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

The integral from 0 to 1 of [(x^2+2)-(-x)] is 17/6

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integral from 0 to 1 of [(x^2+2)-(-x)]

\int_{0}^{1} [(x^{2} + 2) - (- x)] d x

\frac{17}{6}

Decimal Notation

2.83333 \dots

Solution steps

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

Simplify

= \int_{0}^{1} x^{2} + 2 + x 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}^{1} x^{2} d x + \int_{0}^{1} 2 d x + \int_{0}^{1} x d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} 2 d x = 2

\int_{0}^{1} x d x = \frac{1}{2}

= \frac{1}{3} + 2 + \frac{1}{2}

Simplify

\frac{1}{3} + 2 + \frac{1}{2} : \frac{17}{6}

= \frac{17}{6}

\int x (x - 3)^{7} d x

\int_{1}^{t} \frac{33}{x ^{3}} d x

\frac{d}{d x} (lo g_{6} (2 x^{2} + 4))

\int (\frac{10 x}{1 + 5 x ^{2}}) d x

\frac{2 x ^{3}}{3}

What is the integral from 0 to 1 of [(x^2+2)-(-x)] ?
The integral from 0 to 1 of [(x^2+2)-(-x)] is 17/6