Question

Помогите решить пожалуйста)

Answer 1

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

Answer 2

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

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

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

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

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