What is the integral from-1 to 1 of (9x^2+6x-3) ?

The integral from-1 to 1 of (9x^2+6x-3) is 0

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integral from-1 to 1 of (9x^2+6x-3)

\int_{- 1}^{1} (9 x^{2} + 6 x - 3) d x

0

Solution steps

\int_{- 1}^{1} 9 x^{2} + 6 x - 3 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} 9 x^{2} d x + \int_{- 1}^{1} 6 x d x - \int_{- 1}^{1} 3 d x

\int_{- 1}^{1} 9 x^{2} d x = 6

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

\int_{- 1}^{1} 3 d x = 6

= 6 + 0 - 6

Simplify

= 0

t e^{4 t}

tan^{2} (sin (θ))

\int u e^{u} d u

\frac{\partial}{\partial y} (4 x^{3} y^{2} + 4)

4 - 4 x^{2}, x^{8} - 1

What is the integral from-1 to 1 of (9x^2+6x-3) ?
The integral from-1 to 1 of (9x^2+6x-3) is 0