The Area of a Region under a Curve and
Riemann Sum
The Area of a Region under a Curve
The Area of a Region
Definition of the Area of a Region
•
Let f be continuous and nonnegative on the interval 𝑎, 𝑏 .
The area of the region bounded by the graph of f, the x-axis,
and the vertical lines x=a and x=b is
𝑛
𝐴𝑟𝑒𝑎 = lim
𝑛→∞
𝑓 𝑐1 ∆𝑥,
𝑥𝑖−1 ≤ 𝑐𝑖 ≤ 𝑥𝑖
𝑖=1
𝑤ℎ𝑒𝑟𝑒∆𝑥 = (𝑏 − 𝑎)/𝑛
The graph above shows the
area under the curve of the
continuous function, f(x) . The
interval, 𝑎, 𝑏 , represents the
vertical bounds of the function.
The region has to be bounded
by the x-axis all the time.
Now, what happens if the curve is found below the x-axis or passes through
above and below the x-axis?
These two graphs are examples of
functions’ curves that are not
completely lying above the horizontal
axis, so when this happens, focus on
finding the region that is bounded by
the horizontal axis.
We can find the actual value of the area found under the curve by evaluating
the integral of the function at the bounds of its interval.
The Area of a Region
𝑏
𝐴𝑟𝑒𝑎 =
𝑓 𝑥 𝑑𝑥
𝑎
𝐴𝑟𝑒𝑎 = 𝐹 𝑏 − 𝐹(𝑎)
Keep in mind that 𝐹 𝑥 represents the antiderivative of 𝑓 𝑥 . This
means that when we want to find the area below 𝑓 𝑥 ′ 𝑠 curve and
bounded by 𝑥 = 𝑎 and 𝑥 = 𝑏 as well as the 𝑥 − 𝑎𝑥𝑖𝑠, simply
evaluate the 𝑓 𝑥 ′ 𝑠 definite integral for the interval 𝑎, 𝑏
How to find the area of a region under a curve?
Step 1: Graph f(x)’s curve and sketch the bounded region.
This step can be skipped when you’re confident with your skills already
Step 2: Set the boundaries for the region at x=a and x=b.
Step 3: Set up the definite integral. Separate the definite integrals
found above and below the x-axis.
Step 4: Evaluate the definite integral. Take the absolute value if
the region is found below the x-axis.
Case 1: When the area below the function’s curve is located entirely above the horizontal axis
•
Set up the definite integral expression.
•
Apply essential properties and antiderivative formula to find
the function’s antiderivative.
•
Evaluate the antiderivative at x = a and x = b then subtract
the results.
Case 2: When the area below the function’s curve is located entirely below the horizontal axis.
•
Set up the definite integral expression.
•
Apply essential properties and antiderivative formula to find
the function’s antiderivative.
•
Evaluate the antiderivative at x = a and x = b then subtract
the results.
•
Take the absolute value of the resulting expression.
Case 3: When the area below the function’s curve is located entirely below the horizontal axis.
•
.
•
•
Identify the intervals where the area is found below and
above the x-axis
For the definite integrals representing the area below the xaxis, enclose them with an absolute value.
Apply the same steps as with Case 1 then add the resulting
values to find the total area.
Example 1: Find the area bounded by the curve of 𝑓 𝑥 = 4 − 𝑥 2 from 𝑥 = −2 to 𝑥 = 2
Solution:
Since the graph confirms that the entire
region we need to account for is located
above the x-axis, we simply evaluate the
definite integral of f(x) from 𝑥 =
− 2 to 𝑥 = 2.
Example 1 Solution cont.
Apply the integral properties we’ve learned in the past to evaluate
this expression. Once we have the antiderivative of f(x), evaluate it
from 𝑥 = −2 to 𝑥 = 2.
(4 − 𝑥 2 ) 𝑑𝑥 =
𝐴𝑟𝑒𝑎 = 4𝑥 −
4 𝑑𝑥 −
2
3
𝑥
3
−2
2+1
𝑥
𝑥 2 𝑑𝑥 = 4𝑥 −
+𝐶
2+1
𝑥3
= 4𝑥 −
+𝐶
3
23
−2
= 4(2) −
− 4 −2 −
3
3
3
𝟑𝟐
=
𝟑
Example 2: A great example for the second case is by finding the
area bounded by the curve of 𝑔 𝑥 = 𝑥 2 − 9 from 𝑥 = −3 to 𝑥 = 3
Solution:
From this, we can see that the entire region
bounded by the curve, 𝑥 = −3, 𝑥 = 3, and
the horizontal axis is found below the x-axis.
This means that after evaluating the definite
integral, we take the absolute value of the
result to find the area under the curve
Example 2 Solution cont.
