What is the integral from-1 to 1 of (ax^2+bx+c) ?

The integral from-1 to 1 of (ax^2+bx+c) is 2c+(2a)/3

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integral from-1 to 1 of (ax^2+bx+c)

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

2 c + \frac{2 a}{3}

Solution steps

\int_{- 1}^{1} a x^{2} + b x + c 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} a x^{2} d x + \int_{- 1}^{1} b x d x + \int_{- 1}^{1} c d x

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

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

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

= \frac{2 a}{3} + 0 + 2 c

Simplify

= 2 c + \frac{2 a}{3}

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What is the integral from-1 to 1 of (ax^2+bx+c) ?
The integral from-1 to 1 of (ax^2+bx+c) is 2c+(2a)/3