Question

Question: Evaluate the following integrals:

(a)${\mathbf{\int }}_{\mathbf{-}\mathbf{2}}^{\mathbf{2}}\mathbf{(}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{3}\mathbf{)}\mathit{\delta }\mathbf{\left(}\mathbf{3}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}$

(b)${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{2}}\mathbf{(}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{)}\mathit{\delta }\mathbf{(}\mathbf{1}\mathbf{-}\mathit{x}\mathbf{)}\mathit{d}\mathit{x}$

(c)${\mathbf{\int }}_{\mathbf{-}\mathbf{4}}^{\mathbf{1}}\mathbf{9}{\mathit{x}}^{\mathbf{2}}\mathit{\delta }\mathbf{(}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{d}\mathit{x}$

(d)${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{a}}\mathit{\delta }\mathbf{(}\mathit{x}\mathbf{-}\mathit{b}\mathbf{)}\mathit{d}\mathit{x}$

Short answer

(a) The result of $l={\int }_{-2}^2(2x+3)\delta \left(3x\right)dx$in part (a) is 1.

(b) The result of $l={\int }_{-2}^2(x^3+3x+2)\delta (1-x)dx$in part (b) is 6.

(c) The result of $l={\int }_{-1}^1\left(9x^2\right)\delta (3x+1)dx$in part (c) is $\frac{1}{3}$.

(d) For $b<a$, ${\int }_{-\infty }^a\delta (x-b)dx=1$, and for $b<a$,${\int }_{-\infty }^a\delta (x-b)dx=0$ .

Step-by-step solution

Define the Dirac delta function

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

$\mathit{\delta }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}0x\ne 0\\ \infty x=0\end{array}$

The Dirac delta function has the property ${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{\delta }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$, where $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is a continuous containing$\mathit{x}\mathbf{=}\mathbf{0}$

Prove the integral in part (a)

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

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

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

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

Step: 3 Prove the integral in part (a)

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

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

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

Step: 4 Prove the integral in part (c)

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

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

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

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

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

Step: 4 Prove the integral in part (d)

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

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

Thus,${\int }_{-\infty }^a\delta (x-b)dx=0$, for $b>a$$b>a$.