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

The integral from 1 to 8 of sqrt(x-1) is (14sqrt(7))/3

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

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

\frac{14 7}{3}

Decimal Notation

12.34683 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{7} u d u

= \int_{0}^{7} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{7}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{7}

Compute the boundaries:

\frac{14 7}{3}

= \frac{14 7}{3}

x^{2} x^{3} + 1

\int \frac{sec ^{2} ( θ )}{tan ( θ )} d θ

x^{^{''}} - 2 x^{^{'}} - 3 x = 0

\frac{d}{d x} (2 x y)

x^{^{'}} = 6 (x^{2} + 1)

What is the integral from 1 to 8 of sqrt(x-1) ?
The integral from 1 to 8 of sqrt(x-1) is (14sqrt(7))/3