What is the integral from 1 to 4 of 2pix(-3/2 x+6) ?

The integral from 1 to 4 of 2pix(-3/2 x+6) is 27pi

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integral from 1 to 4 of 2pix(-3/2 x+6)

\int_{1}^{4} 2 π x (- \frac{3}{2} x + 6) d x

27 π

Decimal

84.82300 \dots

Solution steps

\int_{1}^{4} 2 π x (- \frac{3}{2} x + 6) 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}^{4} x (- \frac{3}{2} x + 6) d x

- \frac{3}{2} x + 6 = 6 (- \frac{\frac{3}{2} x}{6} + 1)

= 2 π \cdot \int_{1}^{4} x \cdot 6 (- \frac{\frac{3}{2} x}{6} + 1) d x

Take the constant out:

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

= 2 π 6 \cdot \int_{1}^{4} x (- \frac{\frac{3}{2} x}{6} + 1) d x

\frac{\frac{3}{2} x}{6} = \frac{x}{4}

= 2 π 6 \cdot \int_{1}^{4} x (- \frac{x}{4} + 1) d x

Expand

x (- \frac{x}{4} + 1) : - \frac{x ^{2}}{4} + x

= 2 π 6 \cdot \int_{1}^{4} - \frac{x ^{2}}{4} + 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 6 (- \int_{1}^{4} \frac{x ^{2}}{4} d x + \int_{1}^{4} x d x)

\int_{1}^{4} \frac{x ^{2}}{4} d x = \frac{21}{4}

\int_{1}^{4} x d x = \frac{15}{2}

= 2 π 6 (- \frac{21}{4} + \frac{15}{2})

Simplify

2 π 6 (- \frac{21}{4} + \frac{15}{2}) : 27 π

= 27 π

f (x) = x^{3} (2 x - 1)

x \to 0 + lim (\frac{sin ( x )}{x ^{\frac{1}{2}}})

y^{^{'}} = y (x y^{6} + 4)

\int_{- \infty}^{0} x^{2} e^{- x} d x

\frac{d}{d x} ((1 - x) e^{x})

What is the integral from 1 to 4 of 2pix(-3/2 x+6) ?
The integral from 1 to 4 of 2pix(-3/2 x+6) is 27pi