What is the integral from 0 to 1 of 2x(4-x^2) ?

The integral from 0 to 1 of 2x(4-x^2) is 7/2

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integral from 0 to 1 of 2x(4-x^2)

\int_{0}^{1} 2 x (4 - x^{2}) d x

\frac{7}{2}

Decimal Notation

3.5

Solution steps

\int_{0}^{1} 2 x (4 - x^{2}) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} x (4 - x^{2}) d x

Expand

x (4 - x^{2}) : 4 x - x^{3}

= 2 \cdot \int_{0}^{1} 4 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

= 2 (\int_{0}^{1} 4 x d x - \int_{0}^{1} x^{3} d x)

\int_{0}^{1} 4 x d x = 2

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

= 2 (2 - \frac{1}{4})

Simplify

2 (2 - \frac{1}{4}) : \frac{7}{2}

= \frac{7}{2}

\frac{\partial}{\partial x} (10 x y e^{- (x^{2} + y^{2})})

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

f (x) = tan (4 x)

4

- 4 cos (t)

What is the integral from 0 to 1 of 2x(4-x^2) ?
The integral from 0 to 1 of 2x(4-x^2) is 7/2