Bryant Lindsey
07/23/2024 · Middle School
\( \int _{1}^{8}\sqrt {1+4x^{\frac{-2}{3}}} d x \)
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Para resolver la integral \( \int _{1}^{8}\sqrt {1+4x^{\frac{-2}{3}}} \, dx \), primero simplificamos la expresión dentro de la raíz.
La integral se puede reescribir como:
\[
\int_{1}^{8} \sqrt{1 + \frac{4}{x^{\frac{2}{3}}}} \, dx
\]
Ahora, simplificamos la expresión:
\[
\sqrt{1 + \frac{4}{x^{\frac{2}{3}}}} = \sqrt{\frac{x^{\frac{2}{3}} + 4}{x^{\frac{2}{3}}}} = \frac{\sqrt{x^{\frac{2}{3}} + 4}}{\sqrt{x^{\frac{2}{3}}}}
\]
Por lo tanto, la integral se convierte en:
\[
\int_{1}^{8} \frac{\sqrt{x^{\frac{2}{3}} + 4}}{\sqrt{x^{\frac{2}{3}}}} \, dx
\]
Para resolver esta integral, podemos hacer un cambio de variable. Sea \( u = x^{\frac{1}{3}} \), entonces \( x = u^3 \) y \( dx = 3u^2 \, du \). Los límites de integración cambian de \( x = 1 \) a \( x = 8 \) a \( u = 1 \) a \( u = 2 \).
Sustituyendo en la integral, tenemos:
\[
\int_{1}^{2} \frac{\sqrt{(u^3)^{\frac{2}{3}} + 4}}{\sqrt{(u^3)^{\frac{2}{3}}}} \cdot 3u^2 \, du = \int_{1}^{2} \frac{\sqrt{u^2 + 4}}{u} \cdot 3u^2 \, du
\]
Esto se simplifica a:
\[
3 \int_{1}^{2} u \sqrt{u^2 + 4} \, du
\]
Ahora, para resolver \( \int u \sqrt{u^2 + 4} \, du \), podemos usar la sustitución \( v = u^2 + 4 \), lo que implica que \( dv = 2u \, du \) o \( du = \frac{dv}{2u} \).
Los límites de integración cambian de \( u = 1 \) a \( u = 2 \) a \( v = 5 \) a \( v = 8 \).
Sustituyendo, tenemos:
\[
3 \int_{5}^{8} \sqrt{v} \cdot \frac{1}{2} \, dv = \frac{3}{2} \int_{5}^{8} v^{\frac{1}{2}} \, dv
\]
La integral de \( v^{\frac{1}{2}} \) es:
\[
\frac{2}{3} v^{\frac{3}{2}}
\]
Evaluando desde 5 hasta 8:
\[
\frac{3}{2} \cdot \frac{2}{3} \left[ v^{\frac{3}{2}} \right]_{5}^{8} = \left[ v^{\frac{3}{2}} \right]_{5}^{8} = 8^{\frac{3}{2}} - 5^{\frac{3}{2}} = 16 - 5\sqrt{5}
\]
Por lo tanto, el resultado de la integral es:
\[
16 - 5\sqrt{5}
\]
Quick Answer
El resultado de la integral es \( 16 - 5\sqrt{5} \).
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