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

The integral from-1 to 1 of (x^2+x+2) is 14/3

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

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

\frac{14}{3}

Solution steps

\int_{- 1}^{1} x^{2} + 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}^{1} x^{2} d x + \int_{- 1}^{1} x d x + \int_{- 1}^{1} 2 d x

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

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

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

= \frac{2}{3} + 0 + 4

Simplify

\frac{2}{3} + 0 + 4 : \frac{14}{3}

= \frac{14}{3}

\int_{- 1}^{1} \frac{1 + x ^{2}}{1 + x ^{4}} d x

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

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

\int_{- 3}^{3} 12 - x^{2} - x^{2} + 6 d x

\int_{- 6}^{4} 24 - 2 x - x^{2} d x

What is the integral from-1 to 1 of (x^2+x+2) ?
The integral from-1 to 1 of (x^2+x+2) is 14/3