Question

'simple' algebra clarification within integration.

hold, while i make the equations...

Answer 1

\[\int\limits_{2}^{0} x^{2} \sqrt(1+x^{3} ) dx\]

thats the original q

and i have this picure: see attached.

Answer 2

so my actual question is how does this guy get from the So... line to the next one, where u^{1/2}

Answer 3

try substitution:
u=1+x^3, du=3x^2

Answer 4

i get the substitution but not the bit just after that

Answer 5

that is, how does x squared suddenly equate to u to the power of one half?

Answer 6

\[=\int\limits_{ }^{ }u ^{1/2}*1/3 du=2/9*u ^{3/2}=2/9 *(1+x^3)\]

Answer 7

put the value of x -> 2to 0.... calculate

Answer 8

are we good?

Answer 9

no

Answer 10

whats wrong?

Answer 11

after you let u= 1+x cubed
then you've solved the derivative in terms of du

how do you get the u to the power of one half?

Answer 12

you have sq rt in original problem, don't you?

Answer 13

yes...

Answer 14

i feel like your about to say something obvious

Answer 15

ok. when you substitute
\[u=1+x ^{3}, so: \sqrt{1+x ^{3}}=u ^{1/2}\]

Answer 16

and x^2 will be take care off by du=x^2*1/3

Answer 17

is it better now?

Answer 18

almost

Answer 19

do it yourself on a peace of paper - always helps me :)

Answer 20

i am doing it on paper :(

Answer 21

i'll try to re-write differently...

Answer 22

\[if u=1+x^{3}, then du=3x ^{2}dx\]
\[means: x ^{2}dx=1/3 du\]
put this substitution in original problem:

Answer 23

=\[1/3\int\limits_{}^{}u ^{1/2}du=1/3 * 2/3 * u ^{3/2}=2/9 * (1+x ^{3})^{3/2}\]

is it better?

Answer 24

i got that far
and subbed in sqrt (u) = sqrt(1+x cubed)

so now i just have to
\[\int\limits_{1}^{9} 1/3 u ^(1/2) du\]

Answer 25

right!

Answer 26

im just not getting how your

"one third, integral u to the one half wrt du = 1/3 * 2/3 *u^ 3/2

Answer 27

i understand you can just "bring out" theh one third

Answer 28

it's a number... I can put it in a front of integral any time

Answer 29

yeah - thats fine.... but is the integral of u^1/2 = (2u/3)^3/2?

Answer 30

right

Answer 31

integral = add 1 to the power & divide...

Answer 32

oh ok so i typed it out wrong then, it should be = (3u/2)^3/2 * 1/3 from the front

Answer 33

no...

Answer 34

oh no

Answer 35

\[=\int\limits_{}^{}u ^{1/2}=u ^{1/2+1}*1/(3/2)=...\]

Answer 36

its cause a division of a fraction is the same as the multiplication of that fraction with the denominator and numerator reversed

Answer 37

I meant: 1+1/2... right :)

Answer 38

sigh

Answer 39

i know... confusing sometime...

Answer 40

remember the rules of integration:
\[\int\limits_{}^{}x ^{n}dx=x ^{(n+1)}/(n+1)\]

Answer 41

hm

well does that mean that i should end in...

Answer 42

18 as the answer?

Answer 43

I got 56/9...let me check

Answer 44

i went from [(2u^3/2)/3] from 2 to 0

= (2(1+x^3)^3/2)/3

=(2(1+2^3)^3/2)/3

=18

Answer 45

oups... 52/9

Answer 46

\[1/3 * 2/3 * (1+x ^{3})^{3/2}=2/9 (9)^{3/2}-2/9=...\]
=6-2/9=52/9

Answer 47

can you match/check the answer?

Answer 48

how did you get 1/3 * 2/3 to be 2/9?

Answer 49

(1*2)/(3*3)=2/9

Answer 50

I use * as multiplication sign...confusing?

Answer 51

using * as multiply isnt confusing.. just that a fraction divided by another fraction is the two numerators multiplied by each other, divided by the two denominators multiplied by each other..... how does that work?

Answer 52

i just multiplied one third by 2 third...
I'm lost... what is the question?

Answer 53

i hate to tell you this, but its the crux of why i've failed at maths the last decade

i dont understand division and multiplication of multiple fractions

Answer 54

i've only just worked this out now.

Answer 55

oups.... bummer!

Answer 56

we had (1/3) : (3/2) =( 1/3)* (2/3)=2/9...

I got to go, man :)
Good Luck!!!
you are on a right truck :))

Answer 57

is : divide?

Answer 58

no its times... ok thanks heaps

Answer 59

: is division :)
see ya!

Answer 60

hm ok

Answer 61

: is division
* is multiplication
sorry... gotta run

Answer 62

ok... thanks again