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

x \to 2 lim (x^{2} - x)

x \to \infty lim (\frac{7 ^{12} - 2176782336}{7 ^{11}})

\int \frac{x ^{2}}{( 1 - 4 x ^{2} ) ^{\frac{3}{2}}} d x

\int_{0}^{1} e^{2 x} - e^{x} d x

\int (3 x + 1)^{4} d x

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