French Conner
04/21/2024 · High School
Hallar la longitud de la curva \( 4 y=2 x^{2}-\ln x \) desde \( x=1 \) hasta \( x=e \)
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Para hallar la longitud de la curva dada por la ecuación \( 4y = 2x^2 - \ln x \), primero debemos despejar \( y \):
\[
y = \frac{1}{4}(2x^2 - \ln x) = \frac{1}{2}x^2 - \frac{1}{4}\ln x
\]
La fórmula para la longitud de una curva \( y = f(x) \) desde \( x = a \) hasta \( x = b \) es:
\[
L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]
Primero, calculamos la derivada \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{2}x^2 - \frac{1}{4}\ln x \right) = x - \frac{1}{4x}
\]
Ahora, simplificamos \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = x - \frac{1}{4x} = \frac{4x^2 - 1}{4x}
\]
Ahora, calculamos \( \left( \frac{dy}{dx} \right)^2 \):
\[
\left( \frac{dy}{dx} \right)^2 = \left( \frac{4x^2 - 1}{4x} \right)^2 = \frac{(4x^2 - 1)^2}{16x^2}
\]
Ahora, sumamos 1 a \( \left( \frac{dy}{dx} \right)^2 \):
\[
1 + \left( \frac{dy}{dx} \right)^2 = 1 + \frac{(4x^2 - 1)^2}{16x^2}
\]
Para simplificar esto, llevamos a un denominador común:
\[
1 + \left( \frac{dy}{dx} \right)^2 = \frac{16x^2}{16x^2} + \frac{(4x^2 - 1)^2}{16x^2} = \frac{16x^2 + (4x^2 - 1)^2}{16x^2}
\]
Ahora, expandimos \( (4x^2 - 1)^2 \):
\[
(4x^2 - 1)^2 = 16x^4 - 8x^2 + 1
\]
Por lo tanto,
\[
1 + \left( \frac{dy}{dx} \right)^2 = \frac{16x^2 + 16x^4 - 8x^2 + 1}{16x^2} = \frac{16x^4 + 8x^2 + 1}{16x^2}
\]
Ahora, tomamos la raíz cuadrada:
\[
\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \sqrt{\frac{16x^4 + 8x^2 + 1}{16x^2}} = \frac{\sqrt{16x^4 + 8x^2 + 1}}{4x}
\]
Ahora, la longitud de la curva desde \( x = 1 \) hasta \( x = e \) es:
\[
L = \int_1^e \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx = \int_1^e \frac{\sqrt{16x^4 + 8x^2 + 1}}{4x} \, dx
\]
Para resolver esta integral, podemos simplificar el integrando:
\[
L = \frac{1}{4} \int_1^e \frac{\sqrt{(4x^2 + 1)^2}}{x} \, dx = \frac{1}{4} \int_1^e \frac{4x^2 + 1}{x} \, dx = \frac{1}{4} \int_1^e (4x + \frac{1}{x}) \, dx
\]
Ahora, resolvemos la integral:
\[
L = \frac{1}{4} \left( \int_1^e 4x \, dx + \int_1^e \frac{1}{x} \, dx \right)
\]
Calculamos cada parte:
1. \( \int 4x \, dx = 2x^2
Quick Answer
La longitud de la curva desde \( x = 1 \) hasta \( x = e \) es \( \frac{1}{4} \left( 2e^2 + 2 + \ln e - 1 \right) \).
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