What is the integral from 0 to 1-y of (1-y^2) ?

The integral from 0 to 1-y of (1-y^2) is -y+1-((-y+1)^3)/3

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integral from 0 to 1-y of (1-y^2)

\int_{0}^{1 - y} (1 - y^{2}) d y

- y + 1 - \frac{( - y + 1 ) ^{3}}{3}

Solution steps

\int_{0}^{1 - y} 1 - y^{2} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1 - y} 1 d y - \int_{0}^{1 - y} y^{2} d y

\int_{0}^{1 - y} 1 d y = - y + 1

\int_{0}^{1 - y} y^{2} d y = \frac{( - y + 1 ) ^{3}}{3}

= - y + 1 - \frac{( - y + 1 ) ^{3}}{3}

\int_{0}^{5} 300 e^{0.2 x - 1} d x

\int_{1}^{4} \frac{3}{2 x + 1} + cos (2 x) d x

\int_{1}^{2} (e^{x} + \frac{1}{x}) d x

\int_{- 1}^{x} \frac{1}{1 + t ^{2}} d t

\int_{0}^{x} t e^{t^{2}} d t

What is the integral from 0 to 1-y of (1-y^2) ?
The integral from 0 to 1-y of (1-y^2) is -y+1-((-y+1)^3)/3