Question

Calc2
evaluating the definite integral of x sqrt (1+2x) from 0 to 4, keep getting around 7.5, but should be around 20
integrating by parts, u=x, dv=sqrt (1+2x)dx, v=2/3(1=2x)^(3/2), du=dx
end up with 2/3 (x)(1+2x)^(3\2) - 4/15 (1+2x)^(5/2)
so good so far or did I miss something?

Answer 1

why not just a simple u-substitution?
\(\large u=1+2x\rightarrow du=2dx\rightarrow \frac{1}{2}du=dx \)
to replace the remaining x, use \(\large \frac{(u-1)}{2}=x \)

Answer 2

I would love to, but instructions say to integrate by parts...

Answer 3

lol... you have to use IBP ?

Answer 4

yes, just to make it difficult

Answer 5

do you also have to write it out using your non-writing hand?

Answer 6

just kidding....

Answer 7

no thank goodness

Answer 8

ok... integration by parts....

Answer 9

i see it... your dv=(1+2x)^(1/2) so
v = (1/3)(1+2x)^(3/2)

Answer 10

\[2/3x(1+2x)^{3/2}-\int\limits_{0}^{4}2/3(1=2X)^{3/2}dx\]

Answer 11

why is dv=(1+2x)^3/2dx?

Answer 12

sorry... that should be a 1/2...

Answer 13

\[\huge dv=(1+2x)^{\frac{1}{2}}dx\rightarrow v=\frac{1}{3}(1+2x)^{\frac{3}{2}} \]

Answer 14

shouldn't that be 2/3(bla bla)? 1/3/2?

Answer 15

same as 2/3?

Answer 16

no because when you take the derivative, of v, you get (1+2x)^(1/2)... i think you may be missing the chain rule...

Answer 17

I think your right, I don't like chains

Answer 18

so your integral should be.,..

Answer 19

\[\large u\cdot v - \int v du = x\cdot \frac{1}{3}(1+2x)^{\frac{3}{2}}-\int \frac{1}{3}(1+2x)^{\frac{3}{2}}dx \]

Answer 20

would the integral of \[x \sqrt{x-1}\] be much different?

Answer 21

as far as v?

Answer 22

not much different....

Answer 23

in fact, that would be easier since you don't need to think about the chain rule in this case...

Answer 24

book shows that as \[2/3 (x-1)^{3/2}\] so i am a little confused

Answer 25

is that v or dv?

Answer 26

v

Answer 27

that's correct because if you integrate \(\large dv=\sqrt{x-1}dx \), \(\large v=\int \sqrt{x-1}dx=\frac{2}{3}(x-1)^{\frac{3}{2}}\)

Answer 28

Ok, so I have to look at the chain rule some more...

Answer 29

yeah... technically this second one does the chain rule but the derivative of the "inside" is 1....

Answer 30

chain says f(g(x))=f'(g(x))*g'(x) for derivatives,

Answer 31

outside then inside right?

Answer 32

yep... do you understand what is meant by derivative of the "inside" function?

Answer 33

inside is what is in the ( )... in this case the derivative of the inside is 2?

Answer 34

(1+2x)

Answer 35

yes... the derivative of the "inside" function is 2 in that original question posted...

Answer 36

that's it^^^

Answer 37

ok so it is 2/3(bla bla)*2, and I was missing the *2 i think, no cuz thats 4/3

Answer 38

Hang on let me think about this for a sec

Answer 39

1 mississippi

Answer 40

lol

Answer 41

2 mississippi

Answer 42

:)

Answer 43

\[\int\limits_{?}^{?}(1+2x)^{1/2}dx=2/3(1+2x)^{3/2}*2\]? close?

Answer 44

still 4/3