What is the integral from 0 to 5 of (sqrt(x-1))^2 ?

The integral from 0 to 5 of (sqrt(x-1))^2 is 15/2

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integral from 0 to 5 of (sqrt(x-1))^2

\int_{0}^{5} (x - 1)^{2} d x

\frac{15}{2}

Decimal Notation

7.5

Solution steps

\int_{0}^{5} (x - 1)^{2} d x

Simplify

(x - 1)^{2} : x - 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{5} x d x = \frac{25}{2}

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

= \frac{25}{2} - 5

Simplify

\frac{25}{2} - 5 : \frac{15}{2}

= \frac{15}{2}

\int_{1}^{5} \frac{1}{x ^{2} - 1} d x

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What is the integral from 0 to 5 of (sqrt(x-1))^2 ?
The integral from 0 to 5 of (sqrt(x-1))^2 is 15/2