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

The integral from 1 to 8 of sqrt(x+1) is 18-(4sqrt(2))/3

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

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

18 - \frac{4 2}{3}

Decimal

16.11438 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{9} u d u

= \int_{2}^{9} u d u

Apply the Power Rule

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

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

Compute the boundaries:

18 - \frac{4 2}{3}

= 18 - \frac{4 2}{3}

f (x) = x^{\frac{3}{2}} + x + 4

x e^{x}

x \to 0 lim (tan (4 x))

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (- 1) = 3, y^{^{'}} (- 1) = 9

\frac{d}{d x} (4 e^{4 e^{x} + x})

What is the integral from 1 to 8 of sqrt(x+1) ?
The integral from 1 to 8 of sqrt(x+1) is 18-(4sqrt(2))/3