ajedrezlaproa6j

2021-12-31

For what values of a is each integral improper?
${\int }_{a}^{4}\frac{x}{3x-1}dx$

Jonathan Burroughs

Consider the integral ${\int }_{a}^{4}\frac{x}{3x-1}dx$.
Check whether the provided function is improper or not, find the point of infinite discontinuity.
Equate the denominator to zero.
3x-1=0
3x=1
$x=\frac{1}{3}$
The function has a discontinuity at $x=\frac{1}{3}$, so the integral is improper on the interval [a,4] for $a\le \frac{1}{3}$.
And the integral is improper when $a=-\mathrm{\infty }$
Thus, the given integral is improper for $a\le \frac{1}{3}$ or $a=-\mathrm{\infty }$.

Matthew Rodriguez

${\int }_{a}^{4}\frac{x}{3x-1}dx={\int }_{a}^{4}\frac{\frac{1}{3}\left(u+1\right)}{3u}du$
$=\frac{\frac{1}{3}}{3}{\int }_{a}^{4}\frac{u+1}{u}du$
$=\frac{1}{9}\left({\int }_{a}^{4}\frac{1}{u}du+{\int }_{a}^{4}1du\right)$...(4)
$=\frac{1}{9}\left[{\left[\mathrm{ln}|u|\right]}_{a}^{4}+{\left[u\right]}_{a}^{4}\right]$
$=\frac{1}{9}\left\{\mathrm{ln}11-\mathrm{ln}\left(3a-1\right)+\left[12-3a\right]\right\}$
$=\frac{1}{9}\left[2.398-\mathrm{ln}3a+12-3a\right]$
$=\frac{1}{9}\left[14.398-\mathrm{ln}3a-3a\right]$
Now is
$1.5998=\frac{1}{9}\left[\mathrm{ln}3a+3a\right]$
$14.398=\mathrm{ln}3a+3a$
a=4

Vasquez

Recall that an integral is improper if one of the limits is $±\mathrm{\infty }$ or it has infinite discontinuities in the interval of integration. In this example, value a is a value at which graph of $y=\frac{x}{3x-1}$ is discontinuous. We can easily determine that graphs of rational functions have a discontinuity when the denominator is equal to 0.
$3x-1=0⇒x=\frac{1}{3}$
We determined that the integral is improper for $a=\frac{1}{3}$

Do you have a similar question?