The average value of the function \( f(x)=\cos \left(\frac{1}{2} x\right) \) on the closed interval \( [-4,0] \) is

To find the average value of the function \( f(x) = \cos\left(\frac{1}{2} x\right) \) on the interval \( [-4, 0] \), we use the formula for the average value of a function over an interval \([a, b]\): \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] In this case, \( a = -4 \) and \( b = 0 \): 1. Calculate \( b - a = 0 - (-4) = 4 \). 2. Now, set up the integral: \[ \int_{-4}^{0} \cos\left(\frac{1}{2} x\right) \, dx \] To solve this integral, we can use a substitution: Let \( u = \frac{1}{2} x \), then \( du = \frac{1}{2} dx \) or \( dx = 2 du \). The limits of integration change accordingly: - When \( x = -4 \), \( u = \frac{1}{2}(-4) = -2 \). - When \( x = 0 \), \( u = \frac{1}{2}(0) = 0 \). Thus, the integral becomes: \[ \int_{-2}^{0} \cos(u) \cdot 2 \, du = 2 \int_{-2}^{0} \cos(u) \, du \] The integral of \(\cos(u)\) is \(\sin(u)\), so we have: \[ 2 \left[ \sin(u) \right]_{-2}^{0} = 2 \left[ \sin(0) - \sin(-2) \right] = 2 \left[ 0 - (-\sin(2)) \right] = 2 \sin(2) \] Now, substituting this back into the average value formula: \[ \text{Average value} = \frac{1}{4} \cdot (2 \sin(2)) = \frac{\sin(2)}{2} \] Therefore, the average value of the function \( f(x) = \cos\left( \frac{1}{2} x \right) \) on the interval \( [-4, 0] \) is: \[ \boxed{\frac{\sin(2)}{2}} \]