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

2024
Views:

292

Pages:

5
Author:

Cristian Santos

Question 1 Integrate 1 ln ⁡ ( 𝑥 ) ln(x) 1 . Answer Using the chain rule: 𝑑 𝑑 𝑥 ( 1 ln ⁡ ( 𝑥 ) ) = − 1 ( ln ⁡ ( 𝑥 ) ) 2 ⋅ 1 𝑥 . dx d ( ln(x) 1 )=− (ln(x)) 2 1 ⋅ x 1 . Simplify: 𝑑 𝑑 𝑥 ( 1 ln ⁡ ( 𝑥 ) ) = − 1 𝑥 ( ln ⁡ ( 𝑥 ) ) 2 . dx d ( ln(x) 1 )=− x(ln(x)) 2 1 . Thus, integrating 1 ln ⁡ ( 𝑥 ) ln(x) 1 involves special techniques or numerical approximations. Question 2 Find the integral of 𝑒 − 𝑥 𝑑 𝑥 e −x dx. Answer Using substitution 𝑢 = − 𝑥 u=−x, 𝑑 𝑢 = − 𝑑 𝑥 du=−dx: ∫ 𝑒 − 𝑥 𝑑 𝑥 = − ∫ 𝑒 𝑢 𝑑 𝑢 = − 𝑒 𝑢 + 𝐶 = − 𝑒 − 𝑥 + 𝐶 . ∫e −x dx=−∫e u du=−e u +C=−e −x +C. Result: − 𝑒 − 𝑥 + 𝐶 −e −x +C. Question 3 How do you find the integration of log ⁡ 𝑥 logx? Answer 1 Use integration by parts: Let 𝑢 = ln ⁡ ( 𝑥 ) u=ln(x), 𝑑 𝑣 = 𝑑 𝑥 dv=dx. Then 𝑑 𝑢 = 1 𝑥 𝑑 𝑥 du= x 1 dx, 𝑣 = 𝑥 v=x: ∫ ln ⁡ ( 𝑥 ) 𝑑 𝑥 = 𝑥 ln ⁡ ( 𝑥 ) − ∫ 𝑥 ⋅ 1 𝑥 𝑑 𝑥 . ∫ln(x)dx=xln(x)−∫x⋅ x 1 dx. Simplify: ∫ ln ⁡ ( 𝑥 ) 𝑑 𝑥 = 𝑥 ln ⁡ ( 𝑥 ) − 𝑥 + 𝐶 . ∫ln(x)dx=xln(x)−x+C. Result: 𝑥 ln ⁡ ( 𝑥 ) − 𝑥 + 𝐶 xln(x)−x+C. Answer 2 For log ⁡ 𝑎 ( 𝑥 ) log a (x), use: ∫ log ⁡ 𝑎 ( 𝑥 ) 𝑑 𝑥 = 𝑥 ( ln ⁡ ( 𝑥 ) − 1 ) ln ⁡ ( 𝑎 ) + 𝐶 . ∫log a (x)dx= ln(a) x(ln(x)−1) +C. Question 4 What is the integral of sec ⁡ 3 ( 𝑥 ) sec 3 (x)? Answer Use integration by parts: Let 𝑢 = sec ⁡ ( 𝑥 ) u=sec(x), 𝑑 𝑣 = sec ⁡ 2 ( 𝑥 ) 𝑑 𝑥 dv=sec 2 (x)dx, then 𝑑 𝑢 = sec ⁡ ( 𝑥 ) tan ⁡ ( 𝑥 ) 𝑑 𝑥 du=sec(x)tan(x)dx, 𝑣 = tan ⁡ ( 𝑥 ) v=tan(x): ∫ sec ⁡ 3 ( 𝑥 ) 𝑑 𝑥 = sec ⁡ ( 𝑥 ) tan ⁡ ( 𝑥 ) − ∫ sec ⁡ ( 𝑥 ) tan ⁡ 2 ( 𝑥 ) 𝑑 𝑥 . ∫sec 3 (x)dx=sec(x)tan(x)−∫sec(x)tan 2 (x)dx. Using tan ⁡ 2 ( 𝑥 ) = sec ⁡ 2 ( 𝑥 ) − 1 tan 2 (x)=sec 2 (x)−1, simplify: ∫ sec ⁡ 3 ( 𝑥 ) 𝑑 𝑥 = 1 2 sec ⁡ ( 𝑥 ) tan ⁡ ( 𝑥 ) + 1 2 ln ⁡ ∣ sec ⁡ ( 𝑥 ) + tan ⁡ ( 𝑥 ) ∣ + 𝐶 . ∫sec 3 (x)dx= 2 1 sec(x)tan(x)+ 2 1 ln∣sec(x)+tan(x)∣+C. Question 5 Find the antiderivative of ( sin ⁡ 𝑥 ) 2 (sinx) 2 . Answer 1 Use the identity sin ⁡ 2 ( 𝑥 ) = 1 − cos ⁡ ( 2 𝑥 ) 2 sin 2 (x)= 2 1−cos(2x) : ∫ ( sin ⁡ 𝑥 ) 2 𝑑 𝑥 = 1 2 ∫ ( 1 − cos ⁡ ( 2 𝑥 ) ) 𝑑 𝑥 . ∫(sinx) 2 dx= 2 1 ∫(1−cos(2x))dx. Simplify: ∫ ( sin ⁡ 𝑥 ) 2 𝑑 𝑥 = 1 2 ( 𝑥 − 1 2 sin ⁡ ( 2 𝑥 ) ) + 𝐶 . ∫(sinx) 2 dx= 2 1 (x− 2 1 sin(2x))+C. Result: 𝑥 2 − sin ⁡ ( 2 𝑥 ) 4 + 𝐶 2 x − 4 sin(2x) +C.