Find the area of the region enclosed by the curves $y=9 \sin x$ and $y=\sin (9 x), 0 \leq x \leq \pi$.
The area of the region enclosed by the curves is (Simplify your answer.)
Answer
Final Answer: The area of the region enclosed by the curves \(y=9 \sin x\) and \(y=\sin (9 x), 0 \leq x \leq \pi\) is \(\boxed{\frac{160}{9}}\).
Step 1 ：The area enclosed by two curves \(f(x)\) and \(g(x)\) from \(a\) to \(b\) is given by the integral \(\int_a^b |f(x) - g(x)| dx\). In this case, we need to find the area enclosed by the curves \(y = 9\sin(x)\) and \(y = \sin(9x)\) from \(0\) to \(\pi\). We can calculate this area by integrating the absolute difference of the two functions from \(0\) to \(\pi\).
Step 2 ：The area of the region enclosed by the curves is \(\frac{160}{9}\).
Step 3 ：Final Answer: The area of the region enclosed by the curves \(y=9 \sin x\) and \(y=\sin (9 x), 0 \leq x \leq \pi\) is \(\boxed{\frac{160}{9}}\).
