Prove the formula for the integral from a to b of x^(1/n)

Let $n \in \mathbb{Z}_{>0}$ , and let $a,b \in \mathbb{R}_{>0}$ . Prove

$\int_a^b x^{1/n} \, dx = \frac{b^{1 + \frac{1}{n}} - a^{1+\frac{1}{n}}}{1+\frac{1}{n}}.$

Proof. First, we consider $\int_0^a x^{1/n} \, dx$ . The rectangle of base $a$ and height $a^{1/n}$ consists of two components: the ordinate set of $f(x) = x^{1/n}$ from 0 to $a$ , and the ordinate set of $g(y) = y^n$ from 0 to $a^{1/n}$ . Thus,

$\begin{align*} a(R) = a \cdot a^{1/n} = a^{1+\frac{1}{n}} &= \int_0^a x^{\frac{1}{n}} \, dx + \int_0^{a^{\frac{1}{n}}} y^n \, dy \\ \implies \int_0^a x^{\frac{1}{n}} \, dx &= a^{1+\frac{1}{n}} - \left. \frac{y^{n+1}}{n+1}\right|_0^{a^{\frac{1}{n}}} \\ &= a^{1+\frac{1}{n}} - \frac{a^{1+\frac{1}{n}}}{n+1} \\ &= a^{1+\frac{1}{n}} \cdot \left( \frac{1}{1+\frac{1}{n}} \right) \\ &= \frac{a^{1+\frac{1}{n}}}{1+\frac{1}{n}}. \end{align*}$

Then, for $\int_a^b x^{\frac{1}{n}} \, dx$ we have

$\begin{align*} \int_a^b x^{\frac{1}{n}} \, dx &= \int_0^b x^{\frac{1}{n}} \, dx - \int_0^a x^{\frac{1}{n}} \, dx \\ &= \frac{b^{1+\frac{1}{n}}}{1+\frac{1}{n}} - \frac{a^{1+\frac{1}{n}}}{1+\frac{1}{n}} \\ &= \frac{b^{1+\frac{1}{n}} - a^{1+\frac{1}{n}}}{1+\frac{1}{n}}. \end{align*}$

Stumbling Robot

Prove the formula for the integral from a to b of x^(1/n)

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