What is the integral from 0 to 3 of (-2x^2+4x) ?

The integral from 0 to 3 of (-2x^2+4x) is 0

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integral from 0 to 3 of (-2x^2+4x)

\int_{0}^{3} (- 2 x^{2} + 4 x) d x

0

Solution steps

\int_{0}^{3} - 2 x^{2} + 4 x 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}^{3} 2 x^{2} d x + \int_{0}^{3} 4 x d x

\int_{0}^{3} 2 x^{2} d x = 18

\int_{0}^{3} 4 x d x = 18

= - 18 + 18

Simplify

= 0

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

\int_{0}^{2} 3 y d y

\int_{- l n (\frac{1}{2})}^{l n (2)} e^{2 x} d x

\int_{a}^{b} f (a + b - x) d x

\int_{0}^{1} x^{2} - 4 d x

What is the integral from 0 to 3 of (-2x^2+4x) ?
The integral from 0 to 3 of (-2x^2+4x) is 0