Integral of 2/(Y+1) - Solution without Calculator

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Hello,
I need to take the integral of 2/(Y+1) over the interval [2,0]. The only analytical method I know is to take the anti-deriviative of the integrand and subtract the function values at the endpoints of the interval. I know that the anti-deriviate of (x)^-1 is ln(x), but I don't know how to do this problem without using the calculator (and I'd rather not do that). Can someone help?
 
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Whitebread said:
Hello,
I need to take the integral of 2/(Y+1) over the interval [2,0]. The only analytical method I know is to take the anti-deriviative of the integrand and subtract the function values at the endpoints of the interval. I know that the anti-deriviate of (x)^-1 is ln(x), but I don't know how to do this problem without using the calculator (and I'd rather not do that). Can someone help?

well y' = 2/(Y+1) remember that

Integrate:
2/(Y+1) * y'/y'
2/y' * ln(y+1)
substitute y'
(y+1)ln(y+1)|2,0
[3ln3 - 0]
3ln3 is the answer
hope i did it right...
 
Whitebread said:
Hello,
I need to take the integral of 2/(Y+1) over the interval [2,0]. The only analytical method I know is to take the anti-deriviative of the integrand and subtract the function values at the endpoints of the interval. I know that the anti-deriviate of (x)^-1 is ln(x), but I don't know how to do this problem without using the calculator (and I'd rather not do that). Can someone help?


well, clearly if you know that the antiderivative of 1/x is ln (x) then the antiderivative of 1/(x+1) is ln (1 +x) (check! Take the derivative of the result!). and the antiderivative of 1/(Ax+B) with A and B being constants is ln(A +B x) /A (check!).

The way to prove it is to simply do the obvious change of variable, u = 1+x, dx=du and then integrate over u, which you know how to do. (If you do this change of variable to do yoru problem, don't forget to change the limits of integration.)
 
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Remember that dx=d(x+c), c=const.

- Kamataat
 
Hi guys,
the correct answer is 2ln3, thanks for the help guys!
 
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