The average value of a function on the interval [a, b] is given by the formula:

1/(b - a) * ∫ from a to b [f(x) dx]

Here, our function g(x) = 5cos(x) and the interval is [-π/2, π/2].

So, we need to find:

1/(π - (-π/2)) * ∫ from -π/2 to π/2 [5cos(x) dx]

This simplifies to:

2/π * ∫ from -π/2 to π/2 [5cos(x) dx]

The integral of cos(x) is sin(x), so we have:

2/π * [5sin(x)] from -π/2 to π/2

Evaluating this gives:

2/π * [5sin(π/2) - 5sin(-π/2)]

Since sin(π/2) = 1 and sin(-π/2) = -1, this further simplifies to:

2/π * [5*1 - 5*(-1)] = 2/π * [5 + 5] = 2/π * 10 = 20/π

So, the average value of the function g(x) = 5cos(x) on the interval [-π/2, π/2] is 20/π.

Question

The average value of a function on the interval [a, b] is given by the formula:

1/(b - a) * ∫ from a to b [f(x) dx]

Here, our function g(x) = 5cos(x) and the interval is [-π/2, π/2].

So, we need to find:

1/(π - (-π/2)) * ∫ from -π/2 to π/2 [5cos(x) dx]

This simplifies to:

2/π * ∫ from -π/2 to π/2 [5cos(x) dx]

The integral of cos(x) is sin(x), so we have:

2/π * [5sin(x)] from -π/2 to π/2

Evaluating this gives:

2/π * [5sin(π/2) - 5sin(-π/2)]

Since sin(π/2) = 1 and sin(-π/2) = -1, this further simplifies to:

2/π * [5*1 - 5*(-1)] = 2/π * [5 + 5] = 2/π * 10 = 20/π

So, the average value of the function g(x) = 5cos(x) on the interval [-π/2, π/2] is 20/π.

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