# \int_t^5(dt)/((t-4)^2)?

## Answer key says "does not exist" ? My answer was $- 1 + \frac{1}{t - 4}$

Sep 17, 2017

It depends on the range of values for $t$ that you are considering. If $t \setminus \le q 4$, then the integral does not exist. If $t > 4$, then your answer is correct.

#### Explanation:

By the way, notationally-speaking, it would be better to write this integral as ${\int}_{t}^{5} \frac{\mathrm{dx}}{{\left(x - 4\right)}^{2}}$. (If you have a variable as a limit of integration, it's better to use a different letter as your integrand variable).

If $t > 4$, then the integrand is continuous on the interval from $t$ to $5$ (or, for that matter, from 5 to $t$ when $t > 5$). An antiderivative of $f \left(x\right) = \frac{1}{{\left(x - 4\right)}^{2}} = {\left(x - 4\right)}^{- 2}$ is $F \left(x\right) = - {\left(x - 4\right)}^{- 1}$. The Fundamental Theorem of Calculus then implies that ${\int}_{t}^{5} \frac{\mathrm{dx}}{{\left(x - 4\right)}^{2}} = F \left(5\right) - F \left(t\right) = - {\left(1\right)}^{- 1} - \left(- {\left(t - 4\right)}^{- 1}\right) = - 1 + \frac{1}{t - 4}$.

However, if $t \setminus \le q 4$, then $f \left(x\right) = \frac{1}{{\left(x - 4\right)}^{2}}$ is not continuous on the interval from $t$ to 5. The integral could still exist, however, as an improper integral. But, the following limit calculation shows that it does not exist because the integral from 4 to 5 diverges:

${\int}_{4}^{5} \frac{1}{{\left(x - 4\right)}^{2}} \setminus \mathrm{dx} = {\lim}_{t \to 4 +} \left(- 1 + \frac{1}{t - 4}\right)$, but
${\lim}_{t \to 4 +} \frac{1}{t - 4}$ does not exist (some people will write $\setminus {\lim}_{t \to 4 +} \frac{1}{t - 4} = + \infty$ because of the special way this limit fails to exist).