00:01
In part a of the problem, we have to sketch the graph of the function f of x is equal to minus 2x minus 1 on the interval 0 up to 4.
00:27
Here is the sketch of the function we draw using geo -gevra.
00:36
Here in the graph, 0, 1, 2, 3 and 4 are the end points of the subinterverse.
00:52
Here 0, 1, 2 and 3 are the left end points of the subinterverse, where 1, 2, and 4 are the right end points of the subinterference where 1, 2, 3 and 4 are the right end points of the subinterval and midpoints of the subintervals are 0 .5, 1 .5, 2 .5 and 3.
01:39
In part b of this problem, we have to approximate the area using right endpoints, left endpoints and midpoints of the subinterval.
02:02
So first we calculate the net area using left endpoints.
02:10
For this, remunson using left endpoints is the net area using left end points is equal to submission kf round 1 to n f of x k times delta x here f of x k represents heights of the subinterials and delta x is the width of each subinterval since there are four subinteriors so definition k from 1 up to 4 f of x k times delta x this will give delta x times f of x1 plus half of x x plus f of x x plus f of x3 plus f of x4 here in this case delta x is equal to b minus a divided by n times the left end points of the subintervas are f of 0 plus f of 1 plus oph of 2 plus f of 3 and b is the upper limit of the interval which is 4 minus the lower limit is 0 divided by total sub intervals are 4 so this is equal to now we calculate of 0 using 0 in the given function.
05:15
We will get minus 1...