Помогите решить пожалуйста)

Question 1

Question 2

просто ответы?

Question 3

с решением

Question 4

B1. интеграл(x^2)dx = (x^3)/3 = от 3 до 5 = 125/3 - 27/3 = 98/3
B2. интеграл(sinx)dx = -cosx = от π/6 до π/2 = -cos(π/2) + cos(π/6) = √3/2
B3. x^6/6 = от 1 до 2 = 2^6/6 - 1/6 = 63/6
B4. интеграл(sin(6x))dx = (1/6)*интеграл(sin(6x))d(6x) = (1/6)*(-cos(6x)) = -cos(6x) / 6 = от π/24 до π/3 = -cos(6π/3)/6 + cos(6π/24)/6 = -cos(2π)/6 + cos(π/4)/6 = -1/6 + √2/12 = (√2 - 2)/12
B5. π/6

Question 5

1. $\int\limits^5_3 {x^2} \, dx = \frac{x^3}{3} = \frac{5^3-3^3}{3} = \frac{125-27}{3} =32 \frac{2}{3}$

2. $\int\limits^ \frac{ \pi }{2} _ \frac{ \pi }{6} {sinx} \, dx =-cosx= -(cos \frac{ \pi }{2} -cos \frac{ \pi }{6})=-(0- \frac{ \sqrt{3} }{2} )=\frac{ \sqrt{3} }{2}$

3. $\int\limits^2_1 {x^5} \, dx = \frac{x^6}{6} = \frac{2^2-1^6}{6} = \frac{64-1}{6} =10 \frac{1}{2}$

4. $\int\limits^ \frac{ \pi }{3} _ \frac{ \pi }{24} {2cos(3x)sin(3x)} \, dx= \int\limits^ \frac{ \pi }{3} _ \frac{ \pi }{24} {sin(6x)} \, dx= -\frac{cos(6x)}{6}=- \frac{cos(2 \pi )-cos( \frac{ \pi }{4} )}{6}= \\ \\ = \frac{cos( \frac{ \pi }{4})-cos(2 \pi )}{6}= \frac{ \frac{ \sqrt{2} }{2}-1 }{6} = \frac{ \sqrt{2}-2 }{12}$

5. $\int\limits^ \frac{ \sqrt{3} }{2} _ \frac{1}{2} { \frac{dx}{ \sqrt{1-x^2} }=arcsin (x) = arcsin ( \frac{ \sqrt{3} }{2} )-arcsin ( \frac{1}{2} )=\frac{ \pi }{3} -\frac{ \pi }{6}=\frac{ \pi }{6}$

kalbim_zn · Answer 1 · 2018-03-16T14:14:40+0000

B1. интеграл(x^2)dx = (x^3)/3 = от 3 до 5 = 125/3 - 27/3 = 98/3
B2. интеграл(sinx)dx = -cosx = от π/6 до π/2 = -cos(π/2) + cos(π/6) = √3/2
B3. x^6/6 = от 1 до 2 = 2^6/6 - 1/6 = 63/6
B4. интеграл(sin(6x))dx = (1/6)*интеграл(sin(6x))d(6x) = (1/6)*(-cos(6x)) = -cos(6x) / 6 = от π/24 до π/3 = -cos(6π/3)/6 + cos(6π/24)/6 = -cos(2π)/6 + cos(π/4)/6 = -1/6 + √2/12 = (√2 - 2)/12
B5. π/6