if f'(x)=cos^2x and f(1)=0 then f(x)=
cos^2x = (1+cos2x)/2
f(x) = x/2 + sin2x/4 + C
at x=1,
0 = 1/2 + 0 + C
C = -1/2
f(x) = x/2 + sin2x/4 - 1/2
= 1/4 (2x + sin2x - 2)
0 = 1/2 + sin(2)/4 + C
C = -1/4 (2 + sin2)
adjust final answer accordingly
To find the function f(x) given its derivative f'(x), you can integrate f'(x) with respect to x.
In this case, we have f'(x) = cos^2(x). To integrate this function, we need to recall the integral of cos^2(x).
The integral of cos^2(x) can be found using the identity:
∫cos^2(x) dx = (1/2) * (x + sin(x)cos(x)) + C,
where C is the constant of integration.
Now, to determine the value of the constant C, we can use the given condition f(1) = 0.
Plugging x = 1 into the expression obtained above, we get:
f(1) = (1/2) * (1 + sin(1)cos(1)) + C = 0.
Solving this equation for C, we find:
C = -(1/2) * (1 + sin(1)cos(1)).
Therefore, the function f(x) is given by:
f(x) = (1/2) * (x + sin(x)cos(x)) - (1/2) * (1 + sin(1)cos(1)).