What is the integral from 1 to 5 of sqrt(x-1) ?

The integral from 1 to 5 of sqrt(x-1) is 16/3

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integral from 1 to 5 of sqrt(x-1)

\int_{1}^{5} x - 1 d x

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{1}^{5} x - 1 d x

Apply u-substitution

: \int_{0}^{4} u d u

= \int_{0}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= \frac{16}{3}

\int \frac{3}{7} x^{- \frac{1}{7}} d x

\frac{d}{d x} (\frac{x ^{2} - x}{1 + 3 x ^{2}})

k = 0 \sum \infty k! 2^{k} x^{k}

f (x) = - 3 cos (x), at x = \frac{3}{4} π

\int_{0}^{1} (\frac{4 t ^{3}}{1 + t ^{4}}) d t

What is the integral from 1 to 5 of sqrt(x-1) ?
The integral from 1 to 5 of sqrt(x-1) is 16/3