What is the integral of (3y-4)^3 ?

The integral of (3y-4)^3 is 1/12 (3y-4)^4+C

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integral of (3y-4)^3

\int (3 y - 4)^{3} d y

\frac{1}{12} (3 y - 4)^{4} + C

Solution steps

\int (3 y - 4)^{3} d y

Apply u-substitution

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

= \int \frac{u ^{3}}{3} d u

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{u ^{4}}{4}

Substitute back

u = 3 y - 4

= \frac{1}{3} \cdot \frac{( 3 y - 4 ) ^{4}}{4}

Simplify

\frac{1}{3} \cdot \frac{( 3 y - 4 ) ^{4}}{4} : \frac{1}{12} (3 y - 4)^{4}

= \frac{1}{12} (3 y - 4)^{4}

Add a constant to the solution

= \frac{1}{12} (3 y - 4)^{4} + C

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