Question

find the antiderivative of the given derivative.
\\( \frac { d s } { d t } = 14 t \left( 4 t ^ { 2 } - 5 \
ight) ^ { 3 } \\)
\\( s = \\)

Explanation:

Step1: Use substitution

Let \(u = 4t^{2}-5\), then \(du=8tdt\), and \(tdt=\frac{1}{8}du\). The integral \(\int14t(4t^{2} - 5)^{3}dt\) becomes \(\int14u^{3}\times\frac{1}{8}du\).

Step2: Simplify the integral

\(\int14u^{3}\times\frac{1}{8}du=\frac{14}{8}\int u^{3}du=\frac{7}{4}\int u^{3}du\).

Step3: Integrate using the power rule

Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n = 3\), \(\frac{7}{4}\int u^{3}du=\frac{7}{4}\times\frac{u^{4}}{4}+C=\frac{7u^{4}}{16}+C\).

Step4: Substitute back \(u = 4t^{2}-5\)

Substituting \(u = 4t^{2}-5\) back, we get \(s=\frac{7(4t^{2}-5)^{4}}{16}+C\).

Answer:

\(s=\frac{7(4t^{2}-5)^{4}}{16}+C\)