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

The integral from 3 to 1 of (4x-x^2) is -22/3

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

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

- \frac{22}{3}

Decimal

- 7.33333 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{3} 4 x - 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

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

\int_{1}^{3} 4 x d x = 16

\int_{1}^{3} x^{2} d x = \frac{26}{3}

= - (16 - \frac{26}{3})

Join

16 - \frac{26}{3} : \frac{22}{3}

= - \frac{22}{3}

\int_{0}^{1} (x + 1) ln^{2} (x + 1) d x

\int_{2}^{3} \frac{x}{2 x ^{2} - 5} d x

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

\int_{0}^{1} 2 y^{2} - 2 y^{3} d y

\int_{1}^{2} x (x - 1) d x

What is the integral from 3 to 1 of (4x-x^2) ?
The integral from 3 to 1 of (4x-x^2) is -22/3