What is the integral from 0 to 1 of 2x^3-6x^2+4x ?

The integral from 0 to 1 of 2x^3-6x^2+4x is 1/2

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integral from 0 to 1 of 2x^3-6x^2+4x

\int_{0}^{1} 2 x^{3} - 6 x^{2} + 4 x d x

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{0}^{1} 2 x^{3} - 6 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}^{1} 2 x^{3} d x - \int_{0}^{1} 6 x^{2} d x + \int_{0}^{1} 4 x d x

\int_{0}^{1} 2 x^{3} d x = \frac{1}{2}

\int_{0}^{1} 6 x^{2} d x = 2

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

= \frac{1}{2} - 2 + 2

Simplify

= \frac{1}{2}

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What is the integral from 0 to 1 of 2x^3-6x^2+4x ?
The integral from 0 to 1 of 2x^3-6x^2+4x is 1/2