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

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

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

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

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

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

Expand

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

= \int_{0}^{1} 1 - 2 x + x 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} 1 d x - \int_{0}^{1} 2 x d x + \int_{0}^{1} x d x

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

\int_{0}^{1} 2 x d x = \frac{4}{3}

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

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

Simplify

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

= \frac{1}{6}

\int 9 cos^{2} (x) tan^{3} (x) d x

\frac{9}{5 x}

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

\int e^{u t} d t

\frac{d y}{d x} = x y + 3 x - 2 y - 6

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