What is the integral from 1 to 6 of 2pix(-3/2 x+9) ?

The integral from 1 to 6 of 2pix(-3/2 x+9) is 100pi

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integral from 1 to 6 of 2pix(-3/2 x+9)

\int_{1}^{6} 2 π x (- \frac{3}{2} x + 9) d x

100 π

Decimal Notation

314.15926 \dots

Solution steps

\int_{1}^{6} 2 π x (- \frac{3}{2} x + 9) d x

Take the constant out:

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

= 2 π \cdot \int_{1}^{6} x (- \frac{3}{2} x + 9) d x

- \frac{3}{2} x + 9 = 9 (- \frac{\frac{3}{2} x}{9} + 1)

= 2 π \cdot \int_{1}^{6} x \cdot 9 (- \frac{\frac{3}{2} x}{9} + 1) d x

Take the constant out:

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

= 2 π 9 \cdot \int_{1}^{6} x (- \frac{\frac{3}{2} x}{9} + 1) d x

\frac{\frac{3}{2} x}{9} = \frac{x}{6}

= 2 π 9 \cdot \int_{1}^{6} x (- \frac{x}{6} + 1) d x

Expand

x (- \frac{x}{6} + 1) : - \frac{x ^{2}}{6} + x

= 2 π 9 \cdot \int_{1}^{6} - \frac{x ^{2}}{6} + 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 9 (- \int_{1}^{6} \frac{x ^{2}}{6} d x + \int_{1}^{6} x d x)

\int_{1}^{6} \frac{x ^{2}}{6} d x = \frac{215}{18}

\int_{1}^{6} x d x = \frac{35}{2}

= 2 π 9 (- \frac{215}{18} + \frac{35}{2})

Simplify

2 π 9 (- \frac{215}{18} + \frac{35}{2}) : 100 π

= 100 π

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What is the integral from 1 to 6 of 2pix(-3/2 x+9) ?
The integral from 1 to 6 of 2pix(-3/2 x+9) is 100pi