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

The integral from-1 to 1 of (4x^3-6x^2+2x-1) is -6

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

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

- 6

Solution steps

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

\int_{- 1}^{1} 4 x^{3} d x = 0

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

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

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

= 0 - 4 + 0 - 2

Simplify

= - 6

\int_{\frac{π}{2}}^{\frac{3 π}{4}} sin (x) d x

\int_{0}^{\frac{π}{2}} sin (t) cos (t) d t

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

\int_{0}^{x^{3}} sin (t^{2}) d t

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

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