What is the integral from 0 to 1/2 of x(1-4x^2)^{10} ?

The integral from 0 to 1/2 of x(1-4x^2)^{10} is 1/88

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integral from 0 to 1/2 of x(1-4x^2)^{10}

\int_{0}^{\frac{1}{2}} x (1 - 4 x^{2})^{10} d x

\frac{1}{88}

Decimal

0.01136 \dots

Solution steps

\int_{0}^{\frac{1}{2}} x (1 - 4 x^{2})^{10} d x

Apply u-substitution

: \int_{1}^{0} - \frac{u ^{10}}{8} d u

= \int_{1}^{0} - \frac{u ^{10}}{8} d u

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

= - \int_{0}^{1} - \frac{u ^{10}}{8} d u

Take the constant out:

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

= - (- \frac{1}{8} \cdot \int_{0}^{1} u^{10} d u)

Apply the Power Rule

: [\frac{u ^{11}}{11}]_{0}^{1}

= - (- \frac{1}{8} [\frac{u ^{11}}{11}]_{0}^{1})

Simplify

= \frac{1}{8} [\frac{u ^{11}}{11}]_{0}^{1}

Compute the boundaries:

\frac{1}{11}

= \frac{1}{8} \cdot \frac{1}{11}

Simplify

= \frac{1}{88}

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What is the integral from 0 to 1/2 of x(1-4x^2)^{10} ?
The integral from 0 to 1/2 of x(1-4x^2)^{10} is 1/88