How do you evaluate the integral of int (dt)/(t-4)^2 from 1 to 5?

Mar 15, 2016

Substitute $x = t - 4$
Answer is, if you are indeed asked to just find the integral:

$- \frac{4}{3}$

If you seek the area, it's not that simple though.

Explanation:

${\int}_{1}^{5} \frac{\mathrm{dt}}{t - 4} ^ 2$

Set:

$t - 4 = x$

Therefore the differential:

$\frac{d \left(t - 4\right)}{\mathrm{dt}} = \frac{\mathrm{dx}}{\mathrm{dt}}$

$1 = \frac{\mathrm{dx}}{\mathrm{dt}}$

$\mathrm{dt} = \mathrm{dx}$

And the limits:

${x}_{1} = {t}_{1} - 4 = 1 - 4 = - 3$

${x}_{2} = {t}_{2} - 4 = 5 - 4 = 1$

Now substitute these three values found:

${\int}_{1}^{5} \frac{\mathrm{dt}}{t - 4} ^ 2$

${\int}_{- 3}^{1} \frac{\mathrm{dx}}{x} ^ 2$

${\int}_{- 3}^{1} {x}^{-} 2 \mathrm{dx}$

$\frac{1}{- 2 + 1} {\left[{x}^{- 2 + 1}\right]}_{- 3}^{1}$

$- {\left[{x}^{-} 1\right]}_{- 3}^{1}$

$- {\left[\frac{1}{x}\right]}_{- 3}^{1}$

$- \left(\frac{1}{1} - \frac{1}{- 3}\right)$

$- \left(1 + \frac{1}{3}\right)$

$- \frac{4}{3}$

NOTE: DO NOT READ THIS IF YOU HAVEN'T BEEN TAUGHT HOW TO FIND THE AREA. Although this should actually represent the area between the two limits and since it is always positive, it should have been positive. However, this function is not continuous at $x = 4$ so this integral does not represent the area, if that's what you wanted. It's a bit more complicated.

Mar 15, 2016

${\int}_{1}^{5} \frac{d t}{t - 2} ^ 2 = - \frac{4}{3}$

Explanation:

${\int}_{1}^{5} \frac{d t}{t - 2} ^ 2 \text{ " t-2=u" ; } d t = d u$
${\int}_{1}^{5} \frac{d u}{u} ^ 2 = {\int}_{1}^{5} {u}^{-} 2 d u = | {u}^{- 2 + 1} / \left(- 2 + 1\right) {|}_{1}^{5} = | - {u}^{-} 1 {|}_{1}^{5}$
${\int}_{1}^{5} \frac{d t}{t - 2} ^ 2 = | - \frac{1}{u} {|}_{1}^{5} = | - \frac{1}{t - 2} {|}_{1}^{5}$
${\int}_{1}^{5} \frac{d t}{t - 2} ^ 2 = - \frac{1}{\left(5 - 2\right)} + \frac{1}{\left(1 - 2\right)}$
${\int}_{1}^{5} \frac{d t}{t - 2} ^ 2 = - \frac{1}{3} - 1 = - \frac{4}{3}$

Mar 16, 2016

Depending on how much integration you've learned the "best" answer will be either: "the integral is not defined" (yet) or "the integral diverges"

Explanation:

When we try to evaluate ${\int}_{1}^{5} \frac{1}{x - 4} ^ 2 \mathrm{dx}$, we should check that the integrand is defined on the interval over which we are integrating.

$\frac{1}{x - 4} ^ 2$ is not defined at $4$, so it is not defined on the whole interval $\left[1 , 5\right]$.

Early in the study of calculus , we define the integral by starting with

"Let $f$ be define on interval $\left[a , b\right]$ . . . "

So early in our study, the best answer is that
${\int}_{1}^{5} \frac{1}{x - 4} ^ 2 \mathrm{dx}$ $\text{ }$ is not defined (yet?)

Later we extend the definition to what are called "improper integrals"

These include integrals on unbounded intervals ($\left(- \infty , b\right]$, $\left[a , \infty\right)$ and $\left(- \infty , \infty\right)$) and also intervals on which the integrand has points where it is not defined.

To (try) to evaluate ${\int}_{1}^{5} \frac{1}{x - 4} ^ 2 \mathrm{dx}$, we evaluate the two improper integrals ${\int}_{1}^{4} \frac{1}{x - 4} ^ 2 \mathrm{dx} + {\int}_{4}^{5} \frac{1}{x - 4} ^ 2 \mathrm{dx}$.
(Note that the integrand is still not defined on these closed intervals.)

The method is to replace the point where the integrand is undefined by a variable, then take a limit as that variable approaches the number.

${\int}_{1}^{4} \frac{1}{x - 4} ^ 2 \mathrm{dx} = {\lim}_{b \rightarrow {4}^{-}} {\int}_{1}^{b} \frac{1}{x - 4} ^ 2 \mathrm{dx}$

Let's find the integral first:
${\int}_{1}^{b} \frac{1}{x - 4} ^ 2 \mathrm{dx} = {\left[- \frac{1}{x - 4}\right]}_{1}^{b}$

$= \left(- \frac{1}{b - 4}\right) - \left(- \frac{1}{- 3}\right)$

$= - \frac{1}{b - 4} - \frac{1}{3}$

Looking for the limit as $b \rightarrow {4}^{-}$, we see that the limit does not exist. (As $b \rightarrow {4}^{-}$, the value of $- \frac{1}{b - 4}$ increases without bound.)

Therefore the integral over $\left[1 , 4\right]$ does not exist so the integral over $\left[1 , 5\right]$ does not exist.

We say that the integral diverges.

Note
Some would say: we now have a definition of the integral, there just doesn't happen to be any number that satisfies the definition.