What is the integral from 1 to 5 of 9sqrt(3v+1) ?

The integral from 1 to 5 of 9sqrt(3v+1) is 112

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integral from 1 to 5 of 9sqrt(3v+1)

\int_{1}^{5} 9 3 v + 1 d v

112

Solution steps

\int_{1}^{5} 9 3 v + 1 d v

Take the constant out:

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

= 9 \cdot \int_{1}^{5} 3 v + 1 d v

Apply u-substitution

: \int_{4}^{16} \frac{u}{3} d u

= 9 \cdot \int_{4}^{16} \frac{u}{3} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int_{4}^{16} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16}

= 9 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16}

Simplify

9 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16} : 3 [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16}

= 3 [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16}

Compute the boundaries:

\frac{112}{3}

= 3 \cdot \frac{112}{3}

Simplify

= 112

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What is the integral from 1 to 5 of 9sqrt(3v+1) ?
The integral from 1 to 5 of 9sqrt(3v+1) is 112