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Solving integrals/ 1/(1000-2y)

Integral of 1/(1000-2y) dy

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The solution

You have entered [src] 
  1              
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 |      1        
 |  ---------- dy
 |  1000 - 2*y   
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0
∫0111000−2y dy\int\limits_{0}^{1} \frac{1}{1000 - 2 y}\, dy 
Integral(1/(1000 - 2*y), (y, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=1000−2yu = 1000 - 2 y.

      Then let du=−2dydu = - 2 dy and substitute −du2- \frac{du}{2}:

      ∫(−12u) du\int \left(- \frac{1}{2 u}\right)\, du

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫1u du=−∫1u du2\int \frac{1}{u}\, du = - \frac{\int \frac{1}{u}\, du}{2}

        1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

        So, the result is: −log⁡(u)2- \frac{\log{\left(u \right)}}{2}

      Now substitute uu back in:

      −log⁡(1000−2y)2- \frac{\log{\left(1000 - 2 y \right)}}{2}

    Method #2

    1. Rewrite the integrand:

      11000−2y=−12(y−500)\frac{1}{1000 - 2 y} = - \frac{1}{2 \left(y - 500\right)}

    2. The integral of a constant times a function is the constant times the integral of the function:

      ∫(−12(y−500)) dy=−∫1y−500 dy2\int \left(- \frac{1}{2 \left(y - 500\right)}\right)\, dy = - \frac{\int \frac{1}{y - 500}\, dy}{2}

      1. Let u=y−500u = y - 500.

        Then let du=dydu = dy and substitute dudu:

        ∫1u du\int \frac{1}{u}\, du

        1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

        Now substitute uu back in:

        log⁡(y−500)\log{\left(y - 500 \right)}

      So, the result is: −log⁡(y−500)2- \frac{\log{\left(y - 500 \right)}}{2}

    Method #3

    1. Rewrite the integrand:

      11000−2y=−12y−1000\frac{1}{1000 - 2 y} = - \frac{1}{2 y - 1000}

    2. The integral of a constant times a function is the constant times the integral of the function:

      ∫(−12y−1000) dy=−∫12y−1000 dy\int \left(- \frac{1}{2 y - 1000}\right)\, dy = - \int \frac{1}{2 y - 1000}\, dy

      1. Let u=2y−1000u = 2 y - 1000.

        Then let du=2dydu = 2 dy and substitute du2\frac{du}{2}:

        ∫12u du\int \frac{1}{2 u}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫1u du=∫1u du2\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}

          1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

          So, the result is: log⁡(u)2\frac{\log{\left(u \right)}}{2}

        Now substitute uu back in:

        log⁡(2y−1000)2\frac{\log{\left(2 y - 1000 \right)}}{2}

      So, the result is: −log⁡(2y−1000)2- \frac{\log{\left(2 y - 1000 \right)}}{2}

  2. Add the constant of integration:

    −log⁡(1000−2y)2+constant- \frac{\log{\left(1000 - 2 y \right)}}{2}+ \mathrm{constant}

The answer is:

−log⁡(1000−2y)2+constant- \frac{\log{\left(1000 - 2 y \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                   
 |                                    
 |     1               log(1000 - 2*y)
 | ---------- dy = C - ---------------
 | 1000 - 2*y                 2       
 |                                    
/
∫11000−2y dy=C−log⁡(1000−2y)2\int \frac{1}{1000 - 2 y}\, dy = C - \frac{\log{\left(1000 - 2 y \right)}}{2}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.0009980.001004
Plot the graph f(x)
The answer [src] 
log(1000)   log(998)
--------- - --------
    2          2
−log⁡(998)2+log⁡(1000)2- \frac{\log{\left(998 \right)}}{2} + \frac{\log{\left(1000 \right)}}{2} 
=
= 
log(1000)   log(998)
--------- - --------
    2          2
−log⁡(998)2+log⁡(1000)2- \frac{\log{\left(998 \right)}}{2} + \frac{\log{\left(1000 \right)}}{2} 
log(1000)/2 - log(998)/2
Numerical answer [src] 
0.00100100133533654
0.00100100133533654

    Use the examples entering the upper and lower limits of integration.

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