What is the integral from 1 to 3 of (x-6)^2 ?

The integral from 1 to 3 of (x-6)^2 is 98/3

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integral from 1 to 3 of (x-6)^2

\int_{1}^{3} (x - 6)^{2} d x

\frac{98}{3}

Decimal Notation

32.66666 \dots

Solution steps

\int_{1}^{3} (x - 6)^{2} d x

Expand

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

= \int_{1}^{3} x^{2} - 12 x + 36 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_{1}^{3} x^{2} d x - \int_{1}^{3} 12 x d x + \int_{1}^{3} 36 d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 12 x d x = 48

\int_{1}^{3} 36 d x = 72

= \frac{26}{3} - 48 + 72

Simplify

\frac{26}{3} - 48 + 72 : \frac{98}{3}

= \frac{98}{3}

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What is the integral from 1 to 3 of (x-6)^2 ?
The integral from 1 to 3 of (x-6)^2 is 98/3