Question

Решите хоть что-то,пожалуйста умоляю!

Answer 1

1) ∫₂³(1-x)⁴dx=(1/5)*(x-1)⁵ |₂³=32/5-1/5=31/5=6,2.
2) ∫₋π/₂ π/² (3*cosx)dx=3*sinx |-π/₂ π/²=3-(-3)=6.
3) ∫₁²(2x-5)dx=x²-5x |₁²=-6-(-4)=-2
4) ∫₀¹(x+5)⁵=(x+1)⁶/6 |₀¹=(64-1)/6=63/6=21/2=10,5.
5) ∫₀ π/² (3/cos²(x/2))=6*tg(x/1)6-0=6.
6) ∫₀³ x²dx=x³/3 |₀³=27/3-0=9.
7) ∫₀¹ (x²-2x+1)dx=x³/3-x²+x|₀¹=1/3-1+1-0=1/3.
8) ∫₀π/⁴ (4/cos²x)dx=4*tgx=4*tgx |₀π/⁴=4-0=4.
9) ∫₀¹ (x²+4x-1)dx=x³/3+2x²-x |₀¹=1/3+2*1²-1-0=4/3.
10) ∫₀π/¹² (108*sin(6*x)dx=-108*cos(6*x)/6=-18*cos(6*x) |₀π/¹²=-0-(-18)=18.
11) ∫₀π/⁴ (4*cos(2*x)dx=4*sin(2*x)/2=2*sin(2*x) |₀π/⁴=2-0=2.
12) ∫₀π (3*sin(x/2)dx=-3*cos(x/2)/(1/2) |₀π=-6*0-(-6)*1=6.

Answer 2

Да.