What is the integral from 2 to 3 of sqrt(32x-63) ?

The integral from 2 to 3 of sqrt(32x-63) is (33sqrt(33)-1)/(48)

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integral from 2 to 3 of sqrt(32x-63)

\int_{2}^{3} 32 x - 63 d x

\frac{33 33 - 1}{48}

Decimal

3.92855 \dots

Solution steps

\int_{2}^{3} 32 x - 63 d x

Apply u-substitution

= \int_{1}^{33} \frac{u}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int_{1}^{33} u d u

Apply the Power Rule

= \frac{1}{32} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{33}

Compute the boundaries:

2233 - \frac{2}{3}

= \frac{1}{32} (2233 - \frac{2}{3})

Simplify

= \frac{33 33 - 1}{48}

What is the integral from 2 to 3 of sqrt(32x-63) ?
The integral from 2 to 3 of sqrt(32x-63) is (33sqrt(33)-1)/(48)