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

The integral from 0 to 8 of sqrt(1+9x) is (146sqrt(73)-2)/(27)

Popular >

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

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

\frac{146 73 - 2}{27}

Decimal Notation

46.12683 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{73} \frac{u}{9} d u

= \int_{1}^{73} \frac{u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{1}^{73} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{73}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{73}

Compute the boundaries:

\frac{146 73}{3} - \frac{2}{3}

= \frac{1}{9} (\frac{146 73}{3} - \frac{2}{3})

Simplify

= \frac{146 73 - 2}{27}

\frac{\partial}{\partial x} (3 x + 5 y)

\frac{d}{d x} (\frac{x}{4 + x ^{2}})

x \to 0 lim (\frac{sin ( 9 x )}{sin ( 3 x )})

\int - 4 e^{x} sin (4 y) d y

\int - x^{\frac{1}{2}} d x

What is the integral from 0 to 8 of sqrt(1+9x) ?
The integral from 0 to 8 of sqrt(1+9x) is (146sqrt(73)-2)/(27)