For a positive constant number $ p$, find $ \lim_{n\to\infty} \frac {1}{n^{p \plus{} 1}}\sum_{k \equal{} 0}^{n \minus{} 1} \int_{2k\pi}^{(2k \plus{} 1)\pi} x^p\sin ^ 3 x\cos ^ 2x\ dx.$
To find the limit 

\[
\lim_{n\to\infty} \frac {1}{n^{p + 1}}\sum_{k = 0}^{n - 1} \int_{2k\pi}^{(2k + 1)\pi} x^p\sin ^ 3 x\cos ^ 2x\ dx,
\]

we start by simplifying the integrand. Using trigonometric identities, we have:

\[
\sin^3 x \cos^2 x = \sin x \cdot \sin^2 x \cdot \cos^2 x = \sin x \cdot (1 - \cos^2 x) \cdot \cos^2 x = \sin x \cdot (\cos^2 x - \cos^4 x).
\]

Using the identity \(\sin^2 x = \frac{1 - \cos(2x)}{2}\) and \(\cos^2 x = \frac{1 + \cos(2x)}{2}\), we can rewrite \(\sin^3 x \cos^2 x\) as:

\[
\sin^3 x \cos^2 x = \sin x \left( \frac{1 + \cos(2x)}{2} - \left( \frac{1 + \cos(2x)}{2} \right)^2 \right).
\]

Simplifying further, we get:

\[
\sin^3 x \cos^2 x = \sin x \left( \frac{1 + \cos(2x)}{2} - \frac{1 + 2\cos(2x) + \cos^2(2x)}{4} \right).
\]

\[
= \sin x \left( \frac{2(1 + \cos(2x)) - (1 + 2\cos(2x) + \cos^2(2x))}{4} \right).
\]

\[
= \sin x \left( \frac{2 + 2\cos(2x) - 1 - 2\cos(2x) - \cos^2(2x)}{4} \right).
\]

\[
= \sin x \left( \frac{1 - \cos^2(2x)}{4} \right).
\]

\[
= \sin x \left( \frac{\sin^2(2x)}{4} \right).
\]

\[
= \frac{1}{4} \sin x \sin^2(2x).
\]

Using the identity \(\sin^2(2x) = \frac{1 - \cos(4x)}{2}\), we get:

\[
\sin^3 x \cos^2 x = \frac{1}{4} \sin x \cdot \frac{1 - \cos(4x)}{2} = \frac{1}{8} \sin x (1 - \cos(4x)).
\]

Thus, the integral becomes:

\[
\int_{2k\pi}^{(2k + 1)\pi} x^p \sin^3 x \cos^2 x \, dx = \frac{1}{8} \int_{2k\pi}^{(2k + 1)\pi} x^p \sin x (1 - \cos(4x)) \, dx.
\]

This can be split into two integrals:

\[
\frac{1}{8} \left( \int_{2k\pi}^{(2k + 1)\pi} x^p \sin x \, dx - \int_{2k\pi}^{(2k + 1)\pi} x^p \sin x \cos(4x) \, dx \right).
\]

We now evaluate these integrals. The first integral is:

\[
\int_{2k\pi}^{(2k + 1)\pi} x^p \sin x \, dx.
\]

Using integration by parts, let \(u = x^p\) and \(dv = \sin x \, dx\). Then \(du = p x^{p-1} \, dx\) and \(v = -\cos x\). Thus,

\[
\int u \, dv = uv - \int v \, du = -x^p \cos x \bigg|_{2k\pi}^{(2k + 1)\pi} + \int_{2k\pi}^{(2k + 1)\pi} p x^{p-1} \cos x \, dx.
\]

Evaluating the boundary terms, we get:

\[
-x^p \cos x \bigg|_{2k\pi}^{(2k + 1)\pi} = -((2k + 1)\pi)^p \cos((2k + 1)\pi) + (2k\pi)^p \cos(2k\pi).
\]

Since \(\cos((2k + 1)\pi) = -1\) and \(\cos(2k\pi) = 1\), we have:

\[
-((2k + 1)\pi)^p (-1) + (2k\pi)^p (1) = ((2k + 1)\pi)^p + (2k\pi)^p.
\]

The second integral is:

\[
\int_{2k\pi}^{(2k + 1)\pi} x^p \sin x \cos(4x) \, dx.
\]

Using the product-to-sum identities, we have:

\[
\sin x \cos(4x) = \frac{1}{2} (\sin(5x) + \sin(-3x)) = \frac{1}{2} (\sin(5x) - \sin(3x)).
\]

Thus, the integral becomes:

\[
\int_{2k\pi}^{(2k + 1)\pi} x^p \sin x \cos(4x) \, dx = \frac{1}{2} \left( \int_{2k\pi}^{(2k + 1)\pi} x^p \sin(5x) \, dx - \int_{2k\pi}^{(2k + 1)\pi} x^p \sin(3x) \, dx \right).
\]

Using similar integration by parts as before, we find that these integrals also evaluate to terms involving \((2k + 1)\pi\) and \(2k\pi\) raised to the power \(p\).

Combining all these results, we get:

\[
\frac{1}{8} \left( \sum_{k=0}^{n-1} ((2k + 1)\pi)^p + (2k\pi)^p - \frac{1}{2} \left( \sum_{k=0}^{n-1} ((2k + 1)\pi)^p + (2k\pi)^p \right) \right).
\]

Simplifying, we get:

\[
\frac{1}{8} \left( \frac{1}{2} \sum_{k=0}^{n-1} ((2k + 1)\pi)^p + (2k\pi)^p \right).
\]

Thus, the limit becomes:

\[
\lim_{n\to\infty} \frac{1}{n^{p+1}} \cdot \frac{1}{8} \cdot \frac{1}{2} \sum_{k=0}^{n-1} ((2k + 1)\pi)^p + (2k\pi)^p.
\]

Using the Faulhaber's formula for the sum of powers, we get:

\[
\sum_{k=0}^{n-1} k^p \approx \frac{n^{p+1}}{p+1}.
\]

Thus, the limit becomes:

\[
\frac{1}{8} \cdot \frac{1}{2} \cdot \frac{2\pi^p}{p+1} = \frac{\pi^p}{8(p+1)}.
\]

The final answer is \(\boxed{\frac{\pi^p}{8(p+1)}}\).