What is the integral from 0 to 1 of 1/((1+2y)^3) ?

The integral from 0 to 1 of 1/((1+2y)^3) is 2/9

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integral from 0 to 1 of 1/((1+2y)^3)

\int_{0}^{1} \frac{1}{( 1 + 2 y ) ^{3}} d y

\frac{2}{9}

Decimal Notation

0.22222 \dots

Solution steps

\int_{0}^{1} \frac{1}{( 1 + 2 y ) ^{3}} d y

Apply u-substitution

: \int_{1}^{3} \frac{1}{2 u ^{3}} d u

= \int_{1}^{3} \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{3} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{1}^{3}

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{1}^{3}

Compute the boundaries:

\frac{4}{9}

= \frac{1}{2} \cdot \frac{4}{9}

Simplify

= \frac{2}{9}

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What is the integral from 0 to 1 of 1/((1+2y)^3) ?
The integral from 0 to 1 of 1/((1+2y)^3) is 2/9