What is the integral from 0 to 4 of (6-x)^2 ?

The integral from 0 to 4 of (6-x)^2 is 208/3

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integral from 0 to 4 of (6-x)^2

\int_{0}^{4} (6 - x)^{2} d x

\frac{208}{3}

Decimal Notation

69.33333 \dots

Solution steps

\int_{0}^{4} (6 - x)^{2} d x

Expand

(6 - x)^{2} : 36 - 12 x + x^{2}

= \int_{0}^{4} 36 - 12 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} 36 d x - \int_{0}^{4} 12 x d x + \int_{0}^{4} x^{2} d x

\int_{0}^{4} 36 d x = 144

\int_{0}^{4} 12 x d x = 96

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

= 144 - 96 + \frac{64}{3}

Simplify

144 - 96 + \frac{64}{3} : \frac{208}{3}

= \frac{208}{3}

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What is the integral from 0 to 4 of (6-x)^2 ?
The integral from 0 to 4 of (6-x)^2 is 208/3