What is the integral from 0 to 1.5 of (3/2-x)^2 ?

The integral from 0 to 1.5 of (3/2-x)^2 is 9/8

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integral from 0 to 1.5 of (3/2-x)^2

\int_{0}^{1.5} (\frac{3}{2} - x)^{2} d x

\frac{9}{8}

Solution steps

\int_{0}^{1.5} (\frac{3}{2} - x)^{2} d x

Apply u-substitution

: \int_{\frac{3}{2}}^{0} - u^{2} d u

= \int_{\frac{3}{2}}^{0} - u^{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{\frac{3}{2}} - u^{2} d u

Take the constant out:

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

= - (- \int_{0}^{\frac{3}{2}} u^{2} d u)

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{\frac{3}{2}}

= - (- [\frac{u ^{3}}{3}]_{0}^{\frac{3}{2}})

Simplify

= [\frac{u ^{3}}{3}]_{0}^{\frac{3}{2}}

Compute the boundaries:

\frac{9}{8}

= \frac{9}{8}

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What is the integral from 0 to 1.5 of (3/2-x)^2 ?
The integral from 0 to 1.5 of (3/2-x)^2 is 9/8