anonymous
  • anonymous
can someone show me how to find the indefinite integral of 1-y^2/3y
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
do you mean (1-y^2)/3y or 1- (y^2/3y)?
anonymous
  • anonymous
first one
anonymous
  • 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

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anonymous
  • anonymous
does that make sense?
anonymous
  • anonymous
how does 1/3 y dy become y^2/2
anonymous
  • 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.
anonymous
  • anonymous
Yes! You are great! It made perfect sense, thanks a bunch
anonymous
  • anonymous
one more question: Find antiderivative of each derivative that satisfies the given condition: dR/dx = (1-x^4)/x^3
anonymous
  • anonymous
oops left out R(1) = 4
anonymous
  • 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
anonymous
  • anonymous
thanks i understand, but can I express -1/2x^-2 instead of factoring out the negative?
anonymous
  • anonymous
oops.. my bad.. change 2x to be in denominator
anonymous
  • anonymous
i got it now
anonymous
  • anonymous
ok, good ; )
anonymous
  • anonymous
thanks a lot... been great help.. is this open 24/7? I might need to come back again later today
anonymous
  • anonymous
i think so, depends on who or how many people are online here i guess...
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[\int\limits_{4}^{1} (5x+3) dx \] evaluate the integral... I am back already :)
anonymous
  • anonymous
oops the other way around between 1 and 4
anonymous
  • 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
anonymous
  • anonymous
thats right.. I left out the 3x.. thanks!
anonymous
  • anonymous
\[\int\limits_{0}^{1} e^x dx\]
anonymous
  • anonymous
is it e -1 =1.718
anonymous
  • anonymous
correct.
anonymous
  • anonymous
thanks.. one last question for you
anonymous
  • anonymous
\[\int\limits_{1}^{2} (2x^-2 -3)dx\]
anonymous
  • anonymous
evaluate the integral
anonymous
  • anonymous
is that supposed to be 2x^2 - 3?
anonymous
  • anonymous
no its a negative exponent
anonymous
  • anonymous
ok... so integral of 2x^(-2) - 3 = (-2/x) - 3x... then...((-2/2) - 3(2)) - ((-2/1) - 3(1)) = -7 - (-5) = -2
anonymous
  • anonymous
this is where I get confused. since 2/x don't you apply ln|x| to 1/x?
anonymous
  • anonymous
no, the result of integrating 2x^(-2) IS -2/x, you don't want to integrate it again, just evaluate it...
anonymous
  • anonymous
oh i see...
anonymous
  • anonymous
that answered my question..
anonymous
  • anonymous
great.. have a great afternoon and thanks again
anonymous
  • anonymous
1/ \[\sqrt[3]{t}\]

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