What is the integral of 1^2(2x-1)^3 ?

The integral of 1^2(2x-1)^3 is 1/8 (2x-1)^4+C

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integral of 1^2(2x-1)^3

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

\frac{1}{8} (2 x - 1)^{4} + C

Solution steps

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

Take the constant out:

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

= 1^{2} \cdot \int (2 x - 1)^{3} d x

Simplify

= 1 \cdot \int (2 x - 1)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{2} d u

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{4}}{4}

= 1 \cdot \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = 2 x - 1

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

Simplify

1 \cdot \frac{1}{2} \cdot \frac{( 2 x - 1 ) ^{4}}{4} : \frac{1}{8} (2 x - 1)^{4}

= \frac{1}{8} (2 x - 1)^{4}

Add a constant to the solution

= \frac{1}{8} (2 x - 1)^{4} + C

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What is the integral of 1^2(2x-1)^3 ?
The integral of 1^2(2x-1)^3 is 1/8 (2x-1)^4+C