Right-Hand Sum Examples

Let R be the region between the graph f(x) = x² + 1 and the x-axis on the interval [0, 4]:

Use a right-hand sum with two sub-intervals to approximate the area of R.

To take a right-hand sum we first divide the interval in question into sub-intervals of equal size. Since we're looking at the interval [0, 4], each sub-interval will have size 2.

On the first sub-interval, [0,2], we do the following:

• Go to the right endpoint of the sub-interval.
• Go straight up until we hit the function. Then figure out the y-value of the function where we hit it (f (2) = (2)² + 1 = 5).
• Make a rectangle whose base is the subinterval and whose height is the y-value we just found, and calculate the area of the rectangle:

The area of this rectangle is

(height) ⋅ (width) = 5 ⋅ 2 = 10

Now we do the same stuff on the second sub-interval, [2, 4].

• Go to the right endpoint of the sub-interval (4).
• Go straight up until we hit the function. Then figure out the y-value of the function where we hit it (f (4) = (4)² + 1 = 17).
• Make a rectangle whose base is the subinterval and whose height is the y-value you just found, and calculate the area of the rectangle:

The area of this rectangle is

(height) ⋅ (width) = 17 ⋅ 2 = 34

Adding the areas of the rectangles together, we see that the rectangles cover an area of size 10 + 34 = 44:

This is an overestimate for the actual area of R, since the rectangles cover R and then some.

We can get a better estimate by dividing the interval [0, 4] into more sub-intervals and using more rectangles.

Let R be the region between the graph f(x) = x² + 1 and the x-axis on the interval [0, 4]. Use a Right-Hand Sum with 4 sub-intervals to estimate the area of R.

Since we're looking at the interval [0, 4], each sub-interval will have length 1.

First sub-interval, [0,1]:

• Go to the right endpoint of the sub-interval (1).
• Go straight up until you hit the function and figure out the y-value of the function where you hit it (f (1) = (1)² + 1 = 2).
• Make a rectangle whose base is the subinterval and whose height is the y-value you just found, and calculate the area of the rectangle:

The area of this rectangle is
(height) ⋅ (width) = 2 ⋅ 1 = 2

Second sub-interval, [1,2]:

• Go to the right endpoint of the sub-interval (2).
• Go straight up until you hit the function and figure out the y-value of the function where you hit it (f(2) = (2)² + 1 = 5).
• Make a rectangle whose base is the subinterval and whose height is the y-value you just found, and calculate the area of the rectangle. The area of this rectangle is (height) ⋅ (width) = 5 ⋅ 1 = 5

Third sub-interval, [2, 3]:

• Go to the right endpoint of the sub-interval (3).
• Go straight up until you hit the function and figure out the y-value of the function where you hit it (f(3) = (3)² + 1 = 10).
• Make a rectangle whose base is the subinterval and whose height is the y-value you just found, and calculate the area of the rectangle:

The area of this rectangle is

(height) ⋅ (width) = 10 ⋅ 1 = 10

Fourth sub-interval, [3, 4]:

• Go to the right endpoint of the sub-interval (4).
• Go straight up until you hit the function and figure out the y-value of the function where you hit it (f(4) = (4)² + 1 = 17).
• Make a rectangle whose base is the subinterval and whose height is the y-value you just found, and calculate the area of the rectangle:

The area of this rectangle is

(height) ⋅ (width) = 17 ⋅ 1 = 17

The area covered by these four rectangles is

2 + 5 + 10 + 17 = 34.

We're still covering more than just R with the rectangles, but we're getting closer.

Let S be the region between the graph of g and the x-axis on the interval [0, 4].

Use a right-hand sum with 2 sub-intervals to estimate the area of S. Is this an under-estimate or an over-estimate?

Dividing the interval [0, 4] into 2 sub-intervals gives us sub-intervals of length 2.

On the sub-interval [0, 2] the height of the rectangle is g(2), which we can see from the graph is 2. The area of this rectangle is 2(2) = 4.

On the sub-interval [2, 4] the height of the rectangle is g(4), which we can see from the graph is 3. The area of this rectangle is 3(2) = 6.

Adding the areas of the rectangles together, we estimate the area of S is

4 + 6 = 10.

To determine if this is an over-estimate or under-estimate, we need to compare the area we counted but shouldn't have (shaded in pink) with the area we didn't count but should have (shaded in black):

There's more pink shaded area than black shaded area in that picture. This means the area we counted extra is bigger than the area we missed, so we have an overestimate.

Let f(x) = 2 + x² and let R be the region between the graph of f and the x-axis on the interval [0, 8].

Use a right-hand sum with 4 sub-intervals to estimate the area of R.

We cut the original interval into 4 sub-intervals of length 2. On each sub-interval, the height of the rectangle is given by the value of the function at the right-most endpoint of that sub-interval.

On the sub-interval [0, 2] the height of the rectangle is f(2) = 6.

On the sub-interval [2, 4] the height of the rectangle is f(4) = 18.

On the sub-interval [4, 6] the height of the rectangle is f(6) = 38.

On the sub-interval [6, 8] the height of the rectangle is f(8) = 66.

We could find the area covered by all 4 rectangles by finding the area of each rectangle (multiplying each height by 2) and adding them:

f(2)(2) + f(4)(2) + f(6)(2) + f(18)(2)

The distributive property says that we'll get the same answer if we first add all the heights and then multiply their sum by 2:

[f(2) + f(4) +f(6) + f(8)](2).

We think this second option is a lot easier. You only need to hit the multiplication button once!

The final answer is

[6 + 18 + 38 + 66](2) = 256.

To summarize: to quickly find a RHS, take the value of the function at the right endpoint of each sub-interval and find the sum of these values. Then multiply the sum by the width of a sub-interval/rectangle. The value of the function at the left-most endpoint of the original interval will never be used.

Let f (x) = 4x and let R be the region between the graph of f and the x-axis on the interval [1, 2]. Use a right-hand sum with 4 sub-intervals to estimate the area of R.

Dividing the interval [1, 2] into 4 equally-sized sub-intervals gets us sub-intervals of length 0.25. We'll be using the values of f at x = 1.25, 1.5, 1.75, and 2.

Here are the values of f we need:

We add the values of f and multiply by the width of a sub-interval:

[f (1.25) + f (1.5) + f (1.75) + f (2)](width) = [5 + 6 + 7 + 8](.25) = 6.5

Remember, we can also take a right-hand sum when we're given a table or graph for the function instead of a formula.

Let f (x) = 2x on [2, 10]. Find RHS(5). That is, use a right-hand sum with 5 sub-intervals to estimate the area between the graph of f and the x-axis on [2, 10].

We have a = 2, b = 10, and we want 5 sub-intervals. The width of a sub-interval is

.

Then our endpoints are

We need the value of f at every endpoint but x₀.

Using the formula,

RHS(5) = [f (x₁) + f (x₂) + f (x₃) + f (x₄) + f (x₅)]Δx

= [f(3.6) + f(5.2) + f(6.8) + f(8.4) + f(10)](1.6)

= [7.2 + 10.4 + 13.6 + 16.8 + 20](1.6)

= 108.8

The two important things to remember here:

• For a right-hand sum, we skip x₀ but go all the way to x_n:

• After adding up the values of f at the appropriate endpoints, remember to multiply by the width of a sub-interval: