question find the net signed area between the graph of the function f(x)=-5x/4 - 6 and the x - axis over the…

Explanation:

Step1: Recall the formula for net - signed area

The net - signed area (A) between the graph of (y = f(x)) and the (x) - axis over the interval ([a,b]) is given by (A=\int_{a}^{b}f(x)dx), where (a =-\frac{34}{5}), (b=\frac{11}{5}), and (f(x)=-\frac{5x}{4}-6).

Step2: Calculate the integral

We know that (\int\left(-\frac{5x}{4}-6\right)dx=-\frac{5}{4}\int xdx-6\int dx). Using the power rule (\int x^n dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (-\frac{5}{4}\times\frac{x^{2}}{2}-6x+C=-\frac{5x^{2}}{8}-6x+C).

Step3: Evaluate the definite integral

(\int_{-\frac{34}{5}}^{\frac{11}{5}}\left(-\frac{5x}{4}-6\right)dx=\left[-\frac{5x^{2}}{8}-6x\right]_{-\frac{34}{5}}^{\frac{11}{5}}) First, substitute (x = \frac{11}{5}): (-\frac{5}{8}\times\left(\frac{11}{5}\right)^{2}-6\times\frac{11}{5}=-\frac{5}{8}\times\frac{121}{25}-\frac{66}{5}=-\frac{121}{40}-\frac{66}{5}) (=-\frac{121}{40}-\frac{528}{40}=-\frac{121 + 528}{40}=-\frac{649}{40}) Then substitute (x=-\frac{34}{5}): (-\frac{5}{8}\times\left(-\frac{34}{5}\right)^{2}-6\times\left(-\frac{34}{5}\right)=-\frac{5}{8}\times\frac{1156}{25}+\frac{204}{5}) (=-\frac{1156}{40}+\frac{204}{5}=-\frac{1156}{40}+\frac{1632}{40}=\frac{-1156 + 1632}{40}=\frac{476}{40}) Now, (\left(-\frac{649}{40}\right)-\frac{476}{40}=\frac{-649 - 476}{40}=-\frac{1125}{40}=-\frac{225}{8})

Answer:

(-\frac{225}{8})