Use finite approximations to estimate the area under the...

Use finite approximations to estimate the area under the graph of the function f(x)= 8 - x^2 + 2x between x = -2 and x= 4 for each of the following cases. a. Using a lower sum with two rectangles of equal width b. Using a lower sum with four rectangles of equal width c. Using an upper sum with two rectangles of equal width d. Using an upper sum with four rectangles of equal width a. The area under the graph of f(x) using a lower sum with two rectangles of equal width is ___. Etc.

Transcript

00:01 In this problem we are provided with the function f of x which equals to 8 minus x squared plus 2 times x further we are given that x equals to negative 2 and x equals to 4.

00:18 We are asked to find out the area under the graph.

00:21 In subpart a we are asked to make use of the lower sum where we consider two rectangles which are of equal length.

00:34 So first let us determine the value of del x.

00:37 Del x equals to b minus a divided by n.

00:42 And here we have four minus of minus which becomes plus two divided by n which is two since we require two rectangles.

00:52 So we get 6 over 2 which equals to 3.

00:56 So now let us consider the intervals.

01:00 The intervals would be negative 2 to 1.

01:03 1 to 4 and the values that is f of x for the left end points that is f of negative 2 equals to 0 and f of 1 equals to 9 now we can substitute this in the formula we have l2 equals to del x times f of x not plus f of x 1 so substituting the values we have three times f of negative 2 which is 0 plus 9 and this equals to 3 times 9 which is 27.

01:48 So therefore this is the required answer for subpart a.

01:56 Next in subpart b we are again asked to find out the lower sum and here we are asked to consider four rectangles which are all of equal length.

02:10 So in this case we have del x equals to 4 plus 2 divided by 4 which equals to sorry which equals to 6 divided by 4 which is 1 .5.

02:23 So now let us write down the possible intervals.

02:29 So we have the intervals to be negative 2 up to negative 0 .5 negative 0 .5 up to 1 1 up to 2 .5 and 2 .5 up to 4.

02:47 So now again we calculate the value of the function at the left end points.

02:52 We have f of negative 2 to be 0.

02:54 F of negative 0 .5 equals to 6 .75.

02:59 F of 1 equals to 9 and f of 2 .5 equals to 6 .75...

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