question find the total area between the graph of the function f(x)=|x - 1|-1, graphed below, and the x…
question find the total area between the graph of the function f(x)=|x - 1|-1, graphed below, and the x - axis over the interval -5,3.
Answer
Explanation:
Step1: Split the absolute - value function
The function (y = |x - 1|-1) can be written as a piece - wise function: (y=\begin{cases}(x - 1)-1=x - 2, &x\geq1\-(x - 1)-1=-x, &x<1\end{cases}). We need to split the integral over the interval ([-5,3]) at (x = 1).
Step2: Calculate the integral over sub - intervals
We calculate the integral (\int_{-5}^{3}| |x - 1|-1|dx=\int_{-5}^{0}(-(-x))dx+\int_{0}^{1}-(-x)dx+\int_{1}^{3}(x - 2)dx). First integral: (\int_{-5}^{0}xdx=\left[\frac{x^{2}}{2}\right]{-5}^{0}=0-\frac{(-5)^{2}}{2}=-\frac{25}{2}), but we take the absolute value (\left|\int{-5}^{0}xdx\right|=\frac{25}{2}). Second integral: (\int_{0}^{1}xdx=\left[\frac{x^{2}}{2}\right]{0}^{1}=\frac{1}{2}). Third integral: (\int{1}^{3}(x - 2)dx=\left[\frac{x^{2}}{2}-2x\right]_{1}^{3}=(\frac{3^{2}}{2}-2\times3)-(\frac{1^{2}}{2}-2\times1)=(\frac{9}{2}-6)-(\frac{1}{2}-2)=\frac{9 - 12}{2}-\frac{1 - 4}{2}=-\frac{3}{2}+\frac{3}{2}=0).
Step3: Sum up the absolute values of the integrals
The total area (A=\frac{25}{2}+\frac{1}{2}+0 = 13).
Answer:
13