The area bounded by the curve $\mathrm{y}=x^2+3, \mathrm{y}=x, x=3$ and y -axis is...

Area Between Curves

The area bounded by $y=x^2+3$, $y=x$, $x=3$, and $x=0$ is determined by the definite integral of the difference between the upper and lower functions over the interval $[0, 3]$.

Since $x^2+3 > x$ for all $x$ (as $x^2 - x + 3$ has discriminant $-11$ and is always positive), the area is computed as:

$A = \int_0^3 [(x^2+3) - x] dx = \int_0^3 (x^2 - x + 3) dx$

Evaluating the integral:

$A = \left[ \frac{x^3}{3} - \frac{x^2}{2} + 3x \right]_0^3 = \left( \frac{27}{3} - \frac{9}{2} + 9 \right) - (0) = 9 - \frac{9}{2} + 9 = \frac{27}{2}$

The area is $\frac{27}{2}$ square units, corresponding to option C.