What is the integral from 0 to 2 of (x-3y^2) ?

The integral from 0 to 2 of (x-3y^2) is 2-6y^2

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integral from 0 to 2 of (x-3y^2)

\int_{0}^{2} (x - 3 y^{2}) d x

2 - 6 y^{2}

Solution steps

\int_{0}^{2} x - 3 y^{2} d x

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}^{2} x d x - \int_{0}^{2} 3 y^{2} d x

\int_{0}^{2} x d x = 2

\int_{0}^{2} 3 y^{2} d x = 6 y^{2}

= 2 - 6 y^{2}

\int_{- 2}^{1} x^{2} - 6 x + 12 d x

\int_{1}^{a} \frac{ln ( x )}{x} d x

\int_{0}^{1} ln (1 + 2 x) d x

\int_{0}^{1} \frac{44}{x ^{2} - 121} d x

\int_{0}^{6 - x} (6 - x - z) d z

What is the integral from 0 to 2 of (x-3y^2) ?
The integral from 0 to 2 of (x-3y^2) is 2-6y^2