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\( \left. \begin{array} { l }{ 37 \int \sqrt { x } d x } \\ { x ^ { 1 / 2 } d x } \\ { = \frac { 2 } { 3 } x ^ { 3 / 2 } } \\ { = \frac { 2 } { 3 } \sqrt[ 2 ] { x ^ { 3 } } } \\ { = \frac { 2 } { 3 } x \sqrt { x } + C } \end{array} \right. \)

Ask by Tucker Paul. in Puerto Rico
Jan 24,2025

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Tutor-Verified Answer

Answer

La integral \( 37 \int \sqrt{x} \, dx \) se resuelve como \( \frac{74}{3} x^{3/2} + C \).

Solution

Para resolver la integral \( 37 \int \sqrt{x} \, dx \), sigamos los pasos detalladamente: 1. **Identificar la integral**: La integral que tenemos es \( \int \sqrt{x} \, dx \). 2. **Reescribir la función**: La función \( \sqrt{x} \) se puede reescribir como \( x^{1/2} \). 3. **Aplicar la regla de la integral**: La regla para integrar \( x^n \) es: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] donde \( n \neq -1 \). 4. **Calcular la integral**: En nuestro caso, \( n = \frac{1}{2} \): \[ \int x^{1/2} \, dx = \frac{x^{(1/2) + 1}}{(1/2) + 1} + C = \frac{x^{3/2}}{3/2} + C = \frac{2}{3} x^{3/2} + C \] 5. **Multiplicar por 37**: Ahora, multiplicamos el resultado de la integral por 37: \[ 37 \int \sqrt{x} \, dx = 37 \left( \frac{2}{3} x^{3/2} + C \right) = \frac{74}{3} x^{3/2} + 37C \] 6. **Resultado final**: Por lo tanto, la solución a la integral es: \[ 37 \int \sqrt{x} \, dx = \frac{74}{3} x^{3/2} + C' \] donde \( C' = 37C \) es una constante de integración. Así que la respuesta final es: \[ \frac{74}{3} x^{3/2} + C' \]

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Bonus Knowledge

Did you know that the integral of \( \sqrt{x} \) is commonly seen in calculus, especially in physics when dealing with motion? The process involves applying the power rule for integration, which states you can raise the exponent and divide by the new exponent. So when you see \( \sqrt{x} \), or \( x^{1/2} \), you just need to apply this rule to find that \( \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} + C \). In real-world applications, this integral could be used to calculate the area under a curve representing the relationship between distance and time in physics. If you're plotting distance over time in a graphical representation of free-falling objects, integrating the function gives you the total distance traveled over a specific interval. It's an essential tool for engineers and scientists when dealing with motion derived from acceleration.

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