current objective find the area of a region bounded above by two different functions question find the area…

Explanation:

Step1: Determine which function is greater

First, find the intersection of $f(x)=x^{2}+10x + 25$ and $g(x)=3x + 33$ on the interval $[-11,-5]$. Set $f(x)=g(x)$: [x^{2}+10x + 25=3x + 33] [x^{2}+7x - 8=0] Factor: ((x + 8)(x - 1)=0), solutions are (x=-8) and (x = 1). On the interval ([-11,-5]), we test a value, say (x=-6). (f(-6)=(-6)^{2}+10\times(-6)+25=36-60 + 25=1), (g(-6)=3\times(-6)+33=-18 + 33=15), so (g(x)\geq f(x)) on ([-11,-5]).

Step2: Set up the area - integral formula

The area (A) between two curves (y = f(x)) and (y = g(x)) and the (x) - axis on the interval ([a,b]) is given by (A=\int_{a}^{b}\max{f(x),g(x)}dx). Since (g(x)\geq f(x)) on ([-11,-5]), the area (A=\int_{-11}^{-5}(3x + 33)dx-\int_{-11}^{-5}(x^{2}+10x + 25)dx). We know that (\int(3x + 33)dx=\frac{3}{2}x^{2}+33x+C) and (\int(x^{2}+10x + 25)dx=\frac{1}{3}x^{3}+5x^{2}+25x+C).

Step3: Evaluate the definite - integrals

[ \begin{align*} \int_{-11}^{-5}(3x + 33)dx&=\left[\frac{3}{2}x^{2}+33x\right]{-11}^{-5}\ &=\left(\frac{3}{2}\times(-5)^{2}+33\times(-5)\right)-\left(\frac{3}{2}\times(-11)^{2}+33\times(-11)\right)\ &=\left(\frac{75}{2}-165\right)-\left(\frac{363}{2}-363\right)\ &=\frac{75 - 330}{2}-\frac{363 - 726}{2}\ &=\frac{-255}{2}-\frac{-363}{2}\ &=\frac{-255 + 363}{2}=\frac{108}{2}=54 \end{align*} ] [ \begin{align*} \int{-11}^{-5}(x^{2}+10x + 25)dx&=\left[\frac{1}{3}x^{3}+5x^{2}+25x\right]{-11}^{-5}\ &=\left(\frac{1}{3}\times(-5)^{3}+5\times(-5)^{2}+25\times(-5)\right)-\left(\frac{1}{3}\times(-11)^{3}+5\times(-11)^{2}+25\times(-11)\right)\ &=\left(-\frac{125}{3}+125 - 125\right)-\left(-\frac{1331}{3}+605-275\right)\ &=-\frac{125}{3}-\left(-\frac{1331}{3}+330\right)\ &=-\frac{125}{3}+\frac{1331}{3}-330\ &=\frac{-125 + 1331}{3}-330\ &=\frac{1206}{3}-330\ &=402-330 = 72 \end{align*} ] [A = 54-72+\int{-11}^{-5}(x^{2}+10x + 25)dx-\int_{-11}^{-5}(3x + 33)dx=\int_{-11}^{-5}((3x + 33)-(x^{2}+10x + 25))dx=\int_{-11}^{-5}(-x^{2}-7x + 8)dx] [ \begin{align*} \int_{-11}^{-5}(-x^{2}-7x + 8)dx&=\left[-\frac{1}{3}x^{3}-\frac{7}{2}x^{2}+8x\right]_{-11}^{-5}\ &=\left(-\frac{1}{3}\times(-5)^{3}-\frac{7}{2}\times(-5)^{2}+8\times(-5)\right)-\left(-\frac{1}{3}\times(-11)^{3}-\frac{7}{2}\times(-11)^{2}+8\times(-11)\right)\ &=\left(\frac{125}{3}-\frac{175}{2}-40\right)-\left(\frac{1331}{3}-\frac{847}{2}-88\right)\ &=\frac{125}{3}-\frac{175}{2}-40-\frac{1331}{3}+\frac{847}{2}+88\ &=\left(\frac{125 - 1331}{3}\right)+\left(\frac{-175 + 847}{2}\right)+(88 - 40)\ &=\frac{-1206}{3}+\frac{672}{2}+48\ &=-402 + 336+48\ &=-18 \end{align*} ] Since area is non - negative, we take the absolute value. The area (A = 18).

Answer:

18