Mathematics - Area Under Curves Question with Solution | TestHub

The given curves are $y=x^2+4$ and $y=x$.

The region is bounded by these curves, the Y-axis ($x=0$), and the line $x=3$.

For all $x\in [0,3]$, $x^2+4>x$, so the parabola lies above the line.

The required area is given by the definite integral of the difference between the upper curve and the lower curve from $x=0$ to $x=3$:

$A={\int }_0^3(x^2+4-x)dx$

$A={\left[\frac{x^3}{3}+4x-\frac{x^2}{2}\right]}_0^3$

$A=\left(\frac{3^3}{3}+4(3)-\frac{3^2}{2}\right)-0$

$A=9+12-\frac{9}{2}$

$A=21-\frac{9}{2}$

$A=\frac{42-9}{2}=\frac{33}{2}{\text{unit}}^2$

Answer: $\frac{33}{2}{\text{unit}}^2$