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

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

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

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

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{3}^{4} \frac{1}{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_{3}^{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_{3}^{4} 2 x d x - \int_{3}^{4} 3 d x)

\int_{3}^{4} 2 x d x = 7

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

= \frac{1}{6} (7 - 3)

Simplify

\frac{1}{6} (7 - 3) : \frac{2}{3}

= \frac{2}{3}

\int 25 - 9 x^{2} d x

y = \frac{sin ( x )}{1 + cos ( x )}

\frac{d}{d x} (\frac{1}{3 x ^{3} + e ^{- \frac{2}{x}}})

\int_{0}^{4} ∣ 4 x - 1 ∣ d x

\int \frac{6 x ^{2}}{4 x - x ^{2}} d x

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