What is the integral from 0 to 24 of sqrt(9+3x) ?

The integral from 0 to 24 of sqrt(9+3x) is 156

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integral from 0 to 24 of sqrt(9+3x)

\int_{0}^{24} 9 + 3 x d x

156

Solution steps

\int_{0}^{24} 9 + 3 x d x

Apply u-substitution

: \int_{9}^{81} \frac{u}{3} d u

= \int_{9}^{81} \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{9}^{81} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{81}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{81}

Compute the boundaries:

468

= \frac{1}{3} \cdot 468

Simplify

= 156

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 6)^{9})

\frac{d}{d x} ((x - 1)^{n})

y = 9 - 2 x, y = x^{2} - 39

e^{y} y^{^{'}} = (x + e^{y} - 1)

\int x^{2} 7 x + 8 d x

What is the integral from 0 to 24 of sqrt(9+3x) ?
The integral from 0 to 24 of sqrt(9+3x) is 156