91Ó°ÊÓ

The given expression is

$$\begin{gathered} \left(a\right){∫}_0^3(5x-2)\mathrm{δ}(2-x)dx \\ \left(b\right){∫}_0^n\mathrm{ϕ}\left(x\right)\mathrm{δ}\left(x^2-a^2\right)dx \\ \left(c\right){∫}_{-1}^1\cos x\mathrm{δ}(-2x)dx \\ \left(d\right){∫}_{\frac{2}{2}}^{\frac{\mathrm{π}}{2}}\cos x\mathrm{δ}\left(\sin x\right)dx \end{gathered}$$

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

Given statement is

${∫}_0^3(5x-2)\mathrm{δ}(2-x)dx$

By use the operator equations

${∫}_0^{t_2}x\left(t\right)\mathrm{δ}\left(t-t_0\right)dt·x\left(t_0\right)t_1<t_0<t_1$

Consider the function${∫}_0^3(5x-2)\mathrm{δ}(2-x)dx$

${∫}_0^3(5x-2)\mathrm{δ}(2-x)dx={∫}_0^3(5x-2)\mathrm{δ}(2-x)dx0<x<3$

Therefore

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

Hence,

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

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

Consider the function

$$\begin{gathered} {∫}_0^n\mathrm{Ï•}\left(x\right)\mathrm{δ}\left(x^2-a^2\right)dx \\ at{x}^2=a^2 \\ =Â\pm a \end{gathered}$$

If a is positive then,

role="math" localid="1664276047981" ${∫}_0^{+}\mathrm{ϕ}\left(x\right)\mathrm{δ}\left(x^2-a^2\right)dx=\mathrm{ϕ}\left(a\right)[0<a<=]$

If a is negative then,

${∫}_0^{∞}\mathrm{ϕ}\left(x\right)\mathrm{δ}\left(x^2-a^2\right)dx=\mathrm{ϕ}(-a)[0<-a<a]$

Hence

The evaluation of the integrals ifis positive then,

${∫}_0^{+}\mathrm{ϕ}\left(x\right)\mathrm{δ}\left(x^2-a^2\right)dx$

If a is negative then

role="math" localid="1664276160073" $=\mathrm{ϕ}\left(a\right)[0<a<=]{∫}_0^{-}\mathrm{ϕ}\left(x\right)\mathrm{δ}\left(x^2-a^2\right)dx=\mathrm{ϕ}(-a)[0<-a<a].$

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

Consider the function as${∫}_{-1}^1\cos x\mathrm{δ}(-2x)dx$

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

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

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

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

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

${∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{2}}\cos x\mathrm{δ}\left(\sin x\right)dx$

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

$x=Â\pm (0,\mathrm{Ï€},2\mathrm{Ï€},3\mathrm{Ï€})$

We can also expand the above equation as

$x=Â\pm (0,\mathrm{Ï€},2\mathrm{Ï€},3\mathrm{Ï€})=…..-2\mathrm{Ï€},-\mathrm{Ï€},0,\mathrm{Ï€},2\mathrm{Ï€},3\mathrm{Ï€}……$

But, limits be$-\frac{\mathrm{Ï€}}{2}<0<\frac{\mathrm{Ï€}}{2}$

Therefore${∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{2}}\cos x\mathrm{δ}\left(\sin x\right)dx$

Then,$\cos 0^{∘}=1$

Hence,

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