Use upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width). y=`sqrt(x)` +5 Find the Upper Sum and the Lower Sum Is it simply substituting...
Use upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width).
Find the Upper Sum and the Lower Sum
Is it simply substituting the x values with each 0.25 interval? I'm just having trouble finding which one is the lower sum and which one is the upper sum
The upper sum and lower sum refer to the sums of areas of the rectangles resulting from breaking up the interval into smaller segments.
The lower sum is the sum of the areas of rectangles UNDER the curve (thus, this approximation will be slightly smaller than the actual area under the curve.) In this case, the first rectangle has vertices
(0,0), (0.25, 0), (0, 5) and (0.25, 5). The area of this rectangle is 0.25*5 = 1.25
The second rectangle is next to the first one and has vertices
(0.25, 0), (0.5, 0), (0.25, 5.5) and (0.5, 5.5). (5.5 is the value of the function y = f(0.25)). The area of this rectangle is 0.25*5.5 = 1.375
As you can see, the area of each rectangle can then be found as `0.25*f(x_i)`
where `x_i` are the values of x starting from 0 and increasing by 0.25. There are 8 rectangles all together. Their areas will be
0.25*f(0.5) = 0.25*5.71 = 1.43
0.25*f(0.75) = 1.47
0.25*f(1) = 0.25*6 = 1.5
0.25*f(1.25) = 1.53
Finally, the area of the last rectangle will be 0.25*f(1.75) = 1.58
So the lower sum will be the sum of all these eight areas, which will be approximately 11.695.
The upper sum is the sum of the areas of all the rectangles slightly higher than the curve, so this approximation will be greater than the actual area under the curve. The first such rectangle has vertices
(0, 0), (0, 5.5), (0.25, 0), (0.25, 5.5) and its area is 0.25*f(0.25) = 0.25*5.5 = 1.375.
So, the upper sum can be expressed as
0.25*(f(0.25) + f(0.5) + ...f(2)) - the sum of eight rectangles with width 0.25 and height `f(x_i)` , where `x_i` are the values of x starting with 0.25 increasing by 0.25 until x = 2.
This sum will be approximately 12.05.
So the actual area under the curve will be somewhere between lower sum (11.695) and upper sum (12.05). This can be confirmed by finding the area using the integral
` ` `int_0 ^2 (sqrt(x) + 5) = [2/3 sqrt(x^3) + 5x] |_0 ^2 = 2/3 (sqrt(8)) + 10 =11.89 `
This number is higher than the lower sum but smaller than the upper sum.