91Ó°ÊÓ

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

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

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 $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is a continuous containing$\mathit{x}\mathbf{=}\mathbf{0}$

Let the given integral be$l={∫}_{-2}^2(2x+3)\mathrm{δ}\left(x\right)dx$ . Use the property$\mathrm{δ}\left(kx\right)=\frac{1}{\left|k\right|}\mathrm{δ}x$into the integral l .$l=\frac{1}{3}{∫}_{-2}^2(2x+3)\mathrm{δ}\left(x\right)dx$

Substitute x = 0 into the initial term of$l=\frac{1}{3}{∫}_{-2}^2(2x+3)\mathrm{δ}\left(x\right)dx$ .

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

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

Let the given integral be $l={∫}_{-2}^2(x^3+3x+2)\mathrm{δ}(1-x)dx$. Substitute x = 1 into the initial term of$l={∫}_{-2}^2(x^3+3x+2)\mathrm{δ}(1-x)dx$

$$\begin{gathered} l={∫}_{-2}^2({\left(1\right)}^3+3(1)+2)\mathrm{δ}(1-x)dx \\ l={∫}_{-2}^{2}6\mathrm{δ}(1-x)dx \\ l=6{∫}_{-2}^2\mathrm{δ}(1-x)dx \\ l=6 \end{gathered}$$

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

Let the given integral be $l={∫}_{-1}^1\left(9x^2\right)\mathrm{δ}(3x+1)dx$. Rearrange the integral l , using the property $\mathrm{δ}\left(kx\right)=\frac{1}{\left|k\right|}\mathrm{δ}{\left(x\right)}_1$

$$\begin{gathered} l={∫}_{-1}^1\left(9x^2\right)\mathrm{δ}\left(3\left(x+\frac{1}{3}\right)\right)dx \\ =\frac{1}{3}{∫}_{-1}^1\left(9x^2\right)\mathrm{δ}\left(x+\frac{1}{3}\right)dx \end{gathered}$$

Substitute $x=-\frac{1}{3}$into the initial term of$=\frac{1}{3}{∫}_{-1}^1\left(9x^2\right)\mathrm{δ}\left(x+\frac{1}{3}\right)dx$

$$\begin{gathered} l=\frac{1}{3}{∫}_{-1}^1\left(9{\left(-\frac{1}{3}\right)}^2\right)\mathrm{δ}\left(x+\frac{1}{3}\right)dx \\ =\frac{1}{3}{∫}_{-1}^1\left(\frac{9}{9}\right)\mathrm{δ}\left(x+\frac{1}{3}\right)dx \\ =\frac{1}{3}{∫}_{-1}^1\mathrm{δ}\left(x+\frac{1}{3}\right)dx \\ =\frac{1}{3} \end{gathered}$$

Thus, the result of $l={∫}_{-1}^1\left(9x^2\right)\mathrm{δ}(3x+1)dx$in part (c) is $\frac{1}{3}$.

Let the given integral be $l={∫}_{-∞}^a\mathrm{δ}(x-b)dx$. If $b<a$, the Dirac delta function $\mathrm{δ}(x-b)$is defined within the interval of integration. Thus ${∫}_{-∞}^a\mathrm{δ}(x-b)dx=1$, for $b<a$

If $b<a$, the Dirac delta function$\mathrm{δ}(x-b)$lies outside the interval of integration.

Thus,${∫}_{-∞}^a\mathrm{δ}(x-b)dx=0$, for $b>a$$b>a$.