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

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

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

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

6

Solution steps

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

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

\int_{- 2}^{4} 2 x d x = 12

\int_{- 2}^{4} 3 x^{2} d x = 72

\int_{- 2}^{4} x^{3} d x = 60

= 6 + 12 - 72 + 60

Simplify

= 6

\int \frac{1}{y ^{2} + y} d y

\frac{x}{x ^{2} + 49}

\frac{d}{d x} (\frac{e ^{x}}{4 x ^{2} + 9})

\int 3 e^{6} d x

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

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