What is the integral from 0 to 4 of (2-sqrt(x))^2 ?

The integral from 0 to 4 of (2-sqrt(x))^2 is 8/3

integral from 0 to 4 of (2-sqrt(x))^2

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

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

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

Expand

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

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

\int_{0}^{4} 4 d x = 16

\int_{0}^{4} 4 x d x = \frac{64}{3}

\int_{0}^{4} x d x = 8

= 16 - \frac{64}{3} + 8

Simplify

16 - \frac{64}{3} + 8 : \frac{8}{3}

= \frac{8}{3}

n = 0 \sum \infty \frac{1}{n + 14}

\int t + 2 + \frac{3}{t ^{2} + 1} d t

\frac{d}{d x} (\frac{- 8 x ^{7} + 7 x ^{3} - 2}{- 4 x ^{3}})

\int (a x + b + x^{2})^{2} d x

\frac{d}{d x} (\frac{7 - sec ( x )}{tan ( x )})

What is the integral from 0 to 4 of (2-sqrt(x))^2 ?
The integral from 0 to 4 of (2-sqrt(x))^2 is 8/3