What is the integral from 1 to 6 of sqrt(x-1) ?

The integral from 1 to 6 of sqrt(x-1) is (10sqrt(5))/3

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integral from 1 to 6 of sqrt(x-1)

\int_{1}^{6} x - 1 d x

\frac{10 5}{3}

Decimal

7.45355 \dots

Solution steps

\int_{1}^{6} x - 1 d x

Apply u-substitution

: \int_{0}^{5} u d u

= \int_{0}^{5} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{5}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{5}

Compute the boundaries:

\frac{10 5}{3}

= \frac{10 5}{3}

y = (13 x^{2} - 26 x + 26) e^{- 8 x}

\int 36 - x^{2} d x

\frac{d}{d x} (\frac{( x + 2 )}{2 - x})

y^{^{''}} + 4 y = 12 sin (2 t)

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What is the integral from 1 to 6 of sqrt(x-1) ?
The integral from 1 to 6 of sqrt(x-1) is (10sqrt(5))/3