Q44P - Introduction to Electrodynamics

Question

Question: Evaluate the following integrals:

(a) ${\mathbf{\int }}_{\mathbf{2}}^{\mathbf{6}}\mathbf{(}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{\delta }\mathbf{(}\mathit{x}\mathbf{-}\mathbf{3}\mathbf{)}\mathit{d}\mathit{x}$ 

(b)  ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{5}}\mathbf{(}\mathit{c}\mathit{o}\mathit{s}\mathbf{ }\mathit{x}\mathbf{ }\mathit{\delta }\mathbf{(}\mathit{x}\mathbf{-}\mathit{\pi }\mathbf{)}\mathit{d}\mathit{x}$

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

(d) ${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }\mathbf{ }}^{\mathbf{\infty }}\mathbf{In}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{\delta }\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{dx}$

Step-by-Step Solution

Answer

(a) The result of  $l={\int }_2^6(3x^2-2x-1)\delta (x-3)dx$ in part (a) is 20.

(b) The result of  $l={\int }_0^5(cos x \delta (x-\pi )dx$ in part (b) is  .

(c) The result of  $l={\int }_0^3 x^2 \delta (x+1)dx$ in part (c) is 0.

(d) The result of  $l={\int }_{-\infty }^{\infty }\mathrm{In}(\mathrm{x}+3)\mathrm{\delta }(\mathrm{x}+2)\mathrm{dx}$ in part (d) is 0

Step 1: Define Dirac delta function/.

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

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

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

Step 2: Prove the integral in part (a)

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

 

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

  $$\begin{gathered} l={\int }_{-2}^6\left(3{\left(3\right)}^2-2\left(3\right)-1\right)\delta (x-3)dx \\ ={\int }_{-2}^{6}20\delta (x-3)dx \\ =20{\int }_{-2}^6\delta (x-3)dx \\ =20 \end{gathered}$$

 

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

Step: 3 Prove the integral in part (b)

Let the given integral is $l={\int }_0^5\cos x\delta (x-\pi )dx$  . Substitute  $x=\pi$ into initial term of  $l={\int }_0^5\cos x\delta (x-\pi )dx$ .

 $$\begin{gathered} l={\int }_0^5\cos x\delta (x-\pi )dx \\ ={\int }_0^5(-1)\delta (x-\pi )dx \\ =(-1){\int }_0^5\delta (x-\pi )dx \\ =-1 \end{gathered}$$

 

Thus, the result of $l={\int }_0^5\cos x\delta (x-\pi )dx$ in part (b) is -1  .

Step: 4 Prove the integral in part (c)

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

 

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

$$\begin{gathered} l={\int }_{0 }^3{(-1)}^3\delta (x+1)dx \\ ={\int }_0^3(-1)\delta (x+1)dx \\ =(-1){\int }_0^3\delta (x+1)dx \\ =0 \end{gathered}$$ 

 

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

Step: 5 Prove the integral in part (d)

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

 

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

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

 

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