What is the integral from 0 to 8 of sqrt(1+81/4 x) ?

The integral from 0 to 8 of sqrt(1+81/4 x) is (4(326sqrt(163)-2))/(243)

integral from 0 to 8 of sqrt(1+81/4 x)

\int_{0}^{8} 1 + \frac{81}{4} x d x

\frac{4 ( 326 163 - 2 )}{243}

Decimal

68.47883 \dots

Solution steps

\int_{0}^{8} 1 + \frac{81}{4} x d x

Apply u-substitution

= \int_{1}^{163} \frac{4 u}{81} d u

Take the constant out:

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

= \frac{4}{81} \cdot \int_{1}^{163} u d u

Apply the Power Rule

= \frac{4}{81} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{163}

Compute the boundaries:

\frac{326 163}{3} - \frac{2}{3}

= \frac{4}{81} (\frac{326 163}{3} - \frac{2}{3})

Simplify

= \frac{4 ( 326 163 - 2 )}{243}

d er i v a t i v e 4 x^{2} - 6 x + 2

\frac{d}{d x} (5 sin (x) (\frac{4}{x ^{3}} + 3 x))

d er i v a t i v e arccos (x + 1)

\frac{\partial}{\partial x} (x^{2} + 3 x y - y^{2})

d er i v a t i v e (2 x - 8)^{4} (x^{2} + x + 1)^{5}

What is the integral from 0 to 8 of sqrt(1+81/4 x) ?
The integral from 0 to 8 of sqrt(1+81/4 x) is (4(326sqrt(163)-2))/(243)