use te fundamental theorem of calculus to evaluate the integral
Integral [0, pi/3] sin^2(x)dx
I'm confused on what F(x) should be
well, sin^2(x) = 1/2 (1-cos(2x))
that should be easy to integrate, right?
To evaluate the given integral using the fundamental theorem of calculus, we need to find the antiderivative of the integrand, which is sin^2(x).
To determine the antiderivative, we can use the trigonometric identity:
sin^2(x) = (1 - cos(2x)) / 2
Now, let's rewrite the integral in terms of this identity:
∫[0, pi/3] sin^2(x)dx = ∫[0, pi/3] (1 - cos(2x))/2 dx
To apply the fundamental theorem of calculus, we need to find F(x) such that F'(x) = (1 - cos(2x))/2. We can obtain F(x) by integrating the integrand with respect to x.
First, let's integrate the constant term:
∫(1/2) dx = (1/2)x
Next, let's integrate the cosine term:
∫[0, pi/3] cos(2x)/2 dx
To integrate cos(2x), we use the substitution u = 2x and du = 2dx:
∫(1/2) cos(2x) dx = (1/4) ∫cos(u) du = (1/4) sin(u) + C
Substituting back u = 2x:
(1/4) sin(2x) + C
Now, we can construct F(x) by adding both integrated terms:
F(x) = (1/2)x + (1/4) sin(2x) + C
To evaluate the definite integral, we can simply substitute the limits into the antiderivative:
∫[0, pi/3] (1 - cos(2x))/2 dx = F(pi/3) - F(0)
Evaluate F(pi/3):
F(pi/3) = (1/2)(pi/3) + (1/4) sin(2(pi/3)) + C
= (pi/6) + (1/4)(√3/2) + C
Evaluate F(0):
F(0) = (1/2)(0) + (1/4) sin(2(0)) + C
= 0 + 0 + C
Therefore, the definite integral is:
∫[0, pi/3] sin^2(x)dx = F(pi/3) - F(0)
= (pi/6) + (1/4)(√3/2) + C - C
= (pi/6) + (1/4)(√3/2)
So, the value of the given integral is (pi/6) + (1/4)(√3/2).