Задача 31922 ...

vk215516862

2018-12-14 13:26:46

1. ∫(от -2 до 1) (x^2-4x+4) dx
2.∫(от 0 до π) sin x/4 dx

sova

2018-12-14 13:32:07

★

. ∫^(1)_(-2)(x^2–4x+4) dx=((x^3/3)-4*(x^2/2)+4x)|^(1)_(-2)=

=(1/3)-4*(1/2)+4 - (-8/3)+4*(4/2)-4*(-2)=

=3-2+4+8+8=21

или
∫^(1)_(-2)(x^2–4x+4) dx= ∫^(1)_(-2)(x-2)^2 d(x-2)=((x-2)^3/3)|^(1)_(-2)=

=(-1/3)-(-4)^3/3=63/3=21


2.∫^(π)_(0) sin (x/4) dx=4* ∫^(π)_(0) sin (x/4) d(x/4)=

=4*(- cos(x/4))|^(π)_(0)= 4*(-cos(π/4)+cos0)= 4*((-sqrt(2)/2) + 1)=

=-2sqrt(2) + 4