integral of (3-2x)^3 — step-by-step solution

⚠ 2x is read as the product 2*x: for a power, write 2^x

Let's integrate:

\int({(3 - 2 x)}^{3})dx

We attempt to reverse an integration: starting from

{(3 - 2 x)}^{3}

and

\frac{d}{d x} (a^{n}) = n a^{(n - 1)}

, we differentiate using ^(m+1)

Let's perform the differentiation:

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

\frac{d}{d x} ({(3 - 2 x)}^{4})

Apply the differentiation rule:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{(n - 1)}

with f =

3 - 2 x

4 \cdot \frac{d}{d x} (3 - 2 x) \cdot {(3 - 2 x)}^{3}

4 \cdot (0 + \frac{d}{d x} (- 2 x)) \cdot {(3 - 2 x)}^{3}

4 \cdot (-2) \cdot {(3 - 2 x)}^{3}

-8 \cdot {(3 - 2 x)}^{3}

You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

\frac{1}{-8} \cdot {(3 - 2 x)}^{4}

- \frac{1}{8} \cdot {(3 - 2 x)}^{4}

- \frac{{(3 - 2 x)}^{4}}{8}

Indefinite integrals are defined up to an additive constant C, so this yields:

- \frac{{(3 - 2 x)}^{4}}{8} + C

▶Antiderivative:

- \frac{{(- 2 x + 3)}^{4}}{8} + C

, with C being any constant