What is the integral from 0 to 1 of (6x-9x^2)5x ?

The integral from 0 to 1 of (6x-9x^2)5x is -5/4

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integral from 0 to 1 of (6x-9x^2)5x

\int_{0}^{1} (6 x - 9 x^{2}) 5 x d x

- \frac{5}{4}

Decimal

- 1.25

Solution steps

\int_{0}^{1} (6 x - 9 x^{2}) \cdot 5 x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int_{0}^{1} (6 x - 9 x^{2}) x d x

Expand

(6 x - 9 x^{2}) x : 6 x^{2} - 9 x^{3}

= 5 \cdot \int_{0}^{1} 6 x^{2} - 9 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

= 5 (\int_{0}^{1} 6 x^{2} d x - \int_{0}^{1} 9 x^{3} d x)

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

\int_{0}^{1} 9 x^{3} d x = \frac{9}{4}

= 5 (2 - \frac{9}{4})

Simplify

5 (2 - \frac{9}{4}) : - \frac{5}{4}

= - \frac{5}{4}

- 3 y^{^{'}} + \frac{2 y}{x} = x^{2} y^{4}

\frac{d}{d x} (3 2 x + 3)

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

(e^{2 x} - x^{2} - 1)^{^{'}}

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

What is the integral from 0 to 1 of (6x-9x^2)5x ?
The integral from 0 to 1 of (6x-9x^2)5x is -5/4