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

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

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

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

\frac{29}{9}

Decimal Notation

3.22222 \dots

Solution steps

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

Expand

(2 x - 3) x : 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}

x \to - 2 lim (- x^{2} + 6 x - 1)

f (x) = 2 x + \frac{3}{x}

x \to 4 lim (\frac{5}{x + 2})

\int (\frac{x - 1}{x + 1})^{2} d x

(0, 2), (- 1, - 1)

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