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

The integral from 1 to 2 of (4x^3-3x^2+2x) is 11

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integral from 1 to 2 of (4x^3-3x^2+2x)

\int_{1}^{2} (4 x^{3} - 3 x^{2} + 2 x) d x

11

Solution steps

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

\int_{1}^{2} 4 x^{3} d x = 15

\int_{1}^{2} 3 x^{2} d x = 7

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

= 15 - 7 + 3

Simplify

= 11

\int cos (3 x - 1) d x

\int (3 t e^{t}, ln (t)) d t

\int \frac{3 x ^{2}}{2} - x - 16.5 d x

2 x^{3} \frac{d y}{d x} = y (2 x^{2} - y^{2})

x \to 0 lim (\frac{- 6}{x ^{7}})

What is the integral from 1 to 2 of (4x^3-3x^2+2x) ?
The integral from 1 to 2 of (4x^3-3x^2+2x) is 11