To find the arc length of a curve between two points, we use the formula for arc length in calculus. For a curve defined by \( y = f(x) \), the arc length \( L \) from \( x = a \) to \( x = b \) is given by the integral:
\[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]
In this case, the curve is \( y = x^4 + 2 \). We need to find the derivative \( \frac{dy}{dx} \), substitute it into the formula, and evaluate the integral from \( x = 1 \) to \( x = 2 \).
Dada la curva \( y = x^4 + 2 \), calculamos la derivada con respecto a \( x \):
\[
\frac{dy}{dx} = 4x^3
\]
La fórmula para la longitud de arco es:
\[
L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
\]
Sustituyendo la derivada, el integrando se convierte en:
\[
\sqrt{1 + (4x^3)^2} = \sqrt{1 + 16x^6}
\]
Evaluamos la integral desde \( x = 1 \) hasta \( x = 2 \):
\[
L = \int_{1}^{2} \sqrt{16x^6 + 1} \, dx
\]
El resultado de esta integral es una expresión compleja que involucra funciones gamma e hipergeométricas:
\[
L = -\frac{\Gamma\left(\frac{1}{6}\right) \cdot \text{hyper}\left(\left(-\frac{1}{2}, \frac{1}{6}\right), \left(\frac{7}{6}\right), 16e^{i\pi}\right)}{6\Gamma\left(\frac{7}{6}\right)} + \frac{\Gamma\left(\frac{1}{6}\right) \cdot \text{hyper}\left(\left(-\frac{1}{2}, \frac{1}{6}\right), \left(\frac{7}{6}\right), 1024e^{i\pi}\right)}{3\Gamma\left(\frac{7}{6}\right)}
\]
La longitud de arco entre los puntos \( (1,1) \) y \( (2,16) \) de la curva \( y = x^4 + 2 \) es:
\[
\boxed{-\frac{\Gamma\left(\frac{1}{6}\right) \cdot \text{hyper}\left(\left(-\frac{1}{2}, \frac{1}{6}\right), \left(\frac{7}{6}\right), 16e^{i\pi}\right)}{6\Gamma\left(\frac{7}{6}\right)} + \frac{\Gamma\left(\frac{1}{6}\right) \cdot \text{hyper}\left(\left(-\frac{1}{2}, \frac{1}{6}\right), \left(\frac{7}{6}\right), 1024e^{i\pi}\right)}{3\Gamma\left(\frac{7}{6}\right)}}
\]
Answered
