Question

∫4(3x - 4)^3 dx

Explanation:

Step1: Use substitution method

Let \(u = 3x - 4\), then \(du=3dx\), and \(dx=\frac{1}{3}du\).
The integral \(\int4(3x - 4)^{3}dx\) becomes \(\frac{4}{3}\int u^{3}du\).

Step2: Integrate \(u^{3}\)

Using the power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), for \(n = 3\), \(\int u^{3}du=\frac{u^{4}}{4}+C\).
So \(\frac{4}{3}\int u^{3}du=\frac{4}{3}\times\frac{u^{4}}{4}+C\).

Step3: Substitute back \(u = 3x - 4\)

\(\frac{4}{3}\times\frac{u^{4}}{4}+C=\frac{(3x - 4)^{4}}{3}+C\).

Answer:

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