What is the integral from 0 to 5 of sqrt(x-1) ?

The integral from 0 to 5 of sqrt(x-1) is 16/3+i 2/3

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integral from 0 to 5 of sqrt(x-1)

\int_{0}^{5} x - 1 d x

\frac{16}{3} + i \frac{2}{3}

Solution steps

\int_{0}^{5} x - 1 d x

Apply u-substitution

: \int_{- 1}^{4} u d u

= \int_{- 1}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{- 1}^{4}

= [\frac{2}{3} u^{\frac{3}{2}}]_{- 1}^{4}

Compute the boundaries:

\frac{16}{3} + i \frac{2}{3}

= \frac{16}{3} + i \frac{2}{3}

\int_{0}^{2} x - \frac{x ^{2}}{2} d x

\int_{2}^{6} \frac{5}{x} - \frac{10}{x ^{2}} d x

\int_{0}^{3} 8^{x} d x

\int_{0}^{2} (4 - x) d x

\int_{1}^{4} (1 - x) x d x

What is the integral from 0 to 5 of sqrt(x-1) ?
The integral from 0 to 5 of sqrt(x-1) is 16/3+i 2/3