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Solving integrals/ sqrt(3-2s)ds

Integral of sqrt(3-2s)ds dx

Limits of integration:

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The graph:

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Piecewise:

The solution

You have entered [src] 
  1               
  /               
 |                
 |    _________   
 |  \/ 3 - 2*s  ds
 |                
/                 
0
∫013−2s ds\int\limits_{0}^{1} \sqrt{3 - 2 s}\, ds 
Integral(sqrt(3 - 2*s), (s, 0, 1))
Detail solution

  1. Let u=3−2su = 3 - 2 s.

    Then let du=−2dsdu = - 2 ds and substitute −du2- \frac{du}{2}:

    ∫(−u2) du\int \left(- \frac{\sqrt{u}}{2}\right)\, du

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

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

      1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫u du=2u323\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}

      So, the result is: −u323- \frac{u^{\frac{3}{2}}}{3}

    Now substitute uu back in:

    −(3−2s)323- \frac{\left(3 - 2 s\right)^{\frac{3}{2}}}{3}

  2. Add the constant of integration:

    −(3−2s)323+constant- \frac{\left(3 - 2 s\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}

The answer is:

−(3−2s)323+constant- \frac{\left(3 - 2 s\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                 
 |                               3/2
 |   _________          (3 - 2*s)   
 | \/ 3 - 2*s  ds = C - ------------
 |                           3      
/
∫3−2s ds=C−(3−2s)323\int \sqrt{3 - 2 s}\, ds = C - \frac{\left(3 - 2 s\right)^{\frac{3}{2}}}{3}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
Plot the graph f(x)
The answer [src] 
  1     ___
- - + \/ 3 
  3
−13+3- \frac{1}{3} + \sqrt{3} 
=
= 
  1     ___
- - + \/ 3 
  3
−13+3- \frac{1}{3} + \sqrt{3} 
-1/3 + sqrt(3)
Numerical answer [src] 
1.39871747423554
1.39871747423554

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

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