Question

sqrt(9x+4) integrate from 0 to 1

Answer 1

try to make a u substitution,

u = 9x+4
du =... ?

Answer 2

du=9??

Answer 3

thats correct!

so whats your new integral ?

Answer 4

don't forget to change the limits too

when x = 0, u =....?
when x = 1, u =... ?

Answer 5

when x=0 u=4
when x=1 u=13

Answer 6

\[\sqrt{u}\]

Answer 7

or (u)^(1/2)

Answer 8

correct!
so you can proceed further ?
integral of sqrt u is ?

Answer 9

???

Answer 10

2u?

Answer 11

no,
\(\int x^n dx = x^{n+1}/(n+1)+c\)

so, for integral of u^1/2, plug in n = 1/2 there

Answer 12

\[(2/3)u^(2/3)\]

Answer 13

+c

Answer 14

1/2 +1 is actually 3/2
so,
\(\Large \dfrac{u^{3/2}}{(3/2)} = \dfrac{2 u^{3/2}}{3}+c\)

Answer 15

and then apply your limits for u

Answer 16

duh sorry about that

Answer 17

Thanks so much

Answer 18

wait, haven't you got the wrong fraction multiplying \(u^{3/2}\)?

Answer 19

you're missing the \(9\) from \(du = 9\, dx\) I believe...

Answer 20

\[du = 9\,dx\]\[dx = \frac{1}{9}du\]

Answer 21

oh yes! thanks for pointing that out @whpalmer4 :)

Answer 22

then you have (after moving the constant fraction outside the integral)
\[\frac{1}{9}\int u^{\frac{1}{2}}\,du\] (too lazy to put on the limits)

Answer 23

\[\frac{ 1}{ 9} (\frac{ 2(13)^{\frac{ 2 }{ 3 }}}{ 3 })-(\frac{ 2(4)^{\frac{ 2 }{ 3 }}}{ 3 })\]

Answer 24

is this correct??

Answer 25

Then evaluate?

Answer 26

almost, you've got the exponent fractions upside down.

I'd simplify (after fixing exponents) to
\[\frac{2}{27}(13^{\frac{3}{2}}-4^{\frac{3}{2}})\]rather than do each of those terms as fractions

Answer 27

(still evaluate it, of course!)

Answer 28

I do integrals just often enough to remember where I usually make mistakes doing integrals :-)

Answer 29

Thank you for your help on this I will be back with a answer.

Answer 30

2/27(13sqrt(13)-8)