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

The integral from 0 to 2 of (4x^3-3x^2+2x) is 12

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

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

12

Solution steps

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

\int_{0}^{2} 4 x^{3} d x = 16

\int_{0}^{2} 3 x^{2} d x = 8

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

= 16 - 8 + 4

Simplify

= 12

\int \frac{1}{5} e^{\frac{- y}{5}} d y

\int \frac{ln ( y ^{2} )}{3 y} d y

y = \frac{10 x}{x ^{2} - 6}, (4, 4)

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What is the integral from 0 to 2 of (4x^3-3x^2+2x) ?
The integral from 0 to 2 of (4x^3-3x^2+2x) is 12