What is the integral from 0 to 1/3 of 2x-6x^2 ?

The integral from 0 to 1/3 of 2x-6x^2 is 1/27

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integral from 0 to 1/3 of 2x-6x^2

\int_{0}^{\frac{1}{3}} 2 x - 6 x^{2} d x

\frac{1}{27}

Decimal

0.03703 \dots

Solution steps

\int_{0}^{\frac{1}{3}} 2 x - 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}^{\frac{1}{3}} 2 x d x - \int_{0}^{\frac{1}{3}} 6 x^{2} d x

\int_{0}^{\frac{1}{3}} 2 x d x = \frac{1}{9}

\int_{0}^{\frac{1}{3}} 6 x^{2} d x = \frac{2}{27}

= \frac{1}{9} - \frac{2}{27}

Simplify

\frac{1}{9} - \frac{2}{27} : \frac{1}{27}

= \frac{1}{27}

\int y^{7} 1 - y^{4} d y

\frac{d}{d x} (- x e^{2 x})

x \cdot y^{2} \cdot \frac{d y}{d x} = y^{3} - 4 \cdot x^{3}, y (1) = 2

x^{2} - 8, 7 x

x \to 0 lim (\frac{sin ( 5 x )}{5})

What is the integral from 0 to 1/3 of 2x-6x^2 ?
The integral from 0 to 1/3 of 2x-6x^2 is 1/27