242t.08.02.09 int of rational expression.

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Discussion Overview

The discussion revolves around the evaluation of the integral $\int_5^{10} \frac{3x^5}{x^3-5} \, dx$, focusing on methods for solving the integral and achieving an accurate numerical result to the nearest thousandth. The conversation includes various approaches to integration, substitution techniques, and simplifications.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant presents an initial evaluation of the integral and attempts a substitution with $u = x^3 - 5$, noting difficulties in the process.
  • Another participant reformulates the integral using the substitution and provides limits for $u$, leading to a new expression for the integral.
  • A third participant completes the evaluation of the integral using the substitution method and arrives at the same numerical result, indicating their steps in detail.
  • One participant suggests an alternative evaluation method that simplifies the logarithmic calculation, proposing to combine terms to reduce the number of logarithmic evaluations needed.
  • A final participant expresses appreciation for the tips provided in the discussion.

Areas of Agreement / Disagreement

Participants generally agree on the numerical result of the integral, but there are differing methods presented for achieving that result, with no consensus on which method is superior.

Contextual Notes

Some participants' methods involve different levels of simplification and substitution, which may affect the clarity of the evaluation process. The discussion does not resolve which method is preferable or more efficient.

karush
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$\tiny{242t.08.02.09}$
$\textsf{ Evaluate to nearest thousandth}\\$
\begin{align*}
\displaystyle
I_{8.1.31}
&=\int_5^{10}
\frac{3x^5}{x^3-5} \, dx =885.576\\
\end{align*}
$\textit{ok tried getting to this answer by: }$
$u=x^3-5 \therefore du=3x^2 \, dx$
$\textit{but it didnt seem to go to good}$
 
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karush said:
$\tiny{242t.08.02.09}$
$\textsf{ Evaluate to nearest thousandth}\\$
\begin{align*}
\displaystyle
I_{8.1.31}
&=\int_5^{10}
\frac{3x^5}{x^3-5} \, dx =885.576\\
\end{align*}
$\textit{ok tried getting to this answer by: }$
$u=x^3-5 \therefore du=3x^2 \, dx$
$\textit{but it didnt seem to go to good}$

$\displaystyle \begin{align*} \int_5^{10}{ \frac{3\,x^5}{x^3 - 5}\,\mathrm{d}x } &= \int_5^{10}{ \frac{x^3}{x^3 - 5}\,3\,x^2\,\mathrm{d}x } \end{align*}$

Let $\displaystyle \begin{align*} u = x^3 - 5 \implies \mathrm{d}u = 3\,x^2\,\mathrm{d}x \end{align*}$ and note that $\displaystyle \begin{align*} u(5) = 120 \end{align*}$ and $\displaystyle \begin{align*} u(10) = 995 \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int_5^{10}{ \frac{x^3}{x^3 - 5}\,3\,x^2\,\mathrm{d}x } &= \int_{120}^{995}{ \frac{u + 5}{u}\,\mathrm{d}u } \\ &= \int_{120}^{995}{ \left( 1 + \frac{5}{u}\right) \,\mathrm{d}u } \end{align*}$

Continue.
 
$\tiny{242t.08.02.09}$
$\textsf{ Evaluate to nearest thousandth}\\$
\begin{align*}\displaystyle
I_{8.1.31}&=\int_5^{10}
\frac{3x^5}{x^3-5} \, dx \\
\end{align*}
$u=x^3-5 \therefore du=3x^2 \, dx$
\begin{align*}\displaystyle
I_{8.1.31}&= \int_{120}^{995}{ \frac{u + 5}{u}\,du } \\
&=\left[u+5\ln\left({\left| u \right|}\right)\right]^{995}_{120}\\
&=\left[1029.514-143.938\right] \\
&=885.576
\end{align*}
$\textit{basically}$
 
While there's technically nothing wrong with what you did, I would evaluate as follows:

$$I=\int_{120}^{995} 1+\frac{5}{u}\,du=(995-120)+5\ln\left(\frac{995}{120}\right)=5\left(175+\ln\left(\frac{199}{24}\right)\right)\approx885.576$$

This way, you only have one log to evaluate, and the rounding is done one time.
 
thanks love the tips...
 

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