91Ó°ÊÓ

The Dirac Delta function which is represented as $\mathit{δ}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ , is defined as

$\mathit{δ}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}0≠0\\ ∞=0\end{array}$

The dirac delta function has the property${\mathbf{∫}}_{\mathbf{-}\mathbf{∞}}^{\mathbf{∞}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{δ}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$ , where is a$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ continuous containing$\mathit{x}\mathbf{=}\mathbf{0}$ .

Let the given integral is $l={∫}_{-2}^6\left(3x^2-2x-1\right)\mathrm{δ}(x-3)dx$ .

Substitute $x=3$into initial term of $l={∫}_{-2}^6\left(3x^2-2x-1\right)\mathrm{δ}(x-3)dx$.

role="math" localid="1657521371890" $$\begin{gathered} l={∫}_{-2}^6\left(3{\left(3\right)}^2-2\left(3\right)-1\right)\mathrm{δ}(x-3)dx \\ ={∫}_{-2}^{6}20\mathrm{δ}(x-3)dx \\ =20{∫}_{-2}^6\mathrm{δ}(x-3)dx \\ =20 \end{gathered}$$

Thus, the result of $l={∫}_{-2}^6\left(3x^2-2x-1\right)\mathrm{δ}(x-3)dx$ in part (a) is 20.

Let the given integral is $l={∫}_0^5\cos x\mathrm{δ}(x-\mathrm{π})dx$ . Substitute $x=\mathrm{π}$into initial term of $l={∫}_0^5\cos x\mathrm{δ}(x-\mathrm{π})dx$.

$$\begin{gathered} l={∫}_0^5\cos x\mathrm{δ}(x-\mathrm{π})dx \\ ={∫}_0^5(-1)\mathrm{δ}(x-\mathrm{π})dx \\ =(-1){∫}_0^5\mathrm{δ}(x-\mathrm{π})dx \\ =-1 \end{gathered}$$

Thus, the result of $l={∫}_0^5\cos x\mathrm{δ}(x-\mathrm{π})dx$in part (b) is -1 .

Let the given integral is $l={∫}_0^3{x}^3\mathrm{δ}(x+1)dx$.

Substitute $x=-1$into initial term of $l={∫}_0^3{x}^3\mathrm{δ}(x+1)dx$.

$$\begin{gathered} l={∫}_0^3{(-1)}^3\mathrm{δ}(x+1)dx \\ ={∫}_0^3(-1)\mathrm{δ}(x+1)dx \\ =(-1){∫}_0^3\mathrm{δ}(x+1)dx \\ =0 \end{gathered}$$

Since -1 does not lies in the interval of integration from 0 to 3, thus the result of $l={∫}_0^3{x}^3\mathrm{δ}(x+1)dx$in part (c) is 0.

Let the given integral is $l={∫}_{-∞}^{∞}I\mathrm{n}(x+3)\mathrm{δ}(x+2)dx$.

Substitute $x=-2$ into initial term of $l={∫}_{-∞}^{∞}I\mathrm{n}(x+3)\mathrm{δ}(x+2)dx$.

$$\begin{gathered} l={∫}_{-∞}^{∞}I\mathrm{n}(-2+3)\mathrm{δ}(x+2)dx \\ ={∫}_{-∞}^{∞}I\mathrm{n}\left(1\right)\mathrm{δ}(x+2)dx \\ ={∫}_{-∞}^{∞}\left(0\right)\mathrm{δ}(x+2)dx \\ =0 \end{gathered}$$

Thus the result of $l={∫}_{-∞}^{∞}I\mathrm{n}(x+3)\mathrm{δ}(x+2)dx$in part (d) is 0.