What is the integral from-1 to 2 of (2-5x+(x^2)/2) ?

The integral from-1 to 2 of (2-5x+(x^2)/2) is 0

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integral from-1 to 2 of (2-5x+(x^2)/2)

\int_{- 1}^{2} (2 - 5 x + \frac{x ^{2}}{2}) d x

0

Solution steps

\int_{- 1}^{2} 2 - 5 x + \frac{x ^{2}}{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}^{2} 2 d x - \int_{- 1}^{2} 5 x d x + \int_{- 1}^{2} \frac{x ^{2}}{2} d x

\int_{- 1}^{2} 2 d x = 6

\int_{- 1}^{2} 5 x d x = \frac{15}{2}

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

= 6 - \frac{15}{2} + \frac{3}{2}

Simplify

6 - \frac{15}{2} + \frac{3}{2} : 0

= 0

What is the integral from-1 to 2 of (2-5x+(x^2)/2) ?
The integral from-1 to 2 of (2-5x+(x^2)/2) is 0