What is the integral from 0 to 9 of sqrt(1+x) ?

The integral from 0 to 9 of sqrt(1+x) is (20sqrt(10)-2)/3

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integral from 0 to 9 of sqrt(1+x)

\int_{0}^{9} 1 + x d x

\frac{20 10 - 2}{3}

Decimal Notation

20.41518 \dots

Solution steps

\int_{0}^{9} 1 + x d x

Apply u-substitution

: \int_{1}^{10} u d u

= \int_{1}^{10} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{10}

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{10}

Compute the boundaries:

\frac{20 10}{3} - \frac{2}{3}

= \frac{20 10}{3} - \frac{2}{3}

Simplify

= \frac{20 10 - 2}{3}

n = 1 \sum \infty arctan (n^{2})

x \to 0 lim (\frac{( 2 ^{x} - 2 ^{- x} )}{x})

\frac{d}{d x} (sin^{2} (\frac{x + 2}{x + 1}))

y^{^{'}} = ln (x) y + x^{x}

What is the integral from 0 to 9 of sqrt(1+x) ?
The integral from 0 to 9 of sqrt(1+x) is (20sqrt(10)-2)/3