Question

Anyone know how to do this?

Answer 1

\[\int\limits_{0}^{1}\sqrt[3]{1+7x}dx\]

Answer 2

did you try to put u = 1+7x ?

Answer 3

yea i try to do it but my answer was wrong, I think I did something wrong

Answer 4

can you show me your attempt ?
u = 1+7x should lead you to correct answer.

Answer 5

i'll help you find the error

Answer 6

post one step at a time, so that i can stop you when i get the error

Answer 7

du=7dx
dx=1/7du
\[\int\limits_{0}^{1}u ^{2/3}dx\]\[\int\limits_{0}^{1}(1+7x)^{2/3}(\frac{ 1 }{ 7}du)\]\[\frac{ 1 }{ 7}(\frac{ 1+7x ^{5/3} }{ 5/3 })\]\[\frac{ 3 }{ 35 }(1+7x)^{5/3}\]plug in 1 and 0 for x\[\frac{ 3 }{35 }(1+7)^{5/3}-0\]\[\frac{ 3 }{ 35 }(8)^{5/3}-0=\frac{ 96 }{ 35 }\]

Answer 8

cube root of x is just x^1/3
not x^2.3

Answer 9

not x^2/3

Answer 10

was that the only mistake i did?

Answer 11

you made the substitution but did not change the limits

Answer 12

what do you mean?

Answer 13

i mean when you put
u= 1+7x

the limits which were for x, now needs to be for u
when
x= 0
u = 1
when x = 1
u = 8
so your new limits should ne
1 to 8

Answer 14

oh so i plug 1 and 8 into 1/7(1+7x)^(4/3)/4/3

Answer 15

if you are using the variable 'x' then use the limits for 'x'
put 0 and 1 in 1/7(1+7x)^(4/3)/4/3

if you are using 'u' put limits for 'u'
put 1 and 8 in 1/7(u)^(4/3)/4/3

Answer 16

either way, you get the same answer

Answer 17

for the answer is it 12/7

Answer 18

how did u get 12 ?

Answer 19

would it be 3/28(8)^4/3-0

Answer 20

1st term is correct!
but how -0 ?

Answer 21

because when plugging in 0 for x is all 0

Answer 22

when u plug in x=0 here
1/7(1+7x)^(4/3)/4/3
you get
3/28 only

Answer 23

oh so its 3/28(8)^4/3-3/28

Answer 24

yes

Answer 25

is this correct 8^4/3 = 16

Answer 26

yes!

Answer 27

so the overall answer is 45/28

Answer 28

absolutely correct! :)

Answer 29

thank you it was correct!!!

Answer 30

welcome ^_^