What is the integral from 1 to 4 of (-2x^2+12x-16) ?

The integral from 1 to 4 of (-2x^2+12x-16) is 0

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integral from 1 to 4 of (-2x^2+12x-16)

\int_{1}^{4} (- 2 x^{2} + 12 x - 16) d x

0

Solution steps

\int_{1}^{4} - 2 x^{2} + 12 x - 16 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}^{4} 2 x^{2} d x + \int_{1}^{4} 12 x d x - \int_{1}^{4} 16 d x

\int_{1}^{4} 2 x^{2} d x = 42

\int_{1}^{4} 12 x d x = 90

\int_{1}^{4} 16 d x = 48

= - 42 + 90 - 48

Simplify

= 0

What is the integral from 1 to 4 of (-2x^2+12x-16) ?
The integral from 1 to 4 of (-2x^2+12x-16) is 0