Q. 37 The average value of the functio... [FREE SOLUTION]

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Chapter 4: Q. 37 (page 405)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information

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

Step 2. Calculation

The average value of function $f (x) = \sin 2 x$ on the interval $[0, 4]$ is given by $\overset{炉}{f} = \frac{{鈭�}_{0}^{蟺} \sin 2 x d x}{{鈭�}_{0}^{蟺} d x} 鈬� \overset{炉}{f} = \frac{{鈭�}_{0}^{蟺} \sin 2 x d x}{(蟺 - 0)} = - \frac{1}{2 蟺} {[\cos 2 x]}_{0}^{蟺} = - \frac{1}{2 蟺} [\cos 2 蟺 - \cos 0] = - \frac{1}{2 蟺} [1 - 1] = 0$

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 $[0, \frac{蟺}{2}] a n d [\frac{蟺}{2}, 蟺]$

Exact area $A_{1} = {鈭�}_{0}^{\frac{蟺}{2}} s i n 2 x d x = - \frac{1}{2} {[c o s 2 x]}_{0}^{\frac{蟺}{2}} = - \frac{1}{2} [c o s 2 路 \frac{蟺}{2} - c o s 0] = - \frac{1}{2} [- 1 - 1] = 1$

$A_{2} = {鈭�}_{\frac{蟺}{2}}^{蟺} s i n 2 x d x = - \frac{1}{2} {[c o s 2 x]}_{\frac{蟺}{2}}^{蟺} = - \frac{1}{2} [c o s 2 路蟺 - \cos 2 路 \frac{蟺}{2}] = - \frac{1}{2} [1 - (- 1)] = - 1$

Therefore,

${鈭�}_{0}^{蟺} s i n 2 x d x = A_{1} + A_{2} = 1 - 1 = 0$

Therefore, $\overset{炉}{f} = \frac{A_{1} + A_{2}}{蟺} = \frac{0}{蟺} = 0$

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91影视

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

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

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91影视

The average value of the functionf(x)=sin2xon[0,蟺].

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

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Most popular questions from this chapter

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