What is the integral from 8 to 0 of (2x^2+x+2) ?

The integral from 8 to 0 of (2x^2+x+2) is -1168/3

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integral from 8 to 0 of (2x^2+x+2)

\int_{8}^{0} (2 x^{2} + x + 2) d x

- \frac{1168}{3}

Decimal

- 389.33333 \dots

Solution steps

\int_{8}^{0} 2 x^{2} + x + 2 d x

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

= - \int_{0}^{8} 2 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_{0}^{8} 2 x^{2} d x + \int_{0}^{8} x d x + \int_{0}^{8} 2 d x)

\int_{0}^{8} 2 x^{2} d x = \frac{1024}{3}

\int_{0}^{8} x d x = 32

\int_{0}^{8} 2 d x = 16

= - (\frac{1024}{3} + 32 + 16)

Simplify

- (\frac{1024}{3} + 32 + 16) : - \frac{1168}{3}

= - \frac{1168}{3}

What is the integral from 8 to 0 of (2x^2+x+2) ?
The integral from 8 to 0 of (2x^2+x+2) is -1168/3