What is the integral from 0 to 4 of sqrt(x-1) ?

The integral from 0 to 4 of sqrt(x-1) is 2sqrt(3)+(2i)/3

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integral from 0 to 4 of sqrt(x-1)

\int_{0}^{4} x - 1 d x

23 + \frac{2 i}{3}

Solution steps

\int_{0}^{4} x - 1 d x

Apply u-substitution

: \int_{- 1}^{3} u d u

= \int_{- 1}^{3} u d u

Apply the Power Rule

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

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

Compute the boundaries:

23 + i \frac{2}{3}

= 23 + i \frac{2}{3}

Simplify

= 23 + \frac{2 i}{3}

\int_{- a}^{a} e^{s} - e^{- s} d s

\int_{0}^{1} (15 ln (x)) d x

\int_{2}^{4} 3 x^{2} + 1 d x

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

\int_{0}^{1} x^{2} + 3 d x

What is the integral from 0 to 4 of sqrt(x-1) ?
The integral from 0 to 4 of sqrt(x-1) is 2sqrt(3)+(2i)/3