What is the integral from 0 to 4 of 1/2 (4-x)^2 ?

The integral from 0 to 4 of 1/2 (4-x)^2 is 32/3

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integral from 0 to 4 of 1/2 (4-x)^2

\int_{0}^{4} \frac{1}{2} (4 - x)^{2} d x

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{0}^{4} \frac{1}{2} (4 - x)^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{4} (4 - x)^{2} d x

Expand

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

= \frac{1}{2} \cdot \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

= \frac{1}{2} (\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}

= \frac{1}{2} (64 - 64 + \frac{64}{3})

Simplify

\frac{1}{2} (64 - 64 + \frac{64}{3}) : \frac{32}{3}

= \frac{32}{3}

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What is the integral from 0 to 4 of 1/2 (4-x)^2 ?
The integral from 0 to 4 of 1/2 (4-x)^2 is 32/3