Question

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

(a)${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{2}\mathbf{\pi }\mathbf{/}\mathbf{\omega }}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\omega }\mathit{t}\mathit{d}\mathit{t}$

(b)${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{2}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{2}\mathit{\pi }\mathit{t}\mathit{d}\mathit{t}$

Short answer

a) The solution of the given integral $I=\frac{\pi }{\omega }$.

b) The solution of the given integral $l=1$.

Step-by-step solution

Given

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

a)${\int }_0^{2\pi /\omega }{\sin }^2\omega tdt$

b)${\int }_0^2{\cos }^{2}2\pi tdt$

The concept of the average value of a function over a particular interval

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 $\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$.

From the given information

a)

Evaluate the following integral without a calculation.

$\Rightarrow {\int }_0^{2\pi /\omega }{\sin }^2\omega tdt$

[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.

$$\begin{gathered} I=\frac{1}{2}(\frac{2\pi }{\omega }-0) \\ I=\frac{\pi }{\omega } \end{gathered}$$

Thus, the solution is $I=\frac{\pi }{\omega }$.

Use the average value method for calculation

b)

Evaluate the following integral without a calculation.

$\Rightarrow {\int }_0^2{\cos }^{2}2\pi tdt$

[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.

$$\begin{gathered} I=\frac{1}{2}(2-0) \\ I=1 \end{gathered}$$

Thus, the solution is $I=1$.