let f be a function 0 to 2pi with continuous first and second derivatives and suchthat f"x>0 for 0<x<2pi integral f(x)cosx from 2pm to 0
To find the value of the integral ∫ f(x)cos(x) from 2π to 0, we need to understand the given information about the function f.
The given statement states that f is a function from 0 to 2π with continuous first and second derivatives, and f''(x) > 0 for 0 < x < 2π.
To solve the integral, we will use integration by parts. The integration by parts formula states:
∫ u dv = uv - ∫ v du,
where u and v are differentiable functions.
Let's start by assigning u = f(x) and dv = cos(x). Then, we can differentiate u to find du and integrate dv to find v.
du = f'(x) dx (by differentiating u)
v = ∫ cos(x) dx = sin(x) (by integrating dv)
Now, we can use the integration by parts formula to rewrite the integral:
∫ f(x)cos(x) dx = u*v - ∫ v*du
∫ f(x)cos(x) dx = f(x)sin(x) - ∫ sin(x) * f'(x) dx
We can simplify the integral using the given condition that f''(x) > 0 for 0 < x < 2π. Since the second derivative is positive, the first derivative f'(x) is an increasing function.
Given that the integral ranges from 2π to 0, we can say that f(2π) > f(0). And since sin(x) is positive from 0 to 2π, sin(x) is positive in the given integral limits.
So, f(x)sin(x) is increasing from 2π to 0, and sin(x) is always positive. Therefore, f(x)sin(x) > 0 for 2π > x > 0.
Based on this information, we can conclude that the value of the integral ∫ f(x)cos(x) dx from 2π to 0 is greater than 0.