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

The integral from 2 to 4 of x^2((2x-3)/6) is 32/3

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

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

\frac{32}{3}

Decimal Notation

10.66666 \dots

Solution steps

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

Expand

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

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

\int_{2}^{4} 2 x^{3} d x = 120

\int_{2}^{4} 3 x^{2} d x = 56

= \frac{1}{6} (120 - 56)

Simplify

\frac{1}{6} (120 - 56) : \frac{32}{3}

= \frac{32}{3}

\int_{0}^{1} 6^{2 x} d x

\frac{\partial}{\partial x} (12 x e^{x - 3 y})

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

\frac{d}{d x} (cos^{12} (x))

5 t + \frac{7}{t}

What is the integral from 2 to 4 of x^2((2x-3)/6) ?
The integral from 2 to 4 of x^2((2x-3)/6) is 32/3