Pregunta
Evaluate the integral. \( \int \frac{1}{\sqrt{7 t-9}} d t \) \( \int \frac{1}{\sqrt{7 t-9}} d t=\square \)
Ask by Gough Mcguire. in the United States
Mar 08,2025
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
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\[
\int \frac{1}{\sqrt{7t-9}}\, dt = \frac{2}{7}\sqrt{7t-9} + C
\]
Solución
Let
\[
u = 7t - 9,
\]
so that
\[
\frac{du}{dt} = 7 \quad \Rightarrow \quad dt = \frac{du}{7}.
\]
Substitute into the integral:
\[
\int \frac{1}{\sqrt{7t-9}}\, dt = \int \frac{1}{\sqrt{u}}\, \frac{du}{7} = \frac{1}{7}\int u^{-1/2}\, du.
\]
Integrate:
\[
\int u^{-1/2}\, du = \frac{u^{1/2}}{\frac{1}{2}} = 2u^{1/2}.
\]
Thus,
\[
\frac{1}{7}\int u^{-1/2}\, du = \frac{1}{7}(2u^{1/2}) = \frac{2}{7}\sqrt{u}.
\]
Replace \( u \) with \( 7t-9 \):
\[
\frac{2}{7}\sqrt{7t-9} + C.
\]
So,
\[
\int \frac{1}{\sqrt{7t-9}}\, dt = \frac{2}{7}\sqrt{7t-9} + C.
\]
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Simplifique esta solución Mind Expander
To evaluate the integral \( \int \frac{1}{\sqrt{7 t-9}} d t \), we can use substitution. Let \( u = 7t - 9 \). Thus, \( du = 7 dt \) or \( dt = \frac{du}{7} \). Now, substituting back into the integral, we get: \[ \int \frac{1}{\sqrt{u}} \cdot \frac{du}{7} = \frac{1}{7} \int u^{-1/2} du \] The integral \( \int u^{-1/2} du \) evaluates to \( 2u^{1/2} + C \). Therefore: \[ \frac{1}{7} \cdot 2u^{1/2} + C = \frac{2}{7} \sqrt{u} + C \] Substituting back \( u = 7t - 9 \): \[ \int \frac{1}{\sqrt{7 t-9}} d t = \frac{2}{7} \sqrt{7t - 9} + C \] So, the final answer is: \[ \int \frac{1}{\sqrt{7 t-9}} d t = \frac{2}{7} \sqrt{7t - 9} + C \]
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