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

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In summary, the conversation revolved around finding the integral of 2/(y+1) over the interval [2,0] without using a calculator. One person suggested using the derivative and anti-derivative of ln(x) to solve the problem, while another person provided the correct answer of 2ln3.
  • #1
Whitebread
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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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  • #2
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...
 
  • #3
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 [itex] 1/x [/itex] is [itex] ln (x) [/itex] then the antiderivative of [itex] 1/(x+1) [/itex] is [itex] ln (1 +x) [/itex] (check! Take the derivative of the result!). and the antiderivative of [itex] 1/(Ax+B) [/itex] with A and B being constants is [itex] ln(A +B x) /A [/itex] (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.)
 
Last edited:
  • #4
Remember that dx=d(x+c), c=const.

- Kamataat
 
  • #5
Hi guys,
the correct answer is 2ln3, thanks for the help guys!
 

1. What is the integral of 2/(Y+1)?

The integral of 2/(Y+1) is ln|Y+1| + C, where C is the constant of integration.

2. How do you solve the integral of 2/(Y+1) without a calculator?

To solve the integral of 2/(Y+1) without a calculator, you can use the substitution method. Let u = Y+1, then du = dY. Rewrite the integral as 2/u du. This can be integrated as ln|u| + C, which becomes ln|Y+1| + C after substituting back in for u.

3. Is there a simpler way to solve the integral of 2/(Y+1) without a calculator?

Yes, you can also use the partial fraction decomposition method to simplify the integral. First, rewrite 2/(Y+1) as 2/(Y+1) = A/(Y+1) + B, where A and B are constants. Then, solve for A and B by equating the coefficients of Y and the constants on both sides of the equation. This method may require more algebra, but it can be faster than the substitution method in some cases.

4. Can the integral of 2/(Y+1) be solved using integration by parts?

No, integration by parts is not applicable for solving the integral of 2/(Y+1). This method is used for integrating products of two functions, while the integral of 2/(Y+1) is a single function.

5. What is the significance of the constant of integration in the solution of the integral of 2/(Y+1)?

The constant of integration represents the family of solutions for the integral of 2/(Y+1). This is because the derivative of any constant value is 0, so adding a constant to the solution does not change its derivative. Therefore, the constant of integration allows for the representation of all possible solutions to the integral of 2/(Y+1).

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