Question

Using the substitution rule for integration, evaluate the integration of sqrt(2x+1)dx. Please explain where the 1/2 comes from in front once you get to 1/2 the integral of u^(1/2)du

Answer 1

\[\sqrt{a}=a^{\frac{1}{2}}\]

Answer 2

I know that much but I still don't understand where you go from there

Answer 3

\[\int\sqrt{2x+1}dx\]
\[u=2x1, du=2dx,\frac{1}{2}du=dx\]
\[\int \frac{1}{2}u^{\frac{1}{2}}du\]

Answer 4

sorry first line should have been \(u=2x+1\)

Answer 5

then
\[\frac{1}{2}\times \frac{2}{3}u^{\frac{3}{2}}\] by the power rule backwards

Answer 6

and 1 to 1/2, get 3/2, divide by 3/2 same as multiplying by 2/3

Answer 7

when you are done you should have
\[\frac{1}{3}(2x+1)^{\frac{3}{2}}=\frac{1}{3}\sqrt{2x+1}^3\]

Answer 8

it all makes sense except for the part in the very beginning still. I get the u=2x+1 and du=2dx but once you plug it back in, I get the integral of u^(1/2)du so basically you brought the 1/2 to the front using the chain rule and made it 1/2 integral u^(1/2)du ?

Answer 9

yes

Answer 10

okay thank you so much !

Answer 11

you do not have \(2dx\) you have \(dx\) so you have to adjust by dividing by 2

Answer 12

wait why isn't it 2dx?

Answer 13

once you have \(du=2dx\) you want to divide by 2 to get the thing you want.

Answer 14

lets start at the beginning. you have
\[\int \sqrt{2x+1}dx\] and you make \(u=2x+1\) but then \(du=2dx\) so you have to adjust

Answer 15

you have to divide by 2 to make sure it works out once you take the derivative. so you write
\[\frac{1}{2}du=dx\] and then integrate
\[\frac{1}{2}\int\sqrt{u}du\]

Answer 16

okay that makes so much more sense now I was unaware you had to get dx by itself. thanks again !

Answer 17

yw