What is the integral from 0 to 1 of sqrt(x+16) ?

The integral from 0 to 1 of sqrt(x+16) is (34sqrt(17)-128)/3

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integral from 0 to 1 of sqrt(x+16)

\int_{0}^{1} x + 16 d x

\frac{34 17 - 128}{3}

Solution steps

\int_{0}^{1} x + 16 d x

Apply u-substitution

: \int_{16}^{17} u d u

= \int_{16}^{17} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{17}

= [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{17}

Compute the boundaries:

\frac{34 17}{3} - \frac{128}{3}

= \frac{34 17}{3} - \frac{128}{3}

Simplify

= \frac{34 17 - 128}{3}

sin (x + π)

2 e^{- 4 t}

\frac{d}{d x} (ln (\frac{y}{x}))

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

\frac{x ^{2}}{( x ^{2} - 1 ) ^{4}}

What is the integral from 0 to 1 of sqrt(x+16) ?
The integral from 0 to 1 of sqrt(x+16) is (34sqrt(17)-128)/3