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

The integral from 1 to 2 of (6x^2-3x) is 19/2

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

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

\frac{19}{2}

Decimal Notation

9.5

Solution steps

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

\int_{1}^{2} 6 x^{2} d x = 14

\int_{1}^{2} 3 x d x = \frac{9}{2}

= 14 - \frac{9}{2}

Simplify

14 - \frac{9}{2} : \frac{19}{2}

= \frac{19}{2}

\frac{\partial}{\partial x} (x y - x)

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\int x^{3} 14 + x^{4} d x

What is the integral from 1 to 2 of (6x^2-3x) ?
The integral from 1 to 2 of (6x^2-3x) is 19/2