𝐴𝑟𝑒𝑎 =
3
−3
𝑥 2 − 9 𝑑𝑥
Find the antiderivative of g(x) then evaluate the resulting expression
at the bounds: 𝑥 = −3 𝑎𝑛𝑑 𝑥 = 3
(𝑥 2 − 9) 𝑑𝑥 =
𝑥 2 𝑑𝑥 −
9 𝑑𝑥
𝑥 2+1
=
− 9𝑥 + 𝐶
2+1
𝑥3
=
− 9𝑥 + 𝐶
3
Example 2 Solution cont.
𝐴𝑟𝑒𝑎 =
=
𝑥3
3
3
− 9𝑥
−3
(3)3
− 9(3) −
3
= −36
= 36
−3
3
3
− 9 −3
Example 3: Find the area under the curve of ℎ 𝑥 = 𝑥 3 from 𝑥
= −2 to 𝑥 = 2
Solution:
From the graph, we can see that the area is
below the x-axis from 𝑥 = −2 to 𝑥 = 2 and
above the -axis from and . Enclose the
definite integral from 𝑥 = −2 to 𝑥 = 0 with
an absolute value.
Example 3 Solution cont.
𝐴𝑟𝑒𝑎 =
0
−2
3
𝑥 𝑑𝑥 +
2
0
𝑥 3 𝑑𝑥
Using the power rule for integrals, we have 𝑥 3 𝑑𝑥 =
𝐴𝑟𝑒𝑎 =
=
0
4
𝑥
4
+
2
4
𝑥
−2
04 (−2)4
−
4
4
= −4 + 4 = 8
4
0
04 (−2)4
+
−
4
4
𝑥4
4
+𝐶
Practice Questions:
1.
What is the area under the curve of 𝑓 𝑥 = 64 − 𝑥 2 over the
interval 4 ≤ 𝑥 ≤ 8?
2.
Find the area under the curve of 𝑓 𝑥 = 𝑥 2 − 16 from 𝑥 = −3
to 𝑥 = 3.
Answers:
1.
(64 − 𝑥 2 ) 𝑑𝑥 =
3 8
𝑥
𝐴𝑟𝑒𝑎 = 64𝑥 −
3
4
64 𝑑𝑥 −
2+1
𝑥
𝑥 2 𝑑𝑥 = 64𝑥 −
+𝐶
2+1
𝑥3
= 64𝑥 −
+𝐶
3
(8)3
4 3
= 64(8) −
− 64 4 −
3
3
1024 704
𝟑𝟐𝟎
=
−
=
3
3
𝟑
Answers:
2.
(𝑥 2 − 16) 𝑑𝑥 =
𝑥 2 𝑑𝑥 −
16 𝑑𝑥
𝑥 2+1
𝑥3
=
− 16𝑥 + 𝐶 =
− 16𝑥 + 𝐶
2+1
3
𝐴𝑟𝑒𝑎 =
𝑥3
3
3
− 16𝑥
= −78 = 𝟕𝟖
=
−3
(3)3
− 16(3) −
3
−3
3
3
− 16 −3
Riemann Sum
What is a Riemann Sum?
•
The Riemann sum utilizes a finite number of rectangles to
approximate the value of a given definite integral. We can define
the Riemann sum as the sum of these rectangles’ areas.
•
Let’s take a look at non-negative function, 𝑓(𝑥) , that is
continuous within the closed interval, 𝑎, 𝑏 .
Through the Riemann sum, it’s now possible for us to
approximate the area of the region found right below the curve.
This area is bounded by the 𝑥 − 𝑎𝑥𝑖𝑠 as well as the vertical lines:
𝑥 = 𝑎 and 𝑥 = 𝑏.
•
𝑏
Here’s an example of how a definite integral, 𝑎 𝑓 𝑥 𝑑𝑥 , can be approximated by breaking down the area
under the curve of as a sum of the rectangles’ areas.
Riemann Sum Definition
•
We can divide the region under the curve and bounded by the
interval, 𝑎, 𝑏 . When we use n rectangles to divide the region,
each subinterval is expected to have a width of ∆𝑥 .
The areas of the n rectangles can
then be added to estimate the area
of the entire region.
𝑛−1
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 =
𝑓 𝑥𝑖 ∆𝑥
𝑖=0
The definition of Riemann Sum is…
•
an approximation of the definite integral,
𝑏
𝑓
𝑎
𝑥 𝑑𝑥, by adding the rectangles that divide the region
into n subintervals.
𝑛−1
𝑏
𝑓 𝑥 𝑑𝑥 ≈ lim
𝑎
𝑛→∞
𝑓 𝑥𝑖 ∆𝑥
𝑖=0
≈ 𝑓 𝑥1 ∆𝑥 + 𝑓 𝑥2 ∆𝑥+ 𝑓 𝑥3 ∆𝑥+. . . +𝑓 𝑥𝑛 ∆𝑥
How to do Riemann Sum?
