Question

evaluate the integral
int_{t}^{6}(3 + t^{7})^{7}dt
int_{t}^{6}(3 + t^{7})^{7}dt=square

Explanation:

Step1: Use substitution method

Let \(u = 3 + t^{7}\), then \(du=7t^{6}dt\), and \(t^{6}dt=\frac{1}{7}du\).

Step2: Change the integral limits

When \(t = 0\), \(u = 3+0^{7}=3\); when \(t = t\), \(u = 3 + t^{7}\).
The integral \(\int(3 + t^{7})^{7}t^{6}dt=\frac{1}{7}\int u^{7}du\).

Step3: Integrate \(u^{7}\)

According to the power - rule of integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have \(\frac{1}{7}\int u^{7}du=\frac{1}{7}\times\frac{u^{8}}{8}+C=\frac{u^{8}}{56}+C\).

Step4: Substitute back \(u = 3 + t^{7}\)

\(\frac{(3 + t^{7})^{8}}{56}+C\).

Answer:

\(\frac{(3 + t^{7})^{8}}{56}+C\)