What is the integral from 6 to 8 of 2pix(-3/2 x+12) ?

The integral from 6 to 8 of 2pix(-3/2 x+12) is 40pi

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integral from 6 to 8 of 2pix(-3/2 x+12)

\int_{6}^{8} 2 π x (- \frac{3}{2} x + 12) d x

40 π

Decimal

125.66370 \dots

Solution steps

\int_{6}^{8} 2 π x (- \frac{3}{2} x + 12) d x

Take the constant out:

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

= 2 π \cdot \int_{6}^{8} x (- \frac{3}{2} x + 12) d x

- \frac{3}{2} x + 12 = 12 (- \frac{\frac{3}{2} x}{12} + 1)

= 2 π \cdot \int_{6}^{8} x \cdot 12 (- \frac{\frac{3}{2} x}{12} + 1) d x

Take the constant out:

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

= 2 π 12 \cdot \int_{6}^{8} x (- \frac{\frac{3}{2} x}{12} + 1) d x

\frac{\frac{3}{2} x}{12} = \frac{x}{8}

= 2 π 12 \cdot \int_{6}^{8} x (- \frac{x}{8} + 1) d x

Expand

x (- \frac{x}{8} + 1) : - \frac{x ^{2}}{8} + x

= 2 π 12 \cdot \int_{6}^{8} - \frac{x ^{2}}{8} + x 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

= 2 π 12 (- \int_{6}^{8} \frac{x ^{2}}{8} d x + \int_{6}^{8} x d x)

\int_{6}^{8} \frac{x ^{2}}{8} d x = \frac{37}{3}

\int_{6}^{8} x d x = 14

= 2 π 12 (- \frac{37}{3} + 14)

Simplify

2 π 12 (- \frac{37}{3} + 14) : 40 π

= 40 π

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What is the integral from 6 to 8 of 2pix(-3/2 x+12) ?
The integral from 6 to 8 of 2pix(-3/2 x+12) is 40pi