current attempt in progress find the area of the region between y = sinx + 5 and y = 0.3 for 5 ≤ x ≤ 11…

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 $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, $f(x)=\sin x + 5$, $g(x)=0.3$, $a = 5$, and $b = 11$. Since $\sin x+5>0.3$ for all real $x$ (because the range of $\sin x$ is $[- 1,1]$, so $\sin x+5\geq - 1 + 5=4>0.3$), the area formula becomes $A=\int_{5}^{11}[(\sin x + 5)-0.3]dx=\int_{5}^{11}(\sin x + 4.7)dx$.

Step2: Integrate term - by - term

We know that $\int(\sin x+4.7)dx=-\cos x + 4.7x+C$.

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus $\int_{a}^{b}F'(x)dx=F(b)-F(a)$, where $F(x)=-\cos x + 4.7x$. Then $A=[-\cos(11)+4.7\times11]-[-\cos(5)+4.7\times5]$. We know that $\cos(11)\approx0.999$ and $\cos(5)\approx0.284$. [ \begin{align*} A&=(-0.999 + 51.7)-(-0.284+23.5)\ &=(-0.999 + 51.7)+0.284 - 23.5\ &=(-0.999+0.284)+(51.7 - 23.5)\ &=-0.715 + 28.2\ &=28.889 \end{align*} ]

Answer:

$28.889$