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

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

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

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

\frac{29}{9}

Decimal Notation

3.22222 \dots

Solution steps

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

\int_{- 2}^{2} \frac{12}{x ^{2} - 4 x - 32} d x

\frac{d}{d x} (sin^{2} (x^{3}))

4 x lo g_{10} (x^{9})

\int \frac{5 x ^{7} - 9 x ^{5}}{x ^{3}} d x

f (x) = \frac{2 x}{2 + x}

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