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${鈭�}_a^b{\sin }^{2}kxdx={鈭�}_a^b{\cos }^{2}kxdx=\frac{1}{2}(b-a)$ if $k\left(b-a\right)$is an integral multiples $蟺/2$. Evaluate the following integral without calculations.

a)${鈭�}_{-1/4}^{11/4}{\cos }^2蟺xdx$

b)${鈭�}_{-1}^2{\sin }^2\left(\frac{蟺x}{3}\right)dx$

The average value of a function over a particular interval can be found with an expression involving an integral.

Let's say that interval is $\mathbf{[}\mathit{a}\mathbf{,}\mathit{b}\mathbf{|}$.

Then the average value of f(x)over said interval is $\frac{\mathbf{1}}{\mathbf{b}\mathbf{-}\mathbf{a}}{\mathbf{鈭�}}_{\mathbf{a}}^{\mathbf{b}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}$.

a)

Evaluate the following integral without a calculation.

$鈬�{鈭�}_{-1/4}^{11/4}{\cos }^2蟺xdx$

[Use ${鈭�}_a^b{\sin }^{2}kxdx={鈭�}_a^b{\cos }^{2}kxdx=\frac{1}{2}(b-a)$]

$I_a=\frac{1}{2}(b-a)$

Substitute the value of a and b in the above equation.

$$\begin{gathered} I=\frac{1}{2}(\frac{11}{4}-\frac{\left(-1\right)}{4}) \\ =\frac{3}{2} \\ I=\frac{3}{2} \end{gathered}$$

Thus, the solution is $I=\frac{3}{2}$.

b)

Evaluate the following integral without a calculation.

$鈬�{鈭�}_{-1}^2{\sin }^2\left(\frac{蟺x}{3}\right)dx$

[Use ${鈭�}_a^b{\sin }^{2}kxdx={鈭�}_a^b{\cos }^{2}kxdx=\frac{1}{2}(b-a)$]

$I_a=\frac{1}{2}(b-a)$

Substitute the value of a and b in the above equation.

$$\begin{gathered} I=\frac{1}{2}(2-\left(-1\right)) \\ =\frac{3}{2} \\ I=\frac{3}{2} \end{gathered}$$

Thus, the solution is $I=\frac{3}{2}$.