#1

Suppose we want to find the area under f(x) = \frac{1}{x} between x = 1 and \infty:

\int_1^\infty \frac{1}{x} dx = ln(x) |_1^\infty = ln \infty - ln 1 = \infty

(This is technically bad notation, but we’ll correct it later)

Which indicates the area in infinite.

However, if we wish to find the volume of the solid created by rotating \frac{1}{x} around the x-axis from x= 1 to \infty, we calculate:

\pi \int_1^\infty (\frac{1}{x})^{2} dx

And what do we get?

#2

When we integrate

piint_1^oo(1/x)^2dx

we get the answer pi, which is paradox and intuitively impossible because the value of the area under f(x)=1/x is oo. This is true because:

piint_1^oo(1/x)^2dx = piint_1^oo(1/x^2)dx = piint_1^oo(x^(-2))dx = pi[-x^-1]_|1~|oo = pi(-(1/oo)+(1/1)) = pi

(because (1/oo) approaches 0, the function’s infinite nature is negated.

#3

Hi Rebekah, this is good, but try using \$ symbols around the notation, so that it looks better.

For example you typed:

piint_1^oo(1/x)^2dx

Change this to “dollar symbol” \pi \int_1^\infty (1/x)^2 dx “dollar symbol” and you get:

\pi\int_1^\infty(1/x)^2dx