − 1 + 1 ( 𝑢 − 1 ) 2 . (u−1) 2 u = u−1 1 + (u−1) 2 1 . Integrate: ∫ 2 3 ( 1 𝑢 − 1 + 1 ( 𝑢 − 1 ) 2 ) 𝑑 𝑢 . ∫ 2 3 ( u−1 1 + (u−1) 2 1 )du. Evaluate: ∫ 1 𝑢 − 1 𝑑 𝑢 = ln ⁡ ∣ 𝑢 − 1 ∣ , ∫ 1 ( 𝑢 − 1 ) 2 𝑑 𝑢 = − 1 𝑢 − 1 . ∫ u−1 1 du=ln∣u−1∣,∫ (u−1) 2 1 du=− u−1 1 . Substitute limits 𝑢 = 2 u=2 and 𝑢 = 3 u=3: ∫ ln ⁡ 2 ln ⁡ 3 𝑒 𝑥 𝑒 2 𝑥 − 2 𝑒 𝑥 + 1 𝑑 𝑥 = [ ln ⁡ ∣ 𝑢 − 1 ∣ − 1 𝑢 − 1 ] 2 3 . ∫ ln2 ln3 e 2x −2e x +1 e x dx=[ln∣u−1∣− u−1 1 ] 2 3 . Simplify: = ( ln ⁡ 2 − 1 ) − ( ln ⁡ 1 − ( − 1 ) ) . =(ln2−1)−(ln1−(−1)). = ln ⁡ 2 − 1 + 1 = 7 4 . =ln2−1+1= 4 7 . Result: 7 4 4 7 Question 5 What is the integration of 𝑑 𝑥 dx? Answer ∫ 𝑑 𝑥 = 𝑥 + 𝐶 , ∫dx=x+C, where 𝐶 C is the constant of integration.