Use the results ${\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{k}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{k}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathit{b}\mathbf{-}\mathit{a}\mathbf{)}\mathit{a}$ to evaluate the following integrals without calculation.

(a)${\mathbf{\int }}_{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{4}}^{\mathbf{11}\mathbf{/}\mathbf{4}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\pi }\mathit{x}\mathit{d}\mathit{x}$

(b)${\mathbf{\int }}_{\mathbf{-}\mathbf{1}}^{\mathbf{2}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\left(\frac{\pi x}{3}\right)\mathit{d}\mathit{x}$

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

a)${\int }_{-1/4}^{11/4}{\cos }^2\pi xdx$

b)${\int }_{-1}^2{\sin }^2\left(\frac{\pi x}{3}\right)dx$

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 $\frac{\mathbf{1}}{\mathbf{b}\mathbf{-}\mathbf{a}}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}$.

a)

Evaluate the following integral without a calculation.

$\Rightarrow {\int }_{-1/4}^{11/4}{\cos }^2\pi xdx$

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

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

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.

$\Rightarrow {\int }_{-1}^2{\sin }^2\left(\frac{\pi x}{3}\right)dx$

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

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

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 $I=\frac{3}{2}$.