What is the integral from 0 to 4 of (1+3x-x^2) ?

The integral from 0 to 4 of (1+3x-x^2) is 20/3

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integral from 0 to 4 of (1+3x-x^2)

\int_{0}^{4} (1 + 3 x - x^{2}) d x

\frac{20}{3}

Decimal

6.66666 \dots

Solution steps

\int_{0}^{4} 1 + 3 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_{0}^{4} 1 d x + \int_{0}^{4} 3 x d x - \int_{0}^{4} x^{2} d x

\int_{0}^{4} 1 d x = 4

\int_{0}^{4} 3 x d x = 24

\int_{0}^{4} x^{2} d x = \frac{64}{3}

= 4 + 24 - \frac{64}{3}

Simplify

4 + 24 - \frac{64}{3} : \frac{20}{3}

= \frac{20}{3}

\int_{- \infty}^{0} \frac{1}{3 - 10 x} d x

- 4 x^{5} + 6 x^{7}

x \to + 0 lim (\frac{e ^{x} - 1}{5 x})

\int \frac{10 x ^{2} - 9 x + 8}{x ^{3} + 4 x} d x

f (x) = 19 x - 1.86 x^{2}, at x = 1

What is the integral from 0 to 4 of (1+3x-x^2) ?
The integral from 0 to 4 of (1+3x-x^2) is 20/3