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

The integral from 2 to 4 of (x(2x-3))/6 is 29/9

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

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

\frac{29}{9}

Decimal

3.22222 \dots

Solution steps

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

Take the constant out:

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

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

Expand

x (2 x - 3) : 2 x^{2} - 3 x

= \frac{1}{6} \cdot \int_{2}^{4} 2 x^{2} - 3 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

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

\int_{2}^{4} 2 x^{2} d x = \frac{112}{3}

\int_{2}^{4} 3 x d x = 18

= \frac{1}{6} (\frac{112}{3} - 18)

Simplify

\frac{1}{6} (\frac{112}{3} - 18) : \frac{29}{9}

= \frac{29}{9}

f (x) = lo g_{10} (x^{8} + 9)

e^{s i n (x)}

\frac{d}{d y} (\frac{y ^{3} - 8}{y ^{3} + 8})

\int e^{x} (x^{2} + 2 x) d x

\frac{\partial}{\partial x} x c^{2} \frac{1}{1 - \frac{v ^{2}}{c ^{2}}} - 1

What is the integral from 2 to 4 of (x(2x-3))/6 ?
The integral from 2 to 4 of (x(2x-3))/6 is 29/9