Chapter 4: Q. 37 (page 405)

The average value of the function $f (x) = \sin 2 x o n [0, π] .$

Short Answer

Expert verified

The average of the function $f (x) = \sin 2 x$ on the interval $[0, π] is 0 .$

Step by step solution

Step 1. Given information

The given function $f (x) = \sin 2 x$ and the given interval $[0, π] .$

Step 2. Calculation

The average value of function $f (x) = \sin 2 x$ on the interval $[0, 4]$ is given by $\bar{f} = \frac{\int_{0}^{π} \sin 2 x d x}{\int_{0}^{π} d x} \Rightarrow \bar{f} = \frac{\int_{0}^{π} \sin 2 x d x}{(π - 0)} = - \frac{1}{2 π} {[\cos 2 x]}_{0}^{π} = - \frac{1}{2 π} [\cos 2 π - \cos 0] = - \frac{1}{2 π} [1 - 1] = 0$

The curve of function $f (x) = \sin 2 x$ is shown in Figure-1 here.

Now we calculate area between the curve of the function and x-axis on interval $[0, \frac{π}{2}] a n d [\frac{π}{2}, π]$

Exact area $A_{1} = \int_{0}^{\frac{π}{2}} s i n 2 x d x = - \frac{1}{2} {[c o s 2 x]}_{0}^{\frac{π}{2}} = - \frac{1}{2} [c o s 2 \cdot \frac{π}{2} - c o s 0] = - \frac{1}{2} [- 1 - 1] = 1$

$A_{2} = \int_{\frac{π}{2}}^{π} s i n 2 x d x = - \frac{1}{2} {[c o s 2 x]}_{\frac{π}{2}}^{π} = - \frac{1}{2} [c o s 2 \cdot π - \cos 2 \cdot \frac{π}{2}] = - \frac{1}{2} [1 - (- 1)] = - 1$

Therefore,

$\int_{0}^{π} s i n 2 x d x = A_{1} + A_{2} = 1 - 1 = 0$

Therefore, $\bar{f} = \frac{A_{1} + A_{2}}{π} = \frac{0}{π} = 0$

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The average value of the function $f (x) = \sin 2 x o n [0, π] .$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

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The average value of the function $f (x) = \sin 2 x o n [0, π] .$