What is the integral from 0 to 8 of sqrt(1+36x) ?

The integral from 0 to 8 of sqrt(1+36x) is 2456/27

Popular >

integral from 0 to 8 of sqrt(1+36x)

\int_{0}^{8} 1 + 36 x d x

\frac{2456}{27}

Decimal Notation

90.96296 \dots

Solution steps

\int_{0}^{8} 1 + 36 x d x

Apply u-substitution

: \int_{1}^{289} \frac{u}{36} d u

= \int_{1}^{289} \frac{u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int_{1}^{289} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{289}

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{289}

Compute the boundaries:

\frac{9824}{3}

= \frac{1}{36} \cdot \frac{9824}{3}

Simplify

= \frac{2456}{27}

f (x) = 2 x^{2} + 4 x, at x = - 2

\frac{d}{d x} (4 (3 x^{2} - 2))

\int sin (x) cos (x) + x d x

x \to \infty lim (x sin (\frac{4 π}{x}))

\frac{\partial}{\partial z} (x^{2})

What is the integral from 0 to 8 of sqrt(1+36x) ?
The integral from 0 to 8 of sqrt(1+36x) is 2456/27