Left-Hand Sum Examples

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

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

To take a left-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 has size 2.

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

• Go to the left endpoint of the sub-interval (0).
• Go straight up until you hit the function.
  Figure out the y-value of the function where you hit it (f (0) = (0)² + 1 = 1).
• Make a rectangle whose base is the subinterval and whose height is the y-value you just found:

Finally, calculate the area of this rectangle:

(height) ⋅ (width) = 1 ⋅ 2 = 2

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

• Go to the left endpoint of the sub-interval (2).

• Go straight up until you hit the function.

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 sub-interval and whose height is the y-value you just found:

Finally, calculate the area of this rectangle:

(height) ⋅ (width) = 5 ⋅ 2 = 10

Adding the areas of the rectangles together, we see that the rectangles cover an area of size 12:

This is less than the area we're trying to find, because the two rectangles don't cover all of R. That's okay, because we're only estimating. In order to find a better estimate, we can use more rectangles. We start by dividing the original interval into more sub-intervals (each sub-interval will now be smaller). On each sub-interval we do the same stuff we did before.

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

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

The first sub-interval is [0,1]:

• Go to the left endpoint of the sub-interval (0).

• Go straight up until you hit the function and figure out the y-value of the function where you hit it (f ( 0 ) = (0)² + 1 = 1).

• Make a rectangle whose base is the subinterval and whose height is the y-value you just found:

Calculate the area of the rectangle:

(height) ⋅ (width) = 1 ⋅ 1 = 1

The second sub-interval is [1,2]:

• Go to the left 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:

Calculate the area of the rectangle:

(height) ⋅ (width) = 2 ⋅ 1 = 2

The third sub-interval is [2, 3]:

• Go to the left 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:

Calculate the area of the rectangle:

(height) ⋅ (width) = 5 ⋅ 1 = 5

The last sub-interval is [3, 4]:

• Go to the left 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:

Calculate the area of the rectangle:

(height) ⋅ (width) = 10 ⋅ 1 = 10

The area covered by these four rectangles is

1 + 2 + 5 + 10 = 18.

We're still not quite covering the whole area we want, but we're closer.

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

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

height ⋅ width = 4(2) = 8.

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

0 + 8 = 8.

To determine if this is an over-estimate or under-estimate, we need to compare the area that is part of S that we didn't cover (area we missed) to the area that isn't part of S but got covered anyway (extra area!).

The missed area is shaded in black, the extra area in pink:

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

• Let W be the region between the graph of f and the x-axis on the interval [-4, 0].

• Use a left-hand sum with 2 sub-intervals to estimate the area of Z.
• Use a left-hand sum with 4 sub-intervals to estimate the area of Z.

We divide the interval [-4, 0] into two sub-intervals of size 2, so each rectangle has width 2. On sub-interval [-4, -2] the rectangle has height g(0) = 0, so the area of this rectangle is also 0. On sub-interval [-2, 0] the rectangle has height g (-2) = 4, so the area of this rectangle is 4(2) = 8.

We estimate that the area of W is 0 + 8 = 8. 

• We divide the interval [-4, 0] into four sub-intervals of size 1, so each rectangle has width 1.
    
  On sub-interval [-4, -3] the height of the rectangle is g(-4) = 0 so the area of the rectangle is 0.
    
  On sub-interval [-3, -2] the height of the rectangle is g(-3) = 2 so the area of the rectangle is 2(1) = 2.

On sub-interval [-2, -1] the height of the rectangle is g(-2) = 4 so the area of the rectangle is 4.

On sub-interval [-1, 0] the height of the rectangle is g (-1) = 6 so the area of the rectangle is 6.

Adding the areas of these rectangles, we estimate that the area of Z is

0 + 2 + 4 + 6 = 12.

Let f (x) = x² + 2 and let R be the region between the graph of f and the x-axis on the interval [0, 8]. Use a left-hand sum with 4 sub-intervals to estimate the area of R.

We're going to go through the same process we've been using.

We cut the original interval into 4 sub-intervals of length 2.

On the sub-interval [0, 2] the height of the rectangle is
f(0) = 0² + 2 = 2 so the area of the rectangle is f(0)(2) = 2(2).

(Yes, we know two times two is four, but we don't want to do that just yet. Trust us.)

On the sub-interval [2, 4] the height of the rectangle is f(2) = 2² + 2 = 6 so the area of the rectangle is f (2)(2) = 6(2).

On the sub-interval [4, 6] the height of the rectangle is f(4) = 4² + 2 = 18 so the area of the rectangle is f (4)(2) = 18(2).

On the sub-interval [6, 8] the height of the rectangle is
f (6) = 6² + 2 = 38 so the area of the rectangle is f(6)(2) = 38(2).

Adding the areas of these rectangles, we find
f(0)(2) + f(2)(2) + f(4)(2) + f(6)(2) = 2(2) + 6(2) + 18(2) + 38(2).

We can use the distributive property to pull out the common factor 2 and rewrite this as
[f (0) + f (2) + f (4) + f (6)](2) = [2 + 6 + 18 + 38](2).

In other words, we take the value of the function at the left endpoint of each sub-interval and add these values. Then we multiply by the width of each sub-interval or rectangle.

We never use the value of the function at the right-most endpoint of the original interval.

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 left-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, 1.25, 1.5, and 1.75.

Here are the values of f we need:

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

Remember, we can also take a left-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 LHS(5). That is, use a left-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 the last:

Using the formula,

Use a left-hand sum to estimate the area between the graph of g and the x-axis on the interval [0,10].

We can see from the graph what the value of g is at x = 0, 2, 7, and 8. We have four sub-intervals of lengths 2, 5, 1, and 2:

On the sub-interval [0, 2] we have a rectangle with height g(0) = 1 and width 2, so its area is 2.

On the sub-interval [2, 7] we have a rectangle with height g(2) = 3 and width 5, so its area is 15.

On the sub-interval [7, 8] we have a rectangle with height g(7) = 0, so its area is 0.

On the sub-interval [8, 10] we have a rectangle with height g(8) = 5 and width 2, so its area is 10.

Adding the areas of the rectangles together, we estimate that the area between the graph of g and the x-axis on [0,10] is

2 + 15 + 0 + 10 = 27.