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

2024
Views:

34

Pages:

6
Author:

Cristian Santos

Question 1 Integrate using partial fractions: ∫ 𝑥 𝑥 2 − 3 𝑥 − 4 𝑑 𝑥 ∫ x 2 −3x−4 x dx Answer 1 Factor the denominator: 𝑥 2 − 3 𝑥 − 4 = ( 𝑥 + 1 ) ( 𝑥 − 4 ) . x 2 −3x−4=(x+1)(x−4). Decompose using partial fractions: 𝑥 ( 𝑥 + 1 ) ( 𝑥 − 4 ) = 𝐴 𝑥 + 1 + 𝐵 𝑥 − 4 . (x+1)(x−4) x = x+1 A + x−4 B . Multiply through by ( 𝑥 + 1 ) ( 𝑥 − 4 ) (x+1)(x−4): 𝑥 = 𝐴 ( 𝑥 − 4 ) + 𝐵 ( 𝑥 + 1 ) . x=A(x−4)+B(x+1). Substitute 𝑥 = 4 x=4: 4 = 𝐴 ( 4 − 4 ) + 𝐵 ( 4 + 1 ) ⟹ 𝐵 = 4 5 . 4=A(4−4)+B(4+1)⟹B= 5 4 . Substitute 𝑥 = − 1 x=−1: − 1 = 𝐴 ( − 1 − 4 ) + 𝐵 ( 0 ) ⟹ 𝐴 = 1 5 . −1=A(−1−4)+B(0)⟹A= 5 1 . Rewrite the integral: ∫ 𝑥 ( 𝑥 + 1 ) ( 𝑥 − 4 ) 𝑑 𝑥 = ∫ ( 1 / 5 𝑥 + 1 + 4 / 5 𝑥 − 4 ) 𝑑 𝑥 . ∫ (x+1)(x−4) x dx=∫( x+1 1/5 + x−4 4/5 )dx. Solve each term: ∫ 1 / 5 𝑥 + 1 𝑑 𝑥 + ∫ 4 / 5 𝑥 − 4 𝑑 𝑥 = 1 5 ln ⁡ ∣ 𝑥 + 1 ∣ + 4 5 ln ⁡ ∣ 𝑥 − 4 ∣ + 𝐶 . ∫ x+1 1/5 dx+∫ x−4 4/5 dx= 5 1 ln∣x+1∣+ 5 4 ln∣x−4∣+C. Result: ∫ 𝑥 𝑥 2 − 3 𝑥 − 4 𝑑 𝑥 = 1 5 ln ⁡ ∣ 𝑥 + 1 ∣ + 4 5 ln ⁡ ∣ 𝑥 − 4 ∣ + 𝐶 . ∫ x 2 −3x−4 x dx= 5 1 ln∣x+1∣+ 5 4 ln∣x−4∣+C. Question 2 How to add two compound fractions: 1 2 + 3 𝑥 2 𝑥 2 1 + x 2 3x or fractions with denominators containing fractions. Answer 1 For the first example: 1 2 + 3 𝑥 2 𝑥 . 2 1 + x 2 3x . Simplify the second term: 3 𝑥 2 𝑥 = 3 2 . x 2 3x = 2 3 . Thus: 1 2 + 3 2 = 4 2 = 2. 2 1 + 2 3 = 2 4 =2. For fractions with fractions in the denominator like 𝑥 2 + 1 𝑥 2 x + x 1 : Find a common denominator: 𝑥 2 + 1 𝑥 = 𝑥 2 + 2 2 𝑥 . 