We turn now to the problem of finding area under a curve. Suppose we want to find the area between the curve f(x) = x² and the x-axis on the interval [0, 1], as shown to the right. There is a process for finding this area exactly (we'll get to that next), but to understand that process we're first going to start by approximating this area with rectangles. We're going to approximate the area shown above with four rectangles, using a right endpoint approximation, also called a right Riemann sum. 0 (a) Start by dividing the interval [0, 1] into four equal-width subintervals, each of which has a length of 1/4. The first subinterval is [0, 1]. Write the other three subintervals. (e) Now calculate the area of each of the four rectangles, and add them up. This value is called R4 (the "R" is for "right Riemann sum"), and the 4 means we used four rectangles. (b) Next, give the right endpoint of each subinterval a name of the form xn. For example, *₁ = . Now, assign the correct values to #2, #3, and x4. (c) Now, calculate f(x₁), f(x2), f(x3), and f(x4). (d) Next, draw in the four rectangles. Each rectangle has a width of 1/4 and a height of f(x1), f(x2), f(x3), and f(x4), The rectangles look as shown to the right: (f) The actual area under f(r) 1 2 on the interval [0 11 is exactly 1/3 #1 #2 23 24 How close is R. to

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From Part D onwards please. Part A, B and C was asked already. Thanks !
Math 151 Workshop 13 (Riemann sums)
We turn now to the problem of finding area under a curve. Suppose we want to
find the area between the curve f(x) = x² and the x-axis on the interval [0, 1],
as shown to the right. There is a process for finding this area exactly (we'll
get to that next), but to understand that process we're first going to start by
approximating this area with rectangles.
1. We're going to approximate the area shown above with four rectangles, using a right endpoint
approximation, also called a right Riemann sum.
(a) Start by dividing the interval [0, 1] into four equal-width subintervals, each of which has a
length of 1/4. The first subinterval is [0, 1]. Write the other three subintervals.
u
0
1
(b) Next, give the right endpoint of each subinterval a name of the form xn. For example,
₁. Now, assign the correct values to 22, 23, and 4.
1 =
(c) Now, calculate f(x₁), f(x2), f(x3), and f(x4).
(d) Next, draw in the four rectangles. Each rectangle has a
width of 1/4 and a height of f(x1), f(x2), f(x3), and f(x4),
The rectangles look as shown to the right:
(e) Now calculate the area of each of the four rectangles, and
add them up. This value is called R4 (the "R" is for "right
Riemann sum"), and the 4 means we used four rectangles.
(f) The actual area under f(x)
this area?
=
1
#1 22 23 24
x² on the interval [0,1] is exactly 1/3. How close is R4 to
(g) Now, redo the rectangle approximation with 8 rectangles (that is, calculate R$). How close
is Rs to the actual area?
(h) Geogebra has an online tool for calculating Riemann sums. Go to www.geogebra.org/m/h4P4cjsT
(type it carefully). To use it, you enter a function, an interval [a, b], and a number of rect-
angles n, so enter x² in the function field, a = 0 and b= 1. Then change the checkbox from
Left Riemann Sum to Right Riemann sum. In the field for n, enter 25, 50, and 100 and
record the values for R25, R50, and R100. Are these closer to the actual area of 1/3 than
the values you got for R4 and Rg?
nx
(i) If Rn is the approximation using n rectangles, what do you think the exact value of lim Rn?
Or, put another way, what do you think would happen if we could calculate the approxi-
mation with infinitely many rectangles?
Transcribed Image Text:Math 151 Workshop 13 (Riemann sums) We turn now to the problem of finding area under a curve. Suppose we want to find the area between the curve f(x) = x² and the x-axis on the interval [0, 1], as shown to the right. There is a process for finding this area exactly (we'll get to that next), but to understand that process we're first going to start by approximating this area with rectangles. 1. We're going to approximate the area shown above with four rectangles, using a right endpoint approximation, also called a right Riemann sum. (a) Start by dividing the interval [0, 1] into four equal-width subintervals, each of which has a length of 1/4. The first subinterval is [0, 1]. Write the other three subintervals. u 0 1 (b) Next, give the right endpoint of each subinterval a name of the form xn. For example, ₁. Now, assign the correct values to 22, 23, and 4. 1 = (c) Now, calculate f(x₁), f(x2), f(x3), and f(x4). (d) Next, draw in the four rectangles. Each rectangle has a width of 1/4 and a height of f(x1), f(x2), f(x3), and f(x4), The rectangles look as shown to the right: (e) Now calculate the area of each of the four rectangles, and add them up. This value is called R4 (the "R" is for "right Riemann sum"), and the 4 means we used four rectangles. (f) The actual area under f(x) this area? = 1 #1 22 23 24 x² on the interval [0,1] is exactly 1/3. How close is R4 to (g) Now, redo the rectangle approximation with 8 rectangles (that is, calculate R$). How close is Rs to the actual area? (h) Geogebra has an online tool for calculating Riemann sums. Go to www.geogebra.org/m/h4P4cjsT (type it carefully). To use it, you enter a function, an interval [a, b], and a number of rect- angles n, so enter x² in the function field, a = 0 and b= 1. Then change the checkbox from Left Riemann Sum to Right Riemann sum. In the field for n, enter 25, 50, and 100 and record the values for R25, R50, and R100. Are these closer to the actual area of 1/3 than the values you got for R4 and Rg? nx (i) If Rn is the approximation using n rectangles, what do you think the exact value of lim Rn? Or, put another way, what do you think would happen if we could calculate the approxi- mation with infinitely many rectangles?
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