What is the integral from 0 to 1.5 of (-4x+6)^{3/2} ?

The integral from 0 to 1.5 of (-4x+6)^{3/2} is (18*6^{1/2})/5

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integral from 0 to 1.5 of (-4x+6)^{3/2}

\int_{0}^{1.5} (- 4 x + 6)^{\frac{3}{2}} d x

\frac{18 \cdot 6 ^{\frac{1}{2}}}{5}

Decimal

8.81816 \dots

Solution steps

\int_{0}^{1.5} (- 4 x + 6)^{\frac{3}{2}} d x

Apply u-substitution

= \int_{6}^{0} - \frac{u ^{\frac{3}{2}}}{4} d u

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

= - \int_{0}^{6} - \frac{u ^{\frac{3}{2}}}{4} d u

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int_{0}^{6} u^{\frac{3}{2}} d u)

Apply the Power Rule

= - (- \frac{1}{4} [\frac{2}{5} u^{\frac{5}{2}}]_{0}^{6})

Simplify

= \frac{1}{4} [\frac{2}{5} u^{\frac{5}{2}}]_{0}^{6}

Compute the boundaries:

\frac{72 \cdot 6 ^{\frac{1}{2}}}{5}

= \frac{1}{4} \cdot \frac{72 \cdot 6 ^{\frac{1}{2}}}{5}

Simplify

= \frac{18 \cdot 6 ^{\frac{1}{2}}}{5}

What is the integral from 0 to 1.5 of (-4x+6)^{3/2} ?
The integral from 0 to 1.5 of (-4x+6)^{3/2} is (18*6^{1/2})/5