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

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

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

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

\frac{1}{5}

Decimal Notation

0.2

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - \frac{u ^{4}}{2} d u

= 2 \cdot \int_{1}^{0} - \frac{u ^{4}}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{0}^{1} - \frac{u ^{4}}{2} d u)

Take the constant out:

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

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

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{0}^{1}

= 2 (- (- \frac{1}{2} [\frac{u ^{5}}{5}]_{0}^{1}))

Simplify

2 (- (- \frac{1}{2} [\frac{u ^{5}}{5}]_{0}^{1})) : [\frac{u ^{5}}{5}]_{0}^{1}

= [\frac{u ^{5}}{5}]_{0}^{1}

Compute the boundaries:

\frac{1}{5}

= \frac{1}{5}

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

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

\int_{0}^{2} - x^{2} + 3 x - 2 d x

\int_{- 2}^{2} ∣ x + 1 ∣ d x

\int_{1}^{0} ∣ 2 x - 1 ∣ d x

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