91Ó°ÊÓ

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

a)${∫}_0^{2\mathrm{Ï€}/\mathrm{Ó\neg }}{\sin }^2\mathrm{Ó\neg }tdt$

b)${∫}_0^2{\cos }^{2}2\mathrm{π}tdt$

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

a)

Evaluate the following integral without a calculation.

$⇒{∫}_0^{2\mathrm{Ï€}/\mathrm{Ó\neg }}{\sin }^2\mathrm{Ó\neg }tdt$

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

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

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

$$\begin{gathered} I=\frac{1}{2}(\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}-0) \\ I=\frac{\mathrm{Ï€}}{\mathrm{Ó\neg }} \end{gathered}$$

Thus, the solution is $I=\frac{\mathrm{Ï€}}{\mathrm{Ó\neg }}$.

b)

Evaluate the following integral without a calculation.

$⇒{∫}_0^2{\cos }^{2}2\mathrm{π}tdt$

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

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

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 $I=1$.