Calculate the integral: integral from 0 to 1 mth root

Calculate the integral: $\int_0^1 \sqrt[m]{x^n} \, dx$, where $m, n \in \mathbb{N}$.

We will use the Newton-Leibniz formula after rewriting the integrand in exponential form.

Step 1: Rewrite the integrand using exponents

\[\sqrt[m]{x^n} = (x^n)^{1/m} = x^{n/m}\]

Therefore:

\[\int_0^1 \sqrt[m]{x^n} \, dx = \int_0^1 x^{n/m} \, dx\]

Step 2: Find the antiderivative

Using the power rule $\int x^p \, dx = \frac{x^{p+1}}{p+1} + C$ for $p \neq -1$, with $p = \frac{n}{m}$, so $p+1 = \frac{n+m}{m}$:

\[\int x^{n/m} \, dx = \frac{x^{(n+m)/m}}{(n+m)/m} = \frac{m}{n+m} \cdot x^{(n+m)/m}\]

Step 3: Apply the Newton-Leibniz formula

\[\int_0^1 x^{n/m} \, dx = \left[\frac{m}{n+m} \cdot x^{(n+m)/m}\right]_0^1\]

Step 4: Evaluate at the upper limit $x = 1$

\[F(1) = \frac{m}{n+m} \cdot 1^{(n+m)/m} = \frac{m}{n+m}\]

Step 5: Evaluate at the lower limit $x = 0$

Since $n, m \in \mathbb{N}$ (natural numbers are positive), we have $\frac{n+m}{m} > 0$, so:

\[F(0) = \frac{m}{n+m} \cdot 0^{(n+m)/m} = 0\]

Step 6: Calculate the definite integral

\[\int_0^1 \sqrt[m]{x^n} \, dx = F(1) - F(0) = \frac{m}{n+m} - 0 = \frac{m}{n+m}\]

Alternative form:

\[\int_0^1 \sqrt[m]{x^n} \, dx = \frac{m}{m+n} = \frac{1}{1 + n/m}\]

Special cases:

• For $m = 1$, $n = 1$: $\int_0^1 x \, dx = \frac{1}{1+1} = \frac{1}{2}$
• For $m = 2$, $n = 1$: $\int_0^1 \sqrt{x} \, dx = \int_0^1 x^{1/2} \, dx = \frac{2}{3}$
• For $m = 1$, $n = 2$: $\int_0^1 x^2 \, dx = \frac{1}{3}$
• For $m = 3$, $n = 2$: $\int_0^1 \sqrt[3]{x^2} \, dx = \int_0^1 x^{2/3} \, dx = \frac{3}{5}$

Geometric interpretation:

The integral $\int_0^1 x^{n/m} \, dx$ represents the area under the power function $y = x^{n/m}$ from $x = 0$ to $x = 1$ in the unit square. When $n/m < 1$ (i.e., $n1$, the area is less than $\frac12$. The result $\frac{m}{m+n}$ shows that as the exponent $n/m$ increases, the area decreases.

Pattern observation:

Notice that $\frac{m}{m+n} = \frac{1}{1+n/m} = \frac{1}{\text{exponent}+1}$, consistent with the general formula from Problem 8: $\int_0^1 x^p \, dx = \frac{1}{p+1}$ where $p=n/m$.

Final answer:

\[\boxed{\int_0^1 \sqrt[m]{x^n} \, dx = \frac{m}{m+n} \quad (m, n \in \mathbb{N})}\]