the graph of a function g is shown.\nestimate $int_{-2}^{4}g(x)dx$ with six subintervals using the…

Explanation:

Step1: Calculate the width of sub - intervals

The interval is $[a,b]=[-2,4]$ and $n = 6$. The width of each sub - interval $\Delta x=\frac{b - a}{n}=\frac{4-(-2)}{6}=1$. The sub - intervals are $[-2,-1],[-1,0],[0,1],[1,2],[2,3],[3,4]$.

Step2: Estimate using right - endpoints

The right - endpoints are $x_1=-1,x_2 = 0,x_3=1,x_4=2,x_5=3,x_6=4$. We estimate $\int_{-2}^{4}g(x)dx\approx\sum_{i = 1}^{6}g(x_i)\Delta x$. From the graph, $g(-1)\approx1,g(0)\approx0,g(1)\approx - 1,g(2)\approx1,g(3)\approx - 1,g(4)\approx1$. $\int_{-2}^{4}g(x)dx\approx(1 + 0-1 + 1-1 + 1)\times1=1$.

Step3: Estimate using left - endpoints

The left - endpoints are $x_0=-2,x_1=-1,x_2 = 0,x_3=1,x_4=2,x_5=3$. We estimate $\int_{-2}^{4}g(x)dx\approx\sum_{i = 0}^{5}g(x_i)\Delta x$. From the graph, $g(-2)\approx0,g(-1)\approx1,g(0)\approx0,g(1)\approx - 1,g(2)\approx1,g(3)\approx - 1$. $\int_{-2}^{4}g(x)dx\approx(0 + 1+0 - 1+1 - 1)\times1=0$.

Step4: Estimate using midpoints

The midpoints are $x_1=-1.5,x_2=-0.5,x_3 = 0.5,x_4=1.5,x_5=2.5,x_6=3.5$. We estimate $\int_{-2}^{4}g(x)dx\approx\sum_{i = 1}^{6}g(x_i)\Delta x$. From the graph, $g(-1.5)\approx1,g(-0.5)\approx1,g(0.5)\approx - 1,g(1.5)\approx0,g(2.5)\approx0,g(3.5)\approx0$. $\int_{-2}^{4}g(x)dx\approx(1 + 1-1+0 + 0+0)\times1=1$.

Answer:

(a) 1 (b) 0 (c) 1