Show that x 2 is integrable on any interval [0, k]

To show that $x^2$ is integrable on any interval $[0, k]$, we need to demonstrate that the definite integral of $x^2$ over this interval exists and can be calculated. Here is the step-by-step solution:

1. Break Down the Problem

   We need to evaluate the integral of $x^2$ over the interval $[0, k]$. This involves finding the antiderivative of $x^2$ and then applying the Fundamental Theorem of Calculus.

2. Relevant Concepts

   The antiderivative of a function $f(x)$ is a function $F(x)$ such that $F'(x) = f(x)$. For $x^2$, the antiderivative is found using the power rule for integration, which states that: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ where $C$ is the constant of integration.

3. Analysis and Detail

   • Find the Antiderivative: $\int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3} + C$
   • Apply the Fundamental Theorem of Calculus: This theorem states that if $F(x)$ is an antiderivative of $f(x)$ on an interval $[a, b]$, then: $\int_a^b f(x) \, dx = F(b) - F(a)$
   • Evaluate the Definite Integral: $\int_0^k x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^k = \frac{k^3}{3} - \frac{0^3}{3} = \frac{k^3}{3}$

4. Verify and Summarize

   The calculation shows that the integral of $x^2$ over $[0, k]$ is $\frac{k^3}{3}$. This result is consistent with the properties of polynomial functions, which are continuous and thus integrable over any closed interval.

Final Answer

The function $x^2$ is integrable on any interval $[0, k]$, and the definite integral over this interval is: $\int_0^k x^2 \, dx = \frac{k^3}{3}$