Question

find the definite integral of dx/lnx from e to 6

Answer 1

\(\color{#000000}{ \displaystyle \int\limits_e^6\frac{dx}{\ln x} }\)

Answer 2

yes that's the equation. we have been using substitution to solve these problems but what would i substitute

Answer 3

you will have to use numerical approximation such as simpson's rule, reinmann sums, or other

Answer 4

Just u-sub is not sufficient here

Answer 5

For indefinite integral, this and similar integrals could be done using power-series integration, but for definite integrals, use some numerical method you know.

Answer 6

We can do midpoint sums, heard of those?

Answer 7

yeah We learned that, but the directions to the problem say to use substitution to evaluate the integral

Answer 8

I am not that good at math, but even I know that this is not a u-substitution problem:)

Answer 9

ok i will use midpoint sum

Answer 10

thx for the help

Answer 11

I will show why this is erroneous. Suppose we can try,
\(\color{#000000}{ \displaystyle \int_e^6\frac{1}{\ln x}dx }\)

\(\color{#000000}{ \displaystyle u=\ln x }\)
\(\color{#000000}{ \displaystyle du=dx/x\quad \Longrightarrow \quad x{\tiny~}du=dx\quad \Longrightarrow \quad e^u{\tiny~}du=dx }\)

\(\color{#000000}{ \displaystyle x=e\quad \Longrightarrow \quad u=\ln e =1 }\)
\(\color{#000000}{ \displaystyle x=6\quad \Longrightarrow \quad u=\ln 6 \approx 1.792 }\)

then you would get:
\(\color{#000000}{ \displaystyle \int_1^{\ln 6}\frac{e^u}{u}du }\)

but even by integration by parts this doesn't reduce (not a whole number, power). and even after this sub, you again only have a power-series integration if you really want to integrate.

Answer 12

Yes, so let's do the midpoint:)

Answer 13

We need to evaluate: \(\color{#000000}{ \displaystyle \int_e^6\frac{1}{\ln x}dx }\)

let's assign some sufficient approximation to \(e\), like \(e=2.718\).

Answer 14

Ok, how many rectangles do you want to use?