find the area of the region between y = sinx + 5 and y = 0.3 for 5 ≤ x ≤ 11. round your answer to three…

Explanation:

Step1: Set up the integral

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 $b=11$. So $A=\int_{5}^{11}[(\sin x + 5)-0.3]dx=\int_{5}^{11}(\sin x + 4.7)dx$.

Step2: Integrate term - by - term

We know that $\int\sin xdx=-\cos x+C$ and $\int kdx=kx + C$ (where $k$ is a constant). So $\int_{5}^{11}(\sin x + 4.7)dx=\left[-\cos x+4.7x\right]_{5}^{11}$.

Step3: Evaluate the definite integral

[ \begin{align*} \left(-\cos(11)+4.7\times11\right)-\left(-\cos(5)+4.7\times5\right)&=-\cos(11)+51.7+\cos(5)-23.5\ &=28.2+\cos(5)-\cos(11) \end{align*} ] Using a calculator, $\cos(5)\approx0.284$, $\cos(11)\approx - 0.999$. [ \begin{align*} A&=28.2 + 0.284-(-0.999)\ &=28.2+0.284 + 0.999\ &=29.483 \end{align*} ]

Answer:

$29.483$