Make use of the operator equationsand previous equations to evaluate the following integrals.

(a)${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{3}}\mathbf{(}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}\mathit{\delta }\mathbf{(}\mathbf{2}\mathbf{-}\mathit{x}\mathbf{)}\mathit{d}\mathit{x}$

(b)role="math" localid="1664275233678" ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathit{\varphi }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{\delta }\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}\mathbf{)}\mathit{d}\mathit{x}$

(c)role="math" localid="1664275249399" ${\mathbf{\int }}_{\mathbf{-}\mathbf{1}}^{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathit{\delta }\mathbf{(}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{)}\mathit{d}\mathit{x}$

(d)role="math" localid="1664275262923" ${\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }\mathbf{/}\mathbf{2}}^{\mathbf{\pi }\mathbf{/}\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathit{\delta }\mathbf{\left(}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}$

The given expression is

$$\begin{gathered} \left(a\right){\int }_0^3(5x-2)\delta (2-x)dx \\ \left(b\right){\int }_0^n\varphi \left(x\right)\delta \left(x^2-a^2\right)dx \\ \left(c\right){\int }_{-1}^1\cos x\delta (-2x)dx \\ \left(d\right){\int }_{\frac{2}{2}}^{\frac{\pi }{2}}\cos x\delta \left(\sin x\right)dx \end{gathered}$$

Integration by partsor partial integration is a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandantiderivative.

Given statement is

${\int }_0^3(5x-2)\delta (2-x)dx$

By use the operator equations

${\int }_0^{t_2}x\left(t\right)\delta \left(t-t_0\right)dt·x\left(t_0\right)t_1<t_0<t_1$

Consider the function${\int }_0^3(5x-2)\delta (2-x)dx$

${\int }_0^3(5x-2)\delta (2-x)dx={\int }_0^3(5x-2)\delta (2-x)dx0<x<3$

Therefore

$$\begin{gathered} x=2 \\ =5\left(2\right)-2 \\ =8 \end{gathered}$$

Hence,

The evaluation of the integrals${\int }_0^3(5x-2)\delta (2-x)dx=8$

By using the operator equations${\int }_a^{h}x\left(t\right)\delta \left(t-t_0\right)dt.x\left(t_0\right)t_1<t_0<t_1$

Consider the function

$$\begin{gathered} {\int }_0^n\varphi \left(x\right)\delta \left(x^2-a^2\right)dx \\ at{x}^2=a^2 \\ =\pm a \end{gathered}$$

If a is positive then,

role="math" localid="1664276047981" ${\int }_0^{+}\varphi \left(x\right)\delta \left(x^2-a^2\right)dx=\varphi \left(a\right)[0<a<=]$

If a is negative then,

${\int }_0^{\infty }\varphi \left(x\right)\delta \left(x^2-a^2\right)dx=\varphi (-a)[0<-a<a]$

Hence

The evaluation of the integrals ifis positive then,

${\int }_0^{+}\varphi \left(x\right)\delta \left(x^2-a^2\right)dx$

If a is negative then

role="math" localid="1664276160073" $=\varphi \left(a\right)[0<a<=]{\int }_0^{-}\varphi \left(x\right)\delta \left(x^2-a^2\right)dx=\varphi (-a)[0<-a<a].$

By using the operator equations${\int }_0^{t_1}x\left(t\right)\delta \left(t-t_0\right)dt·x\left(t_0\right)t_1<t_0<t_1$

Consider the function as${\int }_{-1}^1\cos x\delta (-2x)dx$

Where, we can evaluate such integral, and hence comparing we know that

$\varphi \left(x\right)=\frac{1}{2}\cos \left(x\right)x_0=0$

And, since$x_0$lies within the limits of the integral 0 lies between -1 and 1, thus this integral is non-zero and is qual to

$$\begin{gathered} I=\frac{1}{2}\varphi \left(0\right) \\ =\frac{1}{2}\cos \left(0\right) \\ =\frac{1}{2} \end{gathered}$$

By using the operator equations ${\int }_a^{h_2}x\left(t\right)\delta \left(t-t_0\right)dt.x\left(t_0\right)t_1<t_0<t_1$.

${\int }_{\frac{\pi }{2}}^{\frac{\pi }{2}}\cos x\delta \left(\sin x\right)dx$

Calculate the value of x at $\sin x=0$.

$x=\pm (0,\pi ,2\pi ,3\pi )$

We can also expand the above equation as

$x=\pm (0,\pi ,2\pi ,3\pi )=\dots ..-2\pi ,-\pi ,0,\pi ,2\pi ,3\pi \dots \dots$

But, limits be$-\frac{\pi }{2}<0<\frac{\pi }{2}$

Therefore${\int }_{\frac{\pi }{2}}^{\frac{\pi }{2}}\cos x\delta \left(\sin x\right)dx$

Then,$\cos 0^{\circ }=1$

Hence,

The evaluation of the integrals$\cos 0^0=1$