question find the area bounded by the graphs of f(x)=x^2 and g(x)=x + 2 on the interval 0,3. give your…

Explanation:

Step1: Determine the upper - lower function

On the interval $[0,3]$, we need to find which function is greater. Let's find the difference $h(x)=g(x)-f(x)=(x + 2)-x^{2}=-x^{2}+x + 2$. We can find the roots of $h(x)$ using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$. Here, $a=-1$, $b = 1$, $c = 2$. So $x=\frac{-1\pm\sqrt{1-4\times(-1)\times2}}{2\times(-1)}=\frac{-1\pm\sqrt{9}}{-2}=\frac{-1\pm3}{-2}$. The roots are $x=-1$ and $x = 2$. On the interval $[0,2]$, $g(x)\geq f(x)$ and on the interval $[2,3]$, $f(x)\geq g(x)$.

Step2: Calculate the area using definite integrals

The area $A=\int_{0}^{2}[(x + 2)-x^{2}]dx+\int_{2}^{3}[x^{2}-(x + 2)]dx$. First integral: $\int_{0}^{2}(x + 2-x^{2})dx=\left[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}\right]{0}^{2}=\left(\frac{2^{2}}{2}+2\times2-\frac{2^{3}}{3}\right)-\left(0\right)=2 + 4-\frac{8}{3}=\frac{6 + 12-8}{3}=\frac{10}{3}$. Second integral: $\int{2}^{3}(x^{2}-x - 2)dx=\left[\frac{x^{3}}{3}-\frac{x^{2}}{2}-2x\right]_{2}^{3}=\left(\frac{3^{3}}{3}-\frac{3^{2}}{2}-2\times3\right)-\left(\frac{2^{3}}{3}-\frac{2^{2}}{2}-2\times2\right)$ $=(9-\frac{9}{2}-6)-(\frac{8}{3}-2 - 4)$ $=(3-\frac{9}{2})-(\frac{8}{3}-6)$ $=\frac{6 - 9}{2}-\frac{8 - 18}{3}=-\frac{3}{2}+\frac{10}{3}=\frac{-9 + 20}{6}=\frac{11}{6}$. Then $A=\frac{10}{3}+\frac{11}{6}=\frac{20 + 11}{6}=\frac{31}{6}$.

Answer:

$\frac{31}{6}$