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Integral of y-y^2 dy

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The solution

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01(y2+y)dy\int\limits_{0}^{1} \left(- y^{2} + y\right)\, dy
Integral(y - y^2, (y, 0, 1))
Detail solution
  1. Integrate term-by-term:

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

      (y2)dy=y2dy\int \left(- y^{2}\right)\, dy = - \int y^{2}\, dy

      1. The integral of yny^{n} is yn+1n+1\frac{y^{n + 1}}{n + 1} when n1n \neq -1:

        y2dy=y33\int y^{2}\, dy = \frac{y^{3}}{3}

      So, the result is: y33- \frac{y^{3}}{3}

    1. The integral of yny^{n} is yn+1n+1\frac{y^{n + 1}}{n + 1} when n1n \neq -1:

      ydy=y22\int y\, dy = \frac{y^{2}}{2}

    The result is: y33+y22- \frac{y^{3}}{3} + \frac{y^{2}}{2}

  2. Now simplify:

    y2(32y)6\frac{y^{2} \left(3 - 2 y\right)}{6}

  3. Add the constant of integration:

    y2(32y)6+constant\frac{y^{2} \left(3 - 2 y\right)}{6}+ \mathrm{constant}


The answer is:

y2(32y)6+constant\frac{y^{2} \left(3 - 2 y\right)}{6}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                         
 |                    2    3
 | /     2\          y    y 
 | \y - y / dy = C + -- - --
 |                   2    3 
/                           
(y2+y)dy=Cy33+y22\int \left(- y^{2} + y\right)\, dy = C - \frac{y^{3}}{3} + \frac{y^{2}}{2}
The graph
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The answer [src]
1/6
16\frac{1}{6}
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=
1/6
16\frac{1}{6}
1/6
Numerical answer [src]
0.166666666666667
0.166666666666667

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