the graph of h is comprised of a semi - circle and line segments. evaluate the definite integral of ∫₆¹³…

Explanation:

Step1: Split the integral based on function - parts

The integral $\int_{6}^{13}h(x)dx$ can be split into two parts based on the behavior of the function $h(x)$ from $x = 6$ to $x=9$ and from $x = 9$ to $x = 13$. $\int_{6}^{13}h(x)dx=\int_{6}^{9}h(x)dx+\int_{9}^{13}h(x)dx$

Step2: Evaluate $\int_{6}^{9}h(x)dx$

From $x = 6$ to $x = 9$, $h(x)=2$. Using the integral formula $\int_{a}^{b}c dx=c(b - a)$ where $c$ is a constant, we have $\int_{6}^{9}2dx=2\times(9 - 6)=6$.

Step3: Evaluate $\int_{9}^{13}h(x)dx$

The function $h(x)$ from $x = 9$ to $x = 13$ is a line - segment. The line passes through $(9,2)$ and $(11,0)$. The equation of the line using the two - point form $y - y_1=\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$ is $y-2=\frac{0 - 2}{11 - 9}(x - 9)=-(x - 9)$, so $y=-x + 11$. The integral $\int_{9}^{13}h(x)dx=\int_{9}^{11}(-x + 11)dx+\int_{11}^{13}(-x + 11)dx$. First, $\int_{9}^{11}(-x + 11)dx=\left[-\frac{x^{2}}{2}+11x\right]{9}^{11}=\left(-\frac{11^{2}}{2}+11\times11\right)-\left(-\frac{9^{2}}{2}+11\times9\right)=\left(-\frac{121}{2}+121\right)-\left(-\frac{81}{2}+99\right)=\left(\frac{-121 + 242}{2}\right)-\left(\frac{-81+198}{2}\right)=\frac{121}{2}-\frac{117}{2}=2$. Second, for $x\in[11,13]$, $y=-x + 11$, and $\int{11}^{13}(-x + 11)dx=\left[-\frac{x^{2}}{2}+11x\right]{11}^{13}=\left(-\frac{13^{2}}{2}+11\times13\right)-\left(-\frac{11^{2}}{2}+11\times11\right)=\left(-\frac{169}{2}+143\right)-\left(-\frac{121}{2}+121\right)=\left(\frac{-169 + 286}{2}\right)-\left(\frac{-121 + 242}{2}\right)=\frac{117}{2}-\frac{121}{2}=-2$. So, $\int{9}^{13}h(x)dx=2+( - 2)=0$.

Step4: Combine the results

$\int_{6}^{13}h(x)dx=\int_{6}^{9}h(x)dx+\int_{9}^{13}h(x)dx=6 + 0=6$.

Answer:

$6$