Задача 57996 ...
Условие
1
∫ sin^(2)(π*x) dx
0
математика ВУЗ
202
Решение
★
x_(1)=1/6=0,16666...
x_(2)=2/6=1/3=0,33333...
x_(3)=3/6=1/2=0,5
x_(4)=4/6=2/3=0,66666...
x_(5)=5/6=0,83333
x_(6)=6/6=1
f(x)=sin^2πx
y_(o)=sin^2π*0=0^2=0
y_(1)=sin^2π*(1/6)=(1/2)^2=0,25
y_(2)=sin^2π*(1/3)=(\sqrt(3)/2)^2=3/4=0,75
y_(3)=sin^2π*(1/2)=1^2=1
y_(4)=sin^2π*(2/3)=(\sqrt(3)/2)^2=3/4=0,75
y_(5)=sin^2π*(5/6)=(1/2)^2=0,25
y_(6)=sin^2π*(1)=0^2=0
[m] ∫ ^{1}_{0}sin^2πxdx=\frac{1-0}{6}\cdot (\frac{0+0}{2}+0,25+0,75+1+0,75+0,25)=\frac{1}{6}\cdot 3=0,5[/m]
