find the area of the region between the graphs of ( f(x)=11x + 8 ) and ( g(x)=x^{2}+4x + 2 ) over (0,2)…

Explanation:

Step1: Determine the upper and lower function

On the interval ([0,2]), we check (f(x)-g(x)=(11x + 8)-(x^{2}+4x + 2)=-x^{2}+7x + 6). For (x\in[0,2]), (y=-x^{2}+7x + 6) is positive. So (f(x)\geq g(x)) on ([0,2]). The formula for the area (A) between two curves (y = f(x)) and (y = g(x)) on ([a,b]) is (A=\int_{a}^{b}[f(x)-g(x)]dx). Here, (a = 0), (b = 2), (f(x)-g(x)=-x^{2}+7x + 6). So (A=\int_{0}^{2}(-x^{2}+7x + 6)dx).

Step2: Integrate term - by - term

We know that (\int(-x^{2}+7x + 6)dx=-\frac{x^{3}}{3}+\frac{7x^{2}}{2}+6x+C). Using the fundamental theorem of calculus (\int_{a}^{b}F^{\prime}(x)dx=F(b)-F(a)), where (F(x)=-\frac{x^{3}}{3}+\frac{7x^{2}}{2}+6x). (F(2)-F(0)=\left(-\frac{2^{3}}{3}+\frac{7\times2^{2}}{2}+6\times 2\right)-\left(-\frac{0^{3}}{3}+\frac{7\times0^{2}}{2}+6\times0\right)). First, calculate (-\frac{2^{3}}{3}+\frac{7\times2^{2}}{2}+6\times 2): (-\frac{8}{3}+\frac{28}{2}+12=-\frac{8}{3}+14 + 12). (=-\frac{8}{3}+26=\frac{- 8+78}{3}=\frac{70}{3}).

Answer:

(\frac{70}{3})