Answer
$0.3221$
Work Step by Step
Here, the area can be expressed as: $A=\int_{1}^{3.51} [\ln x -(\dfrac{x}{2}-\dfrac{1}{2})] \ dx$
In order to solve the above integral, we will use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Now, we have $\int_{1}^{3.51} [\ln x -\dfrac{x}{2}+\dfrac{1}{2}] \ dx= [x \ln x -x -\dfrac{x^2}{4}+\dfrac{x}{2}]_{1}^{3.51}$
or, $=[3.51 \ln (3.51) -3.51 -\dfrac{(3.51)^2}{4}+\dfrac{3.51}{2}]-[1\ln (1) -1 -\dfrac{(1)^2}{4}+\dfrac{1}{2}]$
Therefore, the required area is: $Area \approx 0.3221$
