Explanation:
To solve the question, we will need to re-express the given function as follow:
[tex]f(x)=e^{-2\ln (x)}[/tex]
Will become
[tex]f(x)=e^{-2\ln (x)}=e^{\ln x^{-2}}[/tex]
Thus
[tex]f(x)=e^{\ln x^{-2}}=x^{-2}[/tex]
This simply means that we will find the area under the curve:
[tex]f(x)=x^{-2}\text{ within the interval \lbrack{}1,2\rbrack}[/tex]
Thus
The area will be
[tex]\int ^2_1f(x)dx=\int ^2_1x^{-2}dx[/tex]
This will then be
[tex]\lbrack\frac{x^{-2+1}}{-2+1}\rbrack^2_1=\lbrack\frac{x^{-1}}{-1}\rbrack^2_1[/tex]
This will be simplified to give
[tex]-\lbrack\frac{1}{x}\rbrack^2_1=-\lbrack(\frac{1}{2})-(\frac{1}{1})\rbrack=-1\lbrack-\frac{1}{2}\rbrack=\frac{1}{2}[/tex]
Therefore, the area under the curve will be
[tex]\frac{1}{2}=0.5[/tex]
Thus, the answer is 0.5
