# Definite Integral for $x^{-n}$

What is the function for calculating the area under a curve for function $x^{-n}$? I know how to integrate and calculate the area under a function which power of $x$ is positive but I don't know how to calculate the area under a curve of a function where $x$ is raised to a negative power.

Examples: I know that area under the curve $x^2$ can be calculated using the integral of $x^2$ which is: $\frac{a^3}{3}$ (starting point $=0$ (the first limit), and the end point $=a$).
But when taking the integral of $x^{-n}$ it makes no sense because it gives a negative area under the curve above $x>0$... Which is totally wrong!

Can someone answer me - What am I missing here? Why is it that when I take the integral of $x$ raised to a negative power it is negative and is no help in finding the area under the curve?!

• what Kind of number is $n$? – Dr. Sonnhard Graubner Jun 28 '17 at 8:32
• The contribution in $0$ is infinite for $-n\le-1$, but if you calculate between $x=\varepsilon$ and $x=a$ then it is positive. – zwim Jun 28 '17 at 8:33
• Do you just mean integrating $f(x) = x^{n}$ for an integer $n$? $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ if $n\neq -1$. $C$ is the integration constant. – Matti P. Jun 28 '17 at 8:36
• You mention that "it gives a negative area under the curve above $x$>0" - $x^{-n}$ is still a positive number, for example $2^{-3} = \frac{1}{8}$. There is no problem ... – Matti P. Jun 28 '17 at 8:41
• I apologize! Sorry for the trouble that I made... When i came back here to look at my question again I understood that I'm a total dummy :D Because . I realized that i was calculating the area the wrong way! As you all have pointed that out. – Grass Jun 28 '17 at 9:21

If $n \neq 1$ then $$\int x^{-n}dx = \frac{x^{-n+1}}{-n+1} + C$$ If $n=1$, then $$\int x^{-1}dx = \ln(x) + C$$

For example you want to find area under the curve from $x=1$ to $x=5$ for $x^{-3}$. Then you have $$\int_1^5 x^{-3}dx = \frac{x^{-3+1}}{-3+1}\bigg \vert_1^5 = \frac{x^{-2}}{-2}\bigg\vert_1^5 = \frac{-5^{-2}}{2} - \frac{-1}{2} = \frac{1}{2}-\frac{1}{50}$$

Your problem was that you were not realizing that if $x > y$ then $1/x$ < $1/y$. This was what was giving you the illusion of negative area. See how the terms get inverted while calculating the area.

• Thank you! This is exactly what I was missing! – Grass Jun 28 '17 at 9:29

It doesn't give a negative area at all (at least on $\mathbf R^{*+}$). You forget that the domain of integration cannot start at $0$, so the formula is slightly less simple than for a positive power.

I'll take the case $n>1$ and, of course, $0<a<b$: $$\int_a^b\frac{\mathrm dx}{x^n}=-\frac{1}{(n-1)x^{n-1}}\Bigg\vert_a^b=\frac1{n-1}\biggl(\frac1{a^{n-1}}-\frac1{b^{n-1}}\biggr),$$ which is positive, since $\dfrac1b<\dfrac1a$.

• $n-1$ goes to the denominator – enzotib Jun 28 '17 at 8:51
• @enzolib: Oh! yes. Thanks for pointing it! I'll fix it. – Bernard Jun 28 '17 at 9:12