Question

if the substitution u=x/2 is made, the integral ∫[2,4] ( 1 - (x/2)^2 ) / x dx = ?

Answer 1

\[\int_2^4\frac{1-(\frac{x}{2})^2}{x}dx\]?

Answer 2

yes

Answer 3

isn't this the same as
\[\int_2^4(\frac{1}{x}-\frac{x}{4})dx\]?

Answer 4

so what happened to the x in x^2/4?

Answer 5

cancelled with the \(x\) in the denominator

Answer 6

ohh ok i see.

Answer 7

if i am reading it right, it is
\[\frac{1-\left(\frac{x}{2}\right)^2}{x}=\frac{1-\frac{x^2}{4}}{x}=\frac{1}{x}-\frac{x}{4}\]

Answer 8

right.

Answer 9

k so no substitution for this one

Answer 10

\[\int\limits_{1}^{2} \frac{ 1-u^2 }{ u } du\]
\[\int\limits_{2}^{4} \frac{ 1-u^2 }{ u } du\]
\[\int\limits_{1}^{2} \frac{ 1-u^2 }{ 2u } du\]
\[\int\limits_{1}^{2} \frac{ 1-u^2 }{ 4u } du\]
\[\int\limits_{2}^{4} \frac{ 1-u^2 }{ 2u } du\]

Answer 11

those are the answer choices. i don't know what to do..

Answer 12

is see

Answer 13

if you put \(u=\frac{x}{2}\) then
\[2du=dx\] and \(x=2u\)
also \(u(2)=1, u(4)=2\)

Answer 14

this gives
\[\int \frac{1-u^2}{2u}2du\] and the 2's cancel leaving you with
\[\int_1^2\frac{1-u^2}{u}du\]

Answer 15

Thank you!