What is the integral from 0 to 3 of 9sqrt(8x+1) ?

The integral from 0 to 3 of 9sqrt(8x+1) is 93

integral from 0 to 3 of 9sqrt(8x+1)

\int_{0}^{3} 9 8 x + 1 d x

93

Solution steps

\int_{0}^{3} 9 8 x + 1 d x

Take the constant out:

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

= 9 \cdot \int_{0}^{3} 8 x + 1 d x

Apply u-substitution

= 9 \cdot \int_{1}^{25} \frac{u}{8} d u

Take the constant out:

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

= 9 \cdot \frac{1}{8} \cdot \int_{1}^{25} u d u

Apply the Power Rule

= 9 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25}

Simplify

9 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25} : \frac{9}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25}

= \frac{9}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25}

Compute the boundaries:

\frac{248}{3}

= \frac{9}{8} \cdot \frac{248}{3}

Simplify

= 93

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What is the integral from 0 to 3 of 9sqrt(8x+1) ?
The integral from 0 to 3 of 9sqrt(8x+1) is 93