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

The integral from 0 to 8 of sqrt(1+x) is 52/3

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

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

\frac{52}{3}

Decimal Notation

17.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{9} u d u

= \int_{1}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Compute the boundaries:

\frac{52}{3}

= \frac{52}{3}

x^{3} - 5 x, 4 x, - 3, 3

\int \frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{2}} d x

- 3 cos (x), at x = - \frac{π}{4}

x \to \infty lim (11 + \frac{4}{x ^{2}})

\frac{d}{d x} (\frac{4}{3} π x^{2})

What is the integral from 0 to 8 of sqrt(1+x) ?
The integral from 0 to 8 of sqrt(1+x) is 52/3