What is the integral from 0 to 6 of sqrt(6x+64) ?

The integral from 0 to 6 of sqrt(6x+64) is 488/9

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integral from 0 to 6 of sqrt(6x+64)

\int_{0}^{6} 6 x + 64 d x

\frac{488}{9}

Decimal Notation

54.22222 \dots

Solution steps

\int_{0}^{6} 6 x + 64 d x

Apply u-substitution

: \int_{64}^{100} \frac{u}{6} d u

= \int_{64}^{100} \frac{u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{64}^{100} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{64}^{100}

= \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{64}^{100}

Compute the boundaries:

\frac{976}{3}

= \frac{1}{6} \cdot \frac{976}{3}

Simplify

= \frac{488}{9}

\int \frac{x}{( x + 1 )} d x

f (x) = \frac{4}{x + 1}, at x = 6

\int \frac{1}{2 y - 1} d y

y = (x^{3} + 2) (3 x^{2} - 4 x + 2), (1, 3)

x^{2} (2.4)

What is the integral from 0 to 6 of sqrt(6x+64) ?
The integral from 0 to 6 of sqrt(6x+64) is 488/9