For what values of a is each integral improper? ∫a4x3x−1dx

Question

For what values of a is each integral improper?  ∫a4x3x−1dx

Jonathan Burroughs · Accepted Answer

Consider the integral ∫a4x3x−1dx.  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=13  The function has a discontinuity at x=13, so the integral is improper on the interval [a,4] for a≤13.  And the integral is improper when a=−∞  Thus, the given integral is improper for a≤13 or a=−∞.

Matthew Rodriguez · Accepted Answer

∫a4x3x−1dx=∫a413(u+1)3udu  =133∫a4u+1udu  =19(∫a41udu+∫a41du)...(4)  =19[[ln|u|]a4+[u]a4]  =19{ln⁡11−ln⁡(3a−1)+[12−3a]}  =19[2.398−ln⁡3a+12−3a]  =19[14.398−ln⁡3a−3a]  Now is  1.5998=19[ln⁡3a+3a]  14.398=ln⁡3a+3a  a=4

Vasquez · Accepted Answer

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

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

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New Questions in Calculus 2