167k views
2 votes
Consider the given function and the given interval.

f(x) = 8 sin x - 4 sin 2x, [0,pi]

(a) Find the average value f ave of f on the given interval.

(b) Find c such that f ave = f(c).

User Ememem
by
8.0k points

1 Answer

2 votes

(a) The average value of f(x) on the closed interval [0, π] is


\displaystyle\frac1{\pi-0}\int_0^\pi f(x)\,\mathrm dx = \frac1\pi\int_0^\pi(8\sin(x)-4\sin(2x))\,\mathrm dx = \boxed{\frac{16}\pi}

(b) By the mean value theorem, there is some c in the open interval (0, π) such that f(c) = 16/π. Solve for c :

8 sin(c) - 4 sin(2c) = 16/π

8 sin(c) - 8 sin(c) cos(c) = 16/π

sin(c) - sin(c) cos(c) = 2/π

Use a calculator to solve this. You should get two solutions, c ≈ 1.2382 and c ≈ 2.8081.

User Nuzhny
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories