What is the integral from 0 to 20 of-0.4x^2+1000-840 ?

The integral from 0 to 20 of-0.4x^2+1000-840 is 2133.33333…

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integral from 0 to 20 of-0.4x^2+1000-840

\int_{0}^{20} - 0.4 x^{2} + 1000 - 840 d x

2133.33333 \dots

Solution steps

\int_{0}^{20} - 0.4 x^{2} + 1000 - 840 d x

Simplify

= \int_{0}^{20} - 0.4 x^{2} + 160 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}^{20} 0.4 x^{2} d x + \int_{0}^{20} 160 d x

\int_{0}^{20} 0.4 x^{2} d x = 1066.66666 \dots

\int_{0}^{20} 160 d x = 3200

= - 1066.66666 \dots + 3200

Simplify

= 2133.33333 \dots

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\int_{- 7}^{4} x^{2} d x

\int_{0}^{1} x + 1 d x

\int_{- 7}^{3} (- x^{2} - 3 x) d x

What is the integral from 0 to 20 of-0.4x^2+1000-840 ?
The integral from 0 to 20 of-0.4x^2+1000-840 is 2133.33333…