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

The integral from 2 to 4 of 2pix(-3/2 x+6) is 16pi

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

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

16 π

Decimal Notation

50.26548 \dots

Solution steps

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

Expand

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

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

\int_{2}^{4} \frac{x ^{2}}{4} d x = \frac{14}{3}

\int_{2}^{4} x d x = 6

= 2 π 6 (- \frac{14}{3} + 6)

Simplify

2 π 6 (- \frac{14}{3} + 6) : 16 π

= 16 π

\int \frac{x}{2} e^{- \frac{x}{2}} d x

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

\int_{0}^{1} \frac{15 5 x + 4}{2} d x

\frac{d y}{d x} - 7 y = e^{x}

y^{^{'}} = \frac{11 x ^{2} - 1}{11 + 4 y}

What is the integral from 2 to 4 of 2pix(-3/2 x+6) ?
The integral from 2 to 4 of 2pix(-3/2 x+6) is 16pi