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

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$\mathit{x}\mathbf{=}\mathbf{0}$ .

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

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

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

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 .

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.

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.