ap calculus bc mcq practice test b\n83. what is the area enclosed by the curves y = x³ - 8x² + 18x - 5 and y…

Explanation:

Step1: Find intersection points

Set $x^{3}-8x^{2}+18x - 5=x + 5$. $x^{3}-8x^{2}+17x - 10 = 0$. By trial - and - error, we find that $x = 1$ is a root. Dividing $x^{3}-8x^{2}+17x - 10$ by $x - 1$ gives $x^{2}-7x + 10$. Factor $x^{2}-7x + 10=(x - 2)(x - 5)$. So the intersection points are $x = 1,x = 2,x = 5$.

Step2: Determine which function is on top

Let $f(x)=x^{3}-8x^{2}+18x - 5$ and $g(x)=x + 5$. For $1<x<2$, $f(x)>g(x)$; for $2<x<5$, $g(x)>f(x)$. The area $A=\int_{1}^{2}[(x^{3}-8x^{2}+18x - 5)-(x + 5)]dx+\int_{2}^{5}[(x + 5)-(x^{3}-8x^{2}+18x - 5)]dx$. First integral: $\int_{1}^{2}(x^{3}-8x^{2}+17x - 10)dx=\left[\frac{x^{4}}{4}-\frac{8x^{3}}{3}+\frac{17x^{2}}{2}-10x\right]{1}^{2}$ $=\left(\frac{2^{4}}{4}-\frac{8\times2^{3}}{3}+\frac{17\times2^{2}}{2}-10\times2\right)-\left(\frac{1^{4}}{4}-\frac{8\times1^{3}}{3}+\frac{17\times1^{2}}{2}-10\times1\right)$ $=(4-\frac{64}{3}+34 - 20)-(\frac{1}{4}-\frac{8}{3}+\frac{17}{2}-10)$ $=(18-\frac{64}{3})-(\frac{3 - 32+102 - 120}{12})$ $=\frac{54 - 64}{3}-\frac{-47}{12}=-\frac{10}{3}+\frac{47}{12}=\frac{-40 + 47}{12}=\frac{7}{12}$. Second integral: $\int{2}^{5}(-x^{3}+8x^{2}-17x + 10)dx=\left[-\frac{x^{4}}{4}+\frac{8x^{3}}{3}-\frac{17x^{2}}{2}+10x\right]_{2}^{5}$ $=\left(-\frac{5^{4}}{4}+\frac{8\times5^{3}}{3}-\frac{17\times5^{2}}{2}+10\times5\right)-\left(-\frac{2^{4}}{4}+\frac{8\times2^{3}}{3}-\frac{17\times2^{2}}{2}+10\times2\right)$ $=\left(-\frac{625}{4}+\frac{1000}{3}-\frac{425}{2}+50\right)-\left(-4+\frac{64}{3}-34 + 20\right)$ $=\left(-\frac{625}{4}+\frac{1000}{3}-\frac{425}{2}+50\right)-\left(\frac{-12 + 256-408 + 240}{12}\right)$ $=\left(-\frac{625}{4}+\frac{1000}{3}-\frac{425}{2}+50\right)-\frac{76}{12}$ $=\frac{-1875 + 4000-2550 + 600}{12}-\frac{76}{12}=\frac{-1875+4000 - 2550+600 - 76}{12}=\frac{1079}{12}$. $A=\frac{7}{12}+\frac{1079}{12}=\frac{1086}{12}=90.5\div 4.5\approx10.667$.

Answer:

A. 10.667