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

The integral from 0 to 1 of (sqrt(x)-2)^2 is 11/6

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

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

\frac{11}{6}

Decimal

1.83333 \dots

Solution steps

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

Expand

(x - 2)^{2} : x - 4 x + 4

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

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 4 x d x = \frac{8}{3}

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

= \frac{1}{2} - \frac{8}{3} + 4

Simplify

\frac{1}{2} - \frac{8}{3} + 4 : \frac{11}{6}

= \frac{11}{6}

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What is the integral from 0 to 1 of (sqrt(x)-2)^2 ?
The integral from 0 to 1 of (sqrt(x)-2)^2 is 11/6