use an integral to find the area between y = cosx + 10 and y = ln(x - 2) for 4 ≤ x ≤ 6. use a calculator if…

Explanation:

Step1: Recall area - between - curves formula

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. For $4\leq x\leq6$, we need to determine which function is greater. Let $f(x)=\cos x + 10$ and $g(x)=\ln(x - 2)$. Since $\cos x\in[- 1,1]$, then $\cos x+10\in[9,11]$, and for $x\in[4,6]$, $\ln(x - 2)\in[\ln(2),\ln(4)]\approx[0.693,1.386]$. So, $f(x)\geq g(x)$ on the interval $[4,6]$, and the area formula becomes $A=\int_{4}^{6}[(\cos x + 10)-\ln(x - 2)]dx$.

Step2: Split the integral

We can split the integral into two parts: $A=\int_{4}^{6}\cos xdx+\int_{4}^{6}10dx-\int_{4}^{6}\ln(x - 2)dx$.

Integrate $\int_{4}^{6}\cos xdx$

The antiderivative of $\cos x$ is $\sin x$. Using the fundamental theorem of calculus, $\int_{4}^{6}\cos xdx=\sin(6)-\sin(4)$.

Integrate $\int_{4}^{6}10dx$

The antiderivative of $10$ is $10x$. So, $\int_{4}^{6}10dx=10x\big|_{4}^{6}=10\times(6 - 4)=20$.

Integrate $\int_{4}^{6}\ln(x - 2)dx$

Let $u=x - 2$, then $du=dx$. When $x = 4$, $u = 2$; when $x = 6$, $u = 4$. The integral becomes $\int_{2}^{4}\ln udu$. Using integration - by - parts with $v=\ln u$ and $dw=du$, then $dv=\frac{1}{u}du$ and $w = u$. By the integration - by - parts formula $\int vdw=vw-\int wdv$, we have $\int_{2}^{4}\ln udu=[u\ln u]2^4-\int{2}^{4}du=4\ln(4)-2\ln(2)-(4 - 2)=4\ln(4)-2\ln(2)-2$.

Step3: Combine the results

$A=\sin(6)-\sin(4)+20-(4\ln(4)-2\ln(2)-2)$. Using a calculator: $\sin(6)\approx - 0.279$, $\sin(4)\approx - 0.757$, $\ln(2)\approx0.693$, $\ln(4)\approx1.386$. $A=-0.279+0.757 + 20-(4\times1.386-2\times0.693-2)$ $A=-0.279+0.757+20-(5.544 - 1.386 - 2)$ $A=-0.279+0.757+20 - 2.158$ $A=18.32$

Answer:

$18.320$