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

The integral from 0 to 4 of (4-x)^2 is 64/3

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

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

\frac{64}{3}

Decimal

21.33333 \dots

Solution steps

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

Expand

(4 - x)^{2} : 16 - 8 x + x^{2}

= \int_{0}^{4} 16 - 8 x + x^{2} 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} 16 d x - \int_{0}^{4} 8 x d x + \int_{0}^{4} x^{2} d x

\int_{0}^{4} 16 d x = 64

\int_{0}^{4} 8 x d x = 64

\int_{0}^{4} x^{2} d x = \frac{64}{3}

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

Simplify

= \frac{64}{3}

y^{^{''}} - (\frac{2}{x}) y^{^{'}} - (\frac{3}{x ^{2}}) y = 0

n = 0 \sum \infty cos (2 n)

r \to 1 lim (\frac{8 r + 1}{r + 3})

d er i v a t i v e 4 x^{3} - 48 x^{2} + 6 x + 4

e^{x} \frac{d y}{d x} + x y^{2} = 0

What is the integral from 0 to 4 of (4-x)^2 ?
The integral from 0 to 4 of (4-x)^2 is 64/3