area between curves that intersect at more than two points (calculator - active)\ngoogle classroom microsoft…

area between curves that intersect at more than two points (calculator - active)\ngoogle classroom microsoft teams\nlet f(x)=-18/x and g(x)=x² + 10x + 27.\nfind the sum of the areas enclosed by the graphs of f and g between x=-6 and x=-1.\nuse a graphing calculator and round your answer to three decimal places.

area between curves that intersect at more than two points (calculator - active)\ngoogle classroom microsoft teams\nlet f(x)=-18/x and g(x)=x² + 10x + 27.\nfind the sum of the areas enclosed by the graphs of f and g between x=-6 and x=-1.\nuse a graphing calculator and round your answer to three decimal places.

Answer

Explanation:

Step1: Determine which function is on top

Use a graphing - calculator to graph $y = f(x)=-\frac{18}{x}$ and $y = g(x)=x^{2}+10x + 27$ on the interval $[-6,-1]$. Observe the relative positions of the two curves on the given interval.

Step2: Set up the integral for the area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ on the interval $[a,b]$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. On the interval $[-6,-1]$, we need to split the interval into sub - intervals where $f(x)\geq g(x)$ and $g(x)\geq f(x)$. After graphing, we set up the integral as $A=\int_{-6}^{-3}\left(-\frac{18}{x}-(x^{2}+10x + 27)\right)dx+\int_{-3}^{-1}\left((x^{2}+10x + 27)+\frac{18}{x}\right)dx$.

Step3: Evaluate the integrals

  1. First integral:
    • $\int_{-6}^{-3}\left(-\frac{18}{x}-x^{2}-10x - 27\right)dx=- 18\int_{-6}^{-3}\frac{1}{x}dx-\int_{-6}^{-3}x^{2}dx-10\int_{-6}^{-3}xdx-27\int_{-6}^{-3}dx$.
    • $-18\int_{-6}^{-3}\frac{1}{x}dx=-18[\ln|x|]_{-6}^{-3}=-18(\ln3-\ln6)=18\ln2$.
    • $-\int_{-6}^{-3}x^{2}dx=-\left[\frac{x^{3}}{3}\right]_{-6}^{-3}=-\left(\frac{(-3)^{3}}{3}-\frac{(-6)^{3}}{3}\right)=-( - 9 + 72)=-63$.
    • $-10\int_{-6}^{-3}xdx=-10\left[\frac{x^{2}}{2}\right]_{-6}^{-3}=-10\left(\frac{9}{2}-\frac{36}{2}\right)=135$.
    • $-27\int_{-6}^{-3}dx=-27(-3 + 6)=-81$.
    • The value of the first integral is $18\ln2-63 + 135-81=18\ln2 - 9$.
  2. Second integral:
    • $\int_{-3}^{-1}\left(x^{2}+10x + 27+\frac{18}{x}\right)dx=\int_{-3}^{-1}x^{2}dx+10\int_{-3}^{-1}xdx+27\int_{-3}^{-1}dx+18\int_{-3}^{-1}\frac{1}{x}dx$.
    • $\int_{-3}^{-1}x^{2}dx=\left[\frac{x^{3}}{3}\right]_{-3}^{-1}=\frac{-1}{3}+\frac{27}{3}=\frac{26}{3}$.
    • $10\int_{-3}^{-1}xdx=10\left[\frac{x^{2}}{2}\right]_{-3}^{-1}=10\left(\frac{1}{2}-\frac{9}{2}\right)=-40$.
    • $27\int_{-3}^{-1}dx=27(-1 + 3)=54$.
    • $18\int_{-3}^{-1}\frac{1}{x}dx=18[\ln|x|]_{-3}^{-1}=18(\ln1-\ln3)=-18\ln3$.
    • The value of the second integral is $\frac{26}{3}-40 + 54-18\ln3=\frac{26}{3}+14-18\ln3=\frac{26 + 42}{3}-18\ln3=\frac{68}{3}-18\ln3$.
  3. Then $A=(18\ln2 - 9)+\left(\frac{68}{3}-18\ln3\right)$.
    • Calculate $18\ln2\approx18\times0.693147 = 12.476646$, $\frac{68}{3}\approx22.666667$, $18\ln3\approx18\times1.098612 = 19.775016$, $9$.
    • $A\approx12.477-9 + 22.667-19.775$.
    • $A\approx6.369$.

Answer:

$6.369$