2 x + x 1 = 2x x 2 +2 . Question 3 What is the integral of ( cos ⁡ 𝑥 ) 2 (cosx) 2 ? Answer 1 Use the double-angle identity: cos ⁡ 2 ( 𝑥 ) = 1 + cos ⁡ ( 2 𝑥 ) 2 . cos 2 (x)= 2 1+cos(2x) . Thus: ∫ cos ⁡ 2 ( 𝑥 ) 𝑑 𝑥 = ∫ 1 + cos ⁡ ( 2 𝑥 ) 2 𝑑 𝑥 . ∫cos 2 (x)dx=∫ 2 1+cos(2x) dx. Split and integrate: ∫ cos ⁡ 2 ( 𝑥 ) 𝑑 𝑥 = 1 2 ∫ 1 𝑑 𝑥 + 1 2 ∫ cos ⁡ ( 2 𝑥 ) 𝑑 𝑥 . ∫cos 2 (x)dx= 2 1 ∫1dx+ 2 1 ∫cos(2x)dx. = 1 2 𝑥 + 1 4 sin ⁡ ( 2 𝑥 ) + 𝐶 . = 2 1 x+ 4 1 sin(2x)+C. Result: ∫ cos ⁡ 2 ( 𝑥 ) 𝑑 𝑥 = 𝑥 2 + sin ⁡ ( 2 𝑥 ) 4 + 𝐶 . ∫cos 2 (x)dx= 2 x + 4 sin(2x) +C. Question 4 How do you integrate ∫ sin ⁡ ( 2 𝑥 ) 𝑑 𝑥 ∫sin(2x)dx? Answer 1 Using substitution: Let 𝑢 = 2 𝑥 u=2x, then 𝑑 𝑢 = 2 𝑑 𝑥 du=2dx or 𝑑 𝑥 = 1 2 𝑑 𝑢 dx= 2 1 du. ∫ sin ⁡ ( 2 𝑥 ) 𝑑 𝑥 = 1 2 ∫ sin ⁡ ( 𝑢 ) 𝑑 𝑢 . ∫sin(2x)dx= 2 1 ∫sin(u)du. Integrate sin ⁡ ( 𝑢 ) sin(u): 1 2 ∫ sin ⁡ ( 𝑢 ) 𝑑 𝑢 = − 1 2 cos ⁡ ( 𝑢 ) + 𝐶 . 2 1 ∫sin(u)du=− 2 1 cos(u)+C. Substitute back 𝑢 = 2 𝑥 u=2x: ∫ sin ⁡ ( 2 𝑥 ) 𝑑 𝑥 = − 1 2 cos ⁡ ( 2 𝑥 ) + 𝐶 . ∫sin(2x)dx=− 2 1 cos(2x)+C. Question 5 Find the antiderivative of ∫ cos ⁡ ( 2 𝑥 ) 𝑑 𝑥 ∫cos(2x)dx. Answer Using substitution: Let 𝑢 = 2 𝑥 u=2x, then 𝑑 𝑢 = 2 𝑑 𝑥 du=2dx or 𝑑 𝑥 = 1 2 𝑑 𝑢 dx= 2 1 du. ∫ cos ⁡ ( 2 𝑥 ) 𝑑 𝑥 = 1 2 ∫ cos ⁡ ( 𝑢 ) 𝑑 𝑢 . ∫cos(2x)dx= 2 1 ∫cos(u)du. Integrate cos ⁡ ( 𝑢 ) cos(u): 1 2 ∫ cos ⁡ ( 𝑢 ) 𝑑 𝑢 = 1 2 sin ⁡ ( 𝑢 ) + 𝐶 . 2 1 ∫cos(u)du= 2 1 sin(u)+C. Substitute back 𝑢 = 2 𝑥 u=2x: ∫ cos ⁡ ( 2 𝑥 ) 𝑑 𝑥 = 1 2 sin ⁡ ( 2 𝑥 ) + 𝐶 . ∫cos(2x)dx= 2 1 sin(2x)+C. Result: ∫ cos ⁡ ( 2 𝑥 ) 𝑑 𝑥 = 1 2 sin ⁡ ( 2 𝑥 ) + 𝐶 . ∫cos(2x)dx= 2 1 sin(2x)+C.