 The right-hand rule and the left-hand rule for the Riemann Sum
•
In the right Riemann sum, we construct the rectangles so
that the curve passes through the top-right right corners.
• For the left Riemann sum, we construct the rectangles so
that the curve passes through the top-left corners.
These two graphs highlight
this difference between the
right and left Riemann
sums. The curve passes
through each of the topright corners for the right
Riemann sum while it
passes through the top-left
corners for the left-hand
Riemann sum.
Right and left Riemann sum’s summation notation:
Right Riemann Sum
𝑛
Left Riemann Sum
𝑛−1
∆𝑥 ∙ 𝑓 𝑥𝑖
𝑖=1
∆𝑥 𝑓 𝑥1 + 𝑓 𝑥2 + 𝑓 𝑥3 +. . . +𝑓 𝑥𝑛
∆𝑥 ∙ 𝑓 𝑥𝑖
𝑖=0
∆𝑥 𝑓 𝑥0 + 𝑓 𝑥1 + 𝑓 𝑥2 +. . . +𝑓 𝑥𝑛−1
Example of writing a Riemann sum formula
2
0
 F𝑜𝑟 𝑒𝑥𝑎𝑚𝑝𝑙𝑒,
4 − 𝑥 2 𝑑𝑥 , 𝑐𝑎𝑛 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑖𝑛 𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑟𝑖𝑔ℎ𝑡
𝑅𝑖emann sum notations with four rectangles
• We first determine the value of ∆𝑥 by dividing 2 − 0 by 4
2−0
1
∆𝑥 =
=
4
2
2
0

4 − 𝑥 2 𝑑𝑥
Right Riemann Sum
4
∆𝑥 ∙ 𝑓 𝑥𝑖
𝑖=1
1
1
3
= 𝑓
+𝑓 1 +𝑓
+ 𝑓(2)
2
2
2
Left Riemann Sum
3
∆𝑥 ∙ 𝑓 𝑥𝑖
𝑖=1
1
1
3
= 𝑓 0 +𝑓
+𝑓 1 +𝑓
2
2
2
1 15
7
=
+3+ +0
2 4
4
1
15
7
=
4+
+3+
2
4
4
17
=
= 4.25
4
25
=
= 6.25
4
Example of sketching a Riemann Sum
Graph
 We can use the values of ∆𝑥 and 𝑓 𝑥𝑖 to represent the Riemann sum.
• Divide the interval, 𝑎, 𝑏 , into n rectangles with a width of ∆𝑥
• Determine the height of each of the rectangle by
evaluating 𝑓 𝑥𝑖
• Double-check if the curve of 𝑓(𝑥) passes through the right
corners of the rectangles.
Example of sketching a Riemann Sum
Graph
 Let’s use
2
0
4 − 𝑥 2 𝑑𝑥. We want 4 rectangles with a width of ½ unit.
Example : Approximate the Riemann sum shown below. Keep in mind that the
graph shows a left-hand approximation of the area under the function 𝑓 𝑥 =
9 − 𝑥 2 𝑑𝑥, 0 ≤ 𝑥 ≤ 3
.Solution:
The graph above shows us that the area
under the region will be divided into four
subintervals. This means that each
rectangle will have a width shown below.
3−0
3
∆𝑥 =
=
4
4
Since we’re performing a left Riemann sum approximation, we’re going to
evaluate f(x) at the following values:
𝑥0 = 0
3
𝑥1 =
4
3
𝑥2 =
2
9
𝑥3 =
4
Approximate the function’s
𝑛−1
𝑖=0 𝑓 𝑥𝑖 ∆𝑥 . 𝑤ℎ𝑒𝑟𝑒 𝑛 = 4.
𝑓(0) =
9− 0
2
3
3
𝑓
= 9−
4
4
≈ 2.9047
=3
left
Riemann
3
𝑓
=
2
sum
using
3
9−
2
the
formula,
2
≈ 2.5981
2
9
𝑓
=
4
9
9−
4
2
≈ 1.9843
Approximate the left Riemann sum by adding these values and multiplying the
3
sum by ∆𝑥 =
4
3
9 − 𝑥 2 𝑑𝑥 ≈ ∆𝑥 𝑓 𝑥1 + 𝑓 𝑥2 + 𝑓 𝑥3
0
3
3
3
9
=
𝑓 0 +𝑓
+𝑓
+𝑓
4
4
2
4
3
=
3 + 2.9047 + 2.5981 + 1.9843
4
≈ 7.8653
Using the graph and the left-hand
rule for the Riemann sum, we have
3
approximated 0 9 − 𝑥 2 𝑑𝑥
as 7.8653
.