Question

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were calculating the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1,\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.

Short answer

Question: Prove that the limit \(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}\) converges to \(\frac{1}{p+1}\) for any positive integer p. Answer: By recognizing the expression as a Riemann sum over the interval [0, 1], we rewrite it as an integral \(\int_0^1 x^p dx\). Evaluating the integral using the power rule, we find it to be \(\frac{1}{p+1}\). Since this matches the given limit in the exercise, we have proven that the limit converges to \(\frac{1}{p+1}\) for any positive integer p.

Step-by-step solution

Recognize Riemann sum

We see that the expression \(\frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}\) is a Riemann sum, with \(dx=\frac{1}{n}\), over the interval \([0, 1]\).

Relate the expression to an integral

Now, let's rewrite the Riemann sum as an integral: $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p} = \int_0^1 x^p dx$$

Evaluate the integral

Now, we need to find the definite integral of the function \(x^p\). Using the power rule, we get: $$\int_0^1 x^p dx = \frac{x^{p+1}}{p+1} \Big|_0^1$$

Substitute the limits

We plug in the limits to find the value of the integral: $$\frac{x^{p+1}}{p+1} \Big|_0^1 = \frac{1^{p+1}}{p+1} - \frac{0^{p+1}}{p+1}$$ As \(1\) raised to any power is \(1\) and \(0\) raised to any power is \(0\), we get: $$\frac{1^{p+1}}{p+1} - \frac{0^{p+1}}{p+1} = \frac{1}{p+1}$$

Check the limit

We have verified that for any positive integer \(p\), the expression \(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}\) converges to \(\frac{1}{p+1}\). It matches the given limit in the exercise, hence proving Wallis, Pascal, and Fermat’s result.

Key concepts

Definite Integrals

The concept of definite integrals is at the heart of calculus, used to find the area under a curve over a specified interval. In this context, the interval is usually defined by two bounds, say from \(a\) to \(b\), which are the lower and upper limits of integration. The symbol for the integral is an elongated "S," representing summation. When we calculate a definite integral, we are essentially summing an infinite number of infinitesimally small values of functions at each point in the interval.

For example, if we integrate the function \(f(x)\) from \(0\) to \(1\), as in our original exercise, we find the total area under the curve \(y = f(x)\), between \(x = 0\) and \(x = 1\). This helps us to verify our Riemann sum calculation and ensures the given expression converges appropriately to \(\frac{1}{p+1}\).

• A definite integral provides a numerical value representing area.
• Limits of integration define the interval under which the function is evaluated.
• Used extensively in fields requiring mathematical analysis of continuous data.

Definite integrals thus bridge the gap between discrete summation and continuous analysis of functions through the Fundamental Theorem of Calculus.

Power Rule

The power rule is a fundamental tool in calculus that simplifies the differentiation and integration of power functions. The rule states that if you have a function of the form \(f(x) = x^n\), the derivative or the integral can be calculated using simple exponent manipulation. This comes particularly handy when dealing with polynomials and integer powers.

This rule simplifies integration significantly, as we see in the solution to the exercise problem. When we integrate \(x^p\) over \([0, 1]\), applying the power rule transforms this integral into a straightforward expression. The formula is:
\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C,\]where \(C\) is the constant of integration, which disappears when computing definite integrals.

In our case, because we have fixed the bounds from \(0\) to \(1\), the calculation becomes:
\[\int_0^1 x^p \, dx = \frac{x^{p+1}}{p+1} \Big|_0^1.\]By substituting the limits \(0\) and \(1\) into this expression, we verify that the result is \(\frac{1}{p+1}\). This simple yet powerful technique highlights the effectiveness of the power rule in solving calculus problems.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation to integration, providing a comprehensive understanding of these core calculus concepts. It shows how an indefinite integral, representing accumulation, has derivatives that return the original function. This theorem is divided into two parts and plays a key role in understanding the relationship between Riemann sums and definite integrals.

• The first part states that if \(f\) is continuous over \([a, b]\) and \(F\) is an antiderivative of \(f\), then the integral of \(f\) from \(a\) to \(b\) is \(F(b) - F(a)\).
• The second part emphasizes that differentiation and integration are inverse processes, implying that the derivative of the integral of a function returns the original function itself.

In our exercise, the theorem's framework allows us to transition from a Riemann sum, which sums discrete values, to a continuous definite integral. By expressing the Riemann sum as \(\int_0^1 x^p dx\), and using the power rule, we apply the Fundamental Theorem to find the exact area under \(x^p\) between \(0\) and \(1\). This precise calculation confirms the limit found by early mathematicians. This theorem thus forms the backbone of calculus, bridging the gap between differentials and integrals with consistent precision.