What is the integral from 2 to 4 of pi(-x^2+6x-8)^2 ?

The integral from 2 to 4 of pi(-x^2+6x-8)^2 is pi 16/15

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integral from 2 to 4 of pi(-x^2+6x-8)^2

\int_{2}^{4} π (- x^{2} + 6 x - 8)^{2} d x

π \frac{16}{15}

Decimal

3.35103 \dots

Solution steps

\int_{2}^{4} π (- x^{2} + 6 x - 8)^{2} d x

Take the constant out:

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

= π \cdot \int_{2}^{4} (- x^{2} + 6 x - 8)^{2} d x

Expand

(- x^{2} + 6 x - 8)^{2} : x^{4} - 12 x^{3} + 52 x^{2} - 96 x + 64

= π \cdot \int_{2}^{4} x^{4} - 12 x^{3} + 52 x^{2} - 96 x + 64 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

= π (\int_{2}^{4} x^{4} d x - \int_{2}^{4} 12 x^{3} d x + \int_{2}^{4} 52 x^{2} d x - \int_{2}^{4} 96 x d x + \int_{2}^{4} 64 d x)

\int_{2}^{4} x^{4} d x = \frac{992}{5}

\int_{2}^{4} 12 x^{3} d x = 720

\int_{2}^{4} 52 x^{2} d x = \frac{2912}{3}

\int_{2}^{4} 96 x d x = 576

\int_{2}^{4} 64 d x = 128

= π (\frac{992}{5} - 720 + \frac{2912}{3} - 576 + 128)

Simplify

π (\frac{992}{5} - 720 + \frac{2912}{3} - 576 + 128) : π \frac{16}{15}

= π \frac{16}{15}

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What is the integral from 2 to 4 of pi(-x^2+6x-8)^2 ?
The integral from 2 to 4 of pi(-x^2+6x-8)^2 is pi 16/15