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

The integral from 0 to 2 of (4-x)^2 is 56/3

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

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

\frac{56}{3}

Decimal Notation

18.66666 \dots

Solution steps

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

Expand

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

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

\int_{0}^{2} 16 d x = 32

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

\int_{0}^{2} x^{2} d x = \frac{8}{3}

= 32 - 16 + \frac{8}{3}

Simplify

32 - 16 + \frac{8}{3} : \frac{56}{3}

= \frac{56}{3}

\frac{- 2 x - y}{x + 3 y ^{2}}

\frac{d y}{d x} = \frac{( x - 3 )}{( y - 3 )}

\int (x + 4) (x - 2) d x

y = 3 sin (x), at x = \frac{π}{6}

\frac{d y}{d x}, e^{x} y = 2

What is the integral from 0 to 2 of (4-x)^2 ?
The integral from 0 to 2 of (4-x)^2 is 56/3