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

The integral from 2 to 4 of (2x-3)/6 is 1

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

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

1

Solution steps

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

\int_{2}^{4} 2 x d x = 12

\int_{2}^{4} 3 d x = 6

= \frac{1}{6} (12 - 6)

Simplify

\frac{1}{6} (12 - 6) : 1

= 1

y = \frac{7}{16 - x ^{2}}, [1, 3]

\int - sin (\frac{x}{2}) d x

\frac{\partial}{\partial y} (3 x^{2} y^{2} - 4)

n = 1 \sum \infty 8 \cdot (0.2)^{n}

x \frac{d y}{d x} = y^{2} - y, y (2) = 1

What is the integral from 2 to 4 of (2x-3)/6 ?
The integral from 2 to 4 of (2x-3)/6 is 1