#### Answer

$ \int_{a}^b f(x) \ dx =- \oint_{C} y \ dx $ (Verified)

#### Work Step by Step

Since, $\int_{a}^b f(x) \ dx =\int_{a}^b \int_{0}^f(x) \ dy \ dx =\iint_{R} \ dx \ dy$
The tangential form for Green's Theorem is given as:
The counterclockwise circulation is:
$\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy$
and $Area = \iint_{R} \ dx \ dy = \iint_{R} 0+1 \ dx \ dy= \iint_{R} (\dfrac{\partial (0)}{\partial x} +\dfrac{\partial (y)}{\partial y}) dx dy \\= \oint_{C} 0 \ dy - y \ dx $
Thus, we have: $ \int_{a}^b f(x) \ dx =- \oint_{C} y \ dx $
The result has been proved.