In a previous article, we learned how to estimate the definite integral with a Riemann sum. This is achieved by doing the following:
 Dividing the area into rectangles where the height of each rectangle comes from the function.
 Calculating the area of each rectangle.
 Adding the individual areas of each rectangle.
This procedure can be summarized using the following formula:
The more rectangles you use, the narrower the rectangles are, and the better your approximation will be.
Eventually, if we use an infinite number of rectangles, that is, if we use an infinite number of subintervals, we will get the exact value for the area under the curve. Therefore, the definite integral of a function f(x) is as follows:
Definition: The Definite Integral
If we take the limit of the Riemann Sum as n (number of subintervals) approaches infinity, we will obtain the exact value of area under the curve – the definite integral:

In this article, you will still be approximating the area under the curve using area formulas and Riemann sums. Various techniques for integration functions will be taught in later sections. The focus for this section will be to represent the area under the curve in definite integral notation, and learn about the signs of definite integrals.
Write the Area under the Curve in Definite Integral Notation:
Example 1: Describe the area under the curve of f(x) = 10 x on the interval of [0,5] as a definite integral.
Solution:
The integrand will be the function itself, f(x) = 10 – x. The lower and upper limits of integration will be a = 0 and b = 5. Using definite integral notation, the area under the curve is:
Looking at the area under the curve in the interval, it is a trapezoid:
Let’s find the area of the trapezoid:
A= 1/2*h *(b_{1}+b_{2})
A= 1/2*5*(10+5)
A= 37.5
Therefore, the area of the curve is:
Example 2: Describe the area under the curve of g(x) = 3x on the interval [0,3] as a definite integral:
Solution:
The integrand will be the function itself, g(x) = 3x. The lower and upper limits of integration will be a = 0 and b = 3. Using definite integral notation, the area under the curve is:
Looking at the area under the curve in the interval, it is a triangle:
Let’s find the area of the triangle:
A= 1/2*h *b
A= 1/2*3*9
A= 13.5
Therefore, the area of the curve is:
Looking at the area under the curve in the interval, it is a semicircle:
Let’s find the area of the semicircle:
A= 1/2*πr^{2}
A= 1/2*π(5)^{2}
A= 25π/2
Therefore, the area of the curve is:
Example 4: Represent the shaded region below as a definite integral.
Solution:
The graph shown is a function is a function f(x) = x (the integrand). The limits of integration are a = 2 and b = 2. Therefore, we can represent the shaded area as the following definite integral:
You can compute the area in this region by combining the areas of two triangles. Coincidentally, both triangles have the same area.
A_{1}=A_{2}=1/2*b*h = 1/2*(2)*(2) = 2
A_{1}+A_{2}=4
Therefore, the area under the curve is:
Signed Areas of Definite Integrals
So far, we have only considered the area under the curve of positive functions, that is, functions that are above the x axis. If the function is above the x axis on the interval [a,b], the definite integral will be positive.
However, let’s say that the function falls below the x axis. In that case, the definite integral on the interval [a,b], that is below the x axis will be negative. Let’s go back to the idea of Riemann sums to prove this.
The area of each rectangle is base times height. If the curve is above the x axis, both the base and height are positive. If the curve falls below the x axis, the height of each rectangle will be negative (because the y value is negative). Therefore, the area will be negative:
Therefore, the area of the curve under the x axis is:
Solution:
Let’s graph the function and shade the region:
The area in the interval [2,1] is positive (function is above x axis). The area in the interval [1,3] is negative (function is below x axis).
The area in the region [2,1] is
A_{1}=1/2*b*h = 1/2*(3)*(6) = +9
The region in the region [1,3] is
A_{2}= – 1/2*b*h = – 1/2*(1)*(2) = 1
The net area under the curve is
Example 6: Evaluate the following integrals using the function of f(x) below:
Since the function falls below the x axis, the area will be negative.
Part B
The integral of the function in the region of [4, 7] can be broken into two regions. The first region is on [4,5] where f(x) is above the x axis. The second region is [5,7], where f(x) is below the x axis.
Region 1:
This region is the area of a triangle.
A_{1}= 1/2*h *b
A_{1}= 1/2*2*1
A_{1}= 1 square unit
Region 2:
This region is the area of a trapezoid.
A_{2}= 1/2*2*(1+2)
A_{2}= 3 square unit
Remember, this area is negative. Therefore, the net area of the region of [4,7] is:
Example 7: The velocity of a bug is modelled by the function below.
 What is the total distance covered by the bug 1 pm and 3 pm?
 What is the total displacement of the bug from 1 pm to 3 pm?
Solution:
It is best to divide the interval into subintervals where the function changes from negative to positive (and vice versa). The first subinterval is [1, 2.5] and the second subinterval is [2.5, 3]
Region [1, 2.5]
The area under the curve in this interval is the area of a trapezoid:
A_{1}= 1/2*10*(1+2.5) = 17.5 feet
Region [2.5, 3]
The area under the curve in this interval is the area of a triangle:
A_{2}= 1/2*0.5*10 = 2.5 feet (this area is negative)
To obtain the total distance the bug travelled, add the areas (ignore the sign of each area):
17.5 + 2.5 = 20 ft
To obtain the total displacement, add the areas but this time include the sign of the bottom area:
17.5 – 2.5 = 14.5 ft
In other words, the displacement is the net change in distance.
Summary of Section
The definite integral can be estimated with a Riemann sum. The more rectangles you use (in other words, as n approaches infinity), the narrower the rectangles are, and the better your estimation of the area under the curve. The integral sign ∫ represents integration. The symbol dx, called the differential of the variable x, indicates that the variable of integration is x. The function f(x) to be integrated is called the integrand. The points a and b are called the limits of the integral. The definite integral gives a numerical value, which is the net area under the curve. 
References:
https://www.khanacademy.org/math/apcalculusab/abintegrationnew/ab63/a/definiteintegralasthelimitofariemannsum
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_4_2_DefInt.pdf
https://tutorial.math.lamar.edu/Classes/CalcI/DefnOfDefiniteIntegral.aspx
https://www.mathsisfun.com/calculus/integrationdefinite.html
http://www.opentextbookstore.com/appcalc/Chapter31.pdf
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