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

The integral from-1 to 2 of 4x(1-x^2) is -9

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

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

- 9

Solution steps

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

Take the constant out:

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

= 4 \cdot \int_{- 1}^{2} x (1 - x^{2}) d x

Expand

x (1 - x^{2}) : x - x^{3}

= 4 \cdot \int_{- 1}^{2} x - 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

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

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

\int_{- 1}^{2} x^{3} d x = \frac{15}{4}

= 4 (\frac{3}{2} - \frac{15}{4})

Simplify

4 (\frac{3}{2} - \frac{15}{4}) : - 9

= - 9

\frac{d}{d x} (2 1^{x})

y^{^{''}} - 4 y^{^{'}} + 26 y = 0

(1 - 4 y + x) \frac{d y}{d x} + y = 4 x - 1

\int 5 e^{3} d x

n = 1 \sum \infty \frac{2 n}{5 n - 9}

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