What is the integral from 0 to 2 of (8x^3-3x^2+6x+2) ?

The integral from 0 to 2 of (8x^3-3x^2+6x+2) is 40

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integral from 0 to 2 of (8x^3-3x^2+6x+2)

\int_{0}^{2} (8 x^{3} - 3 x^{2} + 6 x + 2) d x

40

Solution steps

\int_{0}^{2} 8 x^{3} - 3 x^{2} + 6 x + 2 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} 8 x^{3} d x - \int_{0}^{2} 3 x^{2} d x + \int_{0}^{2} 6 x d x + \int_{0}^{2} 2 d x

\int_{0}^{2} 8 x^{3} d x = 32

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

\int_{0}^{2} 6 x d x = 12

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

= 32 - 8 + 12 + 4

Simplify

= 40

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

\int_{- 2}^{4} ∣ x - 1 ∣ d x

\int_{0}^{2} (\frac{x}{2})^{2} d x

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

\int_{0}^{2} 8 x - 3 + 5 e^{2 - x} d x

What is the integral from 0 to 2 of (8x^3-3x^2+6x+2) ?
The integral from 0 to 2 of (8x^3-3x^2+6x+2) is 40