find the area under the graph of f over the interval 1,5.\n f(x)=\begin{cases}x + 8, &\text{for…

Explanation:

Step1: Split the integral

Since the function is piece - wise, split the integral over the interval $[1,5]$ into two parts: $\int_{1}^{5}f(x)dx=\int_{1}^{4}f(x)dx+\int_{4}^{5}f(x)dx$. For $x\in[1,4]$, $f(x)=x + 8$ and for $x\in(4,5]$, $f(x)=16-\frac{1}{2}x$.

Step2: Calculate $\int_{1}^{4}(x + 8)dx$

Use the power rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$ and $\int kdx=kx + C$ ($k$ is a constant). $\int_{1}^{4}(x + 8)dx=\int_{1}^{4}x dx+\int_{1}^{4}8dx=\left[\frac{x^{2}}{2}\right]{1}^{4}+8\left[x\right]{1}^{4}$. $=\left(\frac{4^{2}}{2}-\frac{1^{2}}{2}\right)+8(4 - 1)=\left(\frac{16}{2}-\frac{1}{2}\right)+24=\frac{15}{2}+24=\frac{15 + 48}{2}=\frac{63}{2}$.

Step3: Calculate $\int_{4}^{5}(16-\frac{1}{2}x)dx$

$\int_{4}^{5}(16-\frac{1}{2}x)dx=\int_{4}^{5}16dx-\frac{1}{2}\int_{4}^{5}x dx$. $=16\left[x\right]{4}^{5}-\frac{1}{2}\left[\frac{x^{2}}{2}\right]{4}^{5}=16(5 - 4)-\frac{1}{4}(5^{2}-4^{2})$. $=16-\frac{1}{4}(25 - 16)=16-\frac{9}{4}=\frac{64 - 9}{4}=\frac{55}{4}$.

Step4: Sum the two results

$\int_{1}^{5}f(x)dx=\frac{63}{2}+\frac{55}{4}=\frac{126+55}{4}=\frac{181}{4}$.

Answer:

$\frac{181}{4}$