Mathematics - Area Under Curves Question with Solution | TestHub

The required area is bounded by the upper curve $y=x^2+3$, the lower curve $y=x$, the y-axis ($x=0$), and the line $x=3$.

The area $A$ is given by the definite integral:

$A={\int }_0^3(y_1-y_2)dx$

$A={\int }_0^3(x^2+3-x)dx$

Integrating with respect to $x$:

$A={\left[\frac{x^3}{3}-\frac{x^2}{2}+3x\right]}_0^3$

Substituting the limits:

$A=\left(\frac{3^3}{3}-\frac{3^2}{2}+3(3)\right)-0$

$A=9-\frac{9}{2}+9$

$A=18-\frac{9}{2}=\frac{27}{2}$ sq. units

Answer: $\frac{27}{2}$ sq. units