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

The integral from 2 to 4 of (x-4)^2+1 is 14/3

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

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

\frac{14}{3}

Decimal Notation

4.66666 \dots

Solution steps

\int_{2}^{4} (x - 4)^{2} + 1 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_{2}^{4} (x - 4)^{2} d x + \int_{2}^{4} 1 d x

\int_{2}^{4} (x - 4)^{2} d x = \frac{8}{3}

\int_{2}^{4} 1 d x = 2

= \frac{8}{3} + 2

Simplify

\frac{8}{3} + 2 : \frac{14}{3}

= \frac{14}{3}

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What is the integral from 2 to 4 of (x-4)^2+1 ?
The integral from 2 to 4 of (x-4)^2+1 is 14/3