Find the average value of the function f(x)=sin6(x)cos3(x) over interval [−π,π]

Question

BlackTom AI · Accepted Answer

Let’s restate the problem to start: Find the average value of f(x) = sin^6(x) cos^3(x) on the interval [-π, π].

Option-by-option reasoning:

- Option 0: '0' — this is the value we should arrive at after evaluating the integral. To see why, observe the integrand f(x) = sin^6(x) cos^3(x) has a particular symmetry: sin^6(x) is an even function because sine raised to an even power satisfies sin^6(-x) = sin^6(x). Likewise, cos^3(x) is also even because cos(-x) = cos(x), so cos^3(-x) = cos^3(x). The product of two even functions is even, so f(x) is even. This means the integral over [-π, π] is twice the integral over [0, π].

- The crucial trick is to compare values around π/2. If we substitute x -> π − x, we get sin(π − x) = sin x and cos(π − x) = −cos x, hence cos^3(π − x) = −cos^3(x). Therefore f(π − x) = sin^6(π − x) cos^3(π − x) = sin^6(x) (−cos^3(x)) = −f(x). This antisymmetry about x = π/2 implies that the integral of f over [0, π] cancels out when paired as (x, π − x). In other words, ∫_0^π f(x) dx = 0, because each pair f(x) and f(π − x) sums to zero.

- Since the interval is [-π, π], the integral ∫_{−π}^{π} f(x) dx equals 2∫_0^{π} f(x) dx, which remains zero in light of the previous cancellation. Consequently, the average value, being (1/(2π)) times this integral, is also zero.

- If one checks the other options conceptually: they correspond to nonzero constants like π, π/5, π/12, and π/6. Those would require a nonzero total integral, which contradicts the observed symmetry argument and cancellation on a symmetric interval for this particular power combination of sine and cosine.

In summary, the key ideas are the even nature of sin^6(x) and cos^3(x), and the antisymmetry under x → π − x that leads to cancellation of contributions over [0, π], and thus over [−π, π].

Find the average value of the function f(x)=sin6(x)cos3(x) over interval [−π,π]Single choice

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