evaluate the integral\n int_{t}^{6}(3 + t^{7})^{7}dt \n int_{t}^{6}(3 + t^{7})^{7}dt=square
evaluate the integral\n int_{t}^{6}(3 + t^{7})^{7}dt \n int_{t}^{6}(3 + t^{7})^{7}dt=square
Answer
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\neq - 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)