The area enclosed by y = x & # 179; and y = x & # 178

The area enclosed by y = x & # 179; and y = x & # 178

Let X & # 179; = x & # 178;
X = 0 or 1
Calculate the definite integral of the function s = x & # 179; - X & # 178; on [0,1]
The definite integral is - 1 / 12
Take the absolute value to get an area of 1 / 12

A mathematical problem: find the maximum and minimum of the function y = 7-4sinx * cosx + 4cosx ^ 2-4cosx ^ 4

Simplification:
y=7+4(cosx^2-sinx*cosx-cosx^4)
=7+4[cosx^2(1-cosx^2)-sinx*cosx)
=7+4[cosx^2*sinx^2-cosx*sinx]
=7+4[sinx*cosx(sinx*cosx-1)]
=7+4{1/2sin(2x)*[1/2sin(2x)-1]}
Let t = 1 / 2Sin (2x)
Then: t ∈ [- 1 / 2,1 / 2]
f(t)=t^2-t
When t = 1 / 2, f (T) has a minimum value of - 1 / 4, then Ymin = 6
When t = - 1 / 2, f (T) has a maximum of 3 / 4, then ymax = 10

Find the minimum value of the function y = - 4cos & # 178; X + 4sinx + 5, and find the set of X when the minimum value is obtained

Y = - 4 (cosx) ^ 2 + 4sinx + 5 = - 4 * [1 - (SiNx) ^ 2] + 4sinx + 5 = 4 (SiNx) ^ 2 + 4sinx + 1 = (2sinx + 1) ^ 2. Therefore, when 2sinx + 1 = 0, that is, SiNx = - 1 / 2, that is, x = 2K π + 3 π / 2 ± π / 3, y has the minimum value of 0