Step-by-step explanation:
To solve the question, we will need to re-express the given function as follow:
![f(x)=e^(-2\ln (x))](https://img.qammunity.org/2023/formulas/mathematics/college/aeylpuqkfqrs32q6o3ixfoepnyqtjtfhpr.png)
Will become
![f(x)=e^(-2\ln (x))=e^{\ln x^(-2)}](https://img.qammunity.org/2023/formulas/mathematics/college/q3xisths2sa0obwxfufchzds7mehcpcjn0.png)
Thus
![f(x)=e^{\ln x^(-2)}=x^(-2)](https://img.qammunity.org/2023/formulas/mathematics/college/x5ph662u8bkz0etgx5lvn7zmrnzkqm2j1h.png)
This simply means that we will find the area under the curve:
![f(x)=x^(-2)\text{ within the interval \lbrack{}1,2\rbrack}](https://img.qammunity.org/2023/formulas/mathematics/college/9reo2jowr4f76lpi1xbi8orgw7841m19m5.png)
Thus
The area will be
![\int ^2_1f(x)dx=\int ^2_1x^(-2)dx](https://img.qammunity.org/2023/formulas/mathematics/college/ewhhn4hfw5mi8uf3hx7pmfmdx3cvezjbry.png)
This will then be
![\lbrack(x^(-2+1))/(-2+1)\rbrack^2_1=\lbrack(x^(-1))/(-1)\rbrack^2_1](https://img.qammunity.org/2023/formulas/mathematics/college/1c6dyg1053de1cm5fkoiltvxj4763bbnjp.png)
This will be simplified to give
![-\lbrack(1)/(x)\rbrack^2_1=-\lbrack((1)/(2))-((1)/(1))\rbrack=-1\lbrack-(1)/(2)\rbrack=(1)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/9leewq8ksab6kgl6by8uv5smw4tebf1071.png)
Therefore, the area under the curve will be
![(1)/(2)=0.5](https://img.qammunity.org/2023/formulas/mathematics/high-school/4h36bsuwugqz27yh65chi8lib8t3ohhqw5.png)
Thus, the answer is 0.5