What is the integral from-1 to y-1 of 3/2 x^2 ?

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

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integral from-1 to y-1 of 3/2 x^2

\int_{- 1}^{y - 1} \frac{3}{2} x^{2} d x

\frac{y ^{3} - 3 y ^{2} + 3 y}{2}

Solution steps

\int_{- 1}^{y - 1} \frac{3}{2} x^{2} d x

Take the constant out:

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

= \frac{3}{2} \cdot \int_{- 1}^{- 1 + y} x^{2} d x

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{- 1}^{- 1 + y}

= \frac{3}{2} [\frac{x ^{3}}{3}]_{- 1}^{- 1 + y}

Compute the boundaries:

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

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

Simplify

= \frac{y ^{3} - 3 y ^{2} + 3 y}{2}

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