What is the integral from 0 to 2 of xy^2 ?

The integral from 0 to 2 of xy^2 is y^2*2

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integral from 0 to 2 of xy^2

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

y^{2} \cdot 2

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{2}

= y^{2} [\frac{x ^{2}}{2}]_{0}^{2}

Compute the boundaries:

2

= y^{2} \cdot 2

\int_{0}^{2} x \frac{3}{4} x (2 - x) d x

\int_{- 1}^{0} (6 x - 1) d x

\int_{0}^{\infty} \frac{x}{x ^{2} + 9} d x

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

\int_{0}^{t} (u^{2} - u + e^{u}) d u

What is the integral from 0 to 2 of xy^2 ?
The integral from 0 to 2 of xy^2 is y^2*2