Question

Find the average value of the function over the given interval and all values of \(x\) in the interval for which the function equals its average value. $$ f(x)=\cos x, \quad[0, \pi / 2] $$

Short answer

The average value of the cosine function over the interval [0, \(\pi / 2\)] is \(\frac{2}{\pi}\). The function equals its average value at \(x = \arccos(\frac{2}{\pi})\), which is within the interval \([0, \pi / 2]\).

Step-by-step solution

Compute the average value of the function

To find the average value of the function \(f(x) = \cos x\) over the interval [0, \(\pi / 2\)], we use the formula for average value of a function: \(Avg = \frac{1}{b - a} \int_a^b f(x) dx\). Substituting the given details, we calculate \(\frac{1}{\pi / 2 - 0} \int_0^{\pi / 2} \cos x dx = \frac{2}{\pi} \int_0^{\pi / 2} \cos x dx\). Evaluating the integral, we get \( \frac{2}{\pi} [sin x]_0^{\pi / 2} = \frac{2}{\pi} (1 - 0) = \frac{2}{\pi}\).

Solve for x

Next, find the value(s) of \(x\) for which \(f(x) = \text{{average value}}\). This means solving the equation \(\cos x = \frac{2}{\pi}\) for \(x\) within the interval \([0, \pi / 2]\). Using the inverse cosine function (also known as arccosine), we can solve for \(x\): \(x = \arccos(\frac{2}{\pi})\). However, we need to make sure that \(x\) is within the given interval. Since \(2/\pi \approx 0.637 < 1\), \(x = \arccos(\frac{2}{\pi})\) is indeed within the interval \([0, \pi / 2]\).