## anonymous 5 years ago can someone show me how to find the indefinite integral of 1-y^2/3y

1. anonymous

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

2. anonymous

first one

3. anonymous

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

4. anonymous

does that make sense?

5. anonymous

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

6. anonymous

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.

7. anonymous

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

8. anonymous

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

9. anonymous

oops left out R(1) = 4

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

i got it now

14. anonymous

ok, good ; )

15. anonymous

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

16. anonymous

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

17. anonymous

ok

18. anonymous

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

19. anonymous

oops the other way around between 1 and 4

20. anonymous

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

21. anonymous

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

22. anonymous

$\int\limits_{0}^{1} e^x dx$

23. anonymous

is it e -1 =1.718

24. anonymous

correct.

25. anonymous

thanks.. one last question for you

26. anonymous

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

27. anonymous

evaluate the integral

28. anonymous

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

29. anonymous

no its a negative exponent

30. anonymous

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

31. anonymous

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

32. anonymous

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

33. anonymous

oh i see...

34. anonymous

35. anonymous

great.. have a great afternoon and thanks again

36. anonymous

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