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

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

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

\frac{2456}{27}

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}

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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