Evaluate the improper integral or state that it diverges: integral from 6 to infinity (1/t^2-5t)dt. I need help on solving this and what does it mean by converges and diverges?
integral = -(2t-5)/(t^2-5)^2
at infinity ---> -2t/t^2 = 2/t = 0
at 6 = (5-2)/31^2
= .00312
It does not diverge but has a finite answer. If t = sqrt(5) were included, we would have had a problem with a zero denominator.
I think you missed a t in the denominator
∫ 1/(t^2-5t) dt = 1/5 log((5-t)/t)
the definite integral is log(6)/5
Whoops, sorry, use Steve' result.
To evaluate the improper integral of an equation, you need to consider whether the integral converges or diverges.
When an improper integral converges, it means that it has a finite value. In other words, you can find a definite numerical value for the integral. On the other hand, if it diverges, it means that the integral does not have a finite value, and it may be infinite or undefined.
Now, let's evaluate the given improper integral step by step.
First, note that the integrand is 1 over a quadratic function (t^2 - 5t). To determine whether the integral converges or diverges, we need to consider the behavior of the integrand as t approaches infinity.
We can rewrite the integrand as follows: 1 / (t^2 - 5t) = 1 / (t(t - 5)).
As t approaches infinity, the denominator becomes increasingly large. However, the term t is dominant in comparison to (t - 5). Hence, we can ignore the (t - 5) term in the denominator for large values of t.
Therefore, for large values of t, the integrand behaves approximately like 1 / t.
Now, we can compare the given integral to a well-known integral. The integral of 1 / t from 6 to infinity is equal to the natural logarithm of t evaluated from 6 to infinity, which can be written as ln(t)│[6, ∞]. This integral is a classic example of an integral that diverges.
Since the integrand of our given improper integral behaves similarly to 1 / t for large values of t, it follows that the given improper integral also diverges.
Hence, we can conclude that the integral from 6 to infinity of (1/t^2-5t)dt diverges.