Question

can someone show me how to find the indefinite integral of 1-y^2/3y

Answer 1

do you mean (1-y^2)/3y or 1- (y^2/3y)?

Answer 2

first one

Answer 3

ok....
first factor out your constant:
(1/3) integral (1-y^2)/y dy
expand the fraction:
(1/3) integral (1/y) - y dy
now integrate term by term:
(1/3) integral (1/y) dy -(1/3) integral y dy
integral of 1/y is ln(y) and integral of y is ((y^2)/2)
solution:
(1/3)ln(y) - (y^2)/6 + C

Answer 4

does that make sense?

Answer 5

how does 1/3 y dy become y^2/2

Answer 6

I factored out the 1/3 first because it's a constant, then integrated y using the power rule, so (Y^2)/2, then multiply by (1/3) which gives (y^2)/6 that you see in the solution.

Answer 7

Yes! You are great! It made perfect sense, thanks a bunch

Answer 8

one more question: Find antiderivative of each derivative that satisfies the given condition: dR/dx = (1-x^4)/x^3

Answer 9

oops left out R(1) = 4

Answer 10

expand fraction in integrand and seperate:
integral 1/x^3 dx - integral x dx
evaluate using power rule on each:
-1/(2x^2) - ((x^2)/2) + C
evaluate using given initial condition of R(1) = 4 and solving for C:
R(x) = -1/(2x^2) - ((x^2)/2) + 5

Answer 11

thanks i understand, but can I express -1/2x^-2 instead of factoring out the negative?

Answer 12

oops.. my bad.. change 2x to be in denominator

Answer 13

i got it now

Answer 14

ok, good ; )

Answer 15

thanks a lot... been great help.. is this open 24/7? I might need to come back again later today

Answer 16

i think so, depends on who or how many people are online here i guess...

Answer 17

ok

Answer 18

\[\int\limits_{4}^{1} (5x+3) dx \] evaluate the integral... I am back already :)

Answer 19

oops the other way around between 1 and 4

Answer 20

ok, so......
integral of 5x + 3 = (5/2)x^2 + 3x...now...
((5/2)(4^2) + 3(4)) - ((5/2)(1^2) + 3(1)) = 93/2

Answer 21

thats right.. I left out the 3x.. thanks!

Answer 22

\[\int\limits_{0}^{1} e^x dx\]

Answer 23

is it e -1 =1.718

Answer 24

correct.

Answer 25

thanks.. one last question for you

Answer 26

\[\int\limits_{1}^{2} (2x^-2 -3)dx\]

Answer 27

evaluate the integral

Answer 28

is that supposed to be 2x^2 - 3?

Answer 29

no its a negative exponent

Answer 30

ok...
so integral of 2x^(-2) - 3 = (-2/x) - 3x...
then...((-2/2) - 3(2)) - ((-2/1) - 3(1)) = -7 - (-5) = -2

Answer 31

this is where I get confused. since 2/x don't you apply ln|x| to 1/x?

Answer 32

no, the result of integrating 2x^(-2) IS -2/x, you don't want to integrate it again, just evaluate it...

Answer 33

oh i see...

Answer 34

that answered my question..

Answer 35

great.. have a great afternoon and thanks again

Answer 36

1/ \[\sqrt[3]{t}\]