1
Calculus answers two very important questions.
The first, how to find the instantaneous rate of change, we answered with our study of the derivative.
We are now ready to answer the second question: how to find the area of irregular regions.
Let us start with simple way how to calculate area under the curve that is not well known shape from geometry
Very often, we don’t even have a function, but only measurements taken from real life. Point to point. And we
have to find area !!!!
Area under the curve:
y
1 2
x 1
8
0 x4
Left-hand Rectangular Approximation Method (LRAM / Ln )
We could estimate the area under the curve by drawing rectangles touching at their left corners.
Approximate area:
Right-hand Rectangular Approximation Method (RRAM / Rn )
Approximate area:
Midpoint Rectangular Approximation Method (MRAM / Mn )
Approximate area:
M4 = 6.625
With 8 subintervals:
M8 = 6.65624
2
Trapezoidal Approximation Method (Tn)
Approximate area:
1  9  1  9 3  1  3 17  1  17

T  1              3 
2 8 28 2 22 8  2 8

T4 = 6.75
Eexact value for area under the curve y 
1 2
x 1
8
0 x4
L4 = 5.75
R4 = 7.75
M4 = 6.625
M8 = 6.65624
T4 = 6.75
If we take the limit as n goes to infinity, the result of each methods gives a definitive integral/area.
Taking the limit as n approaches infinity all methods give the same result: definite
integral
The process of finding the sum of the areas of rectangles to approximate area of a
region is called Riemann Sums, after Bernhard Riemann, who pioneered the
process.
SUM BECOMES INTEGRAL
∆𝑥 =
𝑏−𝑎
𝑛
𝑥𝑖 = 𝑎 + (𝑛 − 1)∆𝑥
𝐼𝑛 𝑙𝑖𝑚𝑖𝑡 𝑎𝑠
𝑛→∞
𝑓(𝑥𝑖 ) → 𝑓(𝑥)
∆𝑥 → 𝑑𝑥
𝑛
𝑏
lim ∑ 𝑓(𝑥𝑖 )∆𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)
∆𝑥→0
𝑖=1
𝑎
Leibnitz introduced a simpler notation for the definite integral:
𝑎
∫𝑏 𝑓(𝑥)
definite integral of
function 𝑓(𝑥) over interval [𝑎, 𝑏].
3
Area under a curve.
When the curve is above the ‘x’ axis,
the area is the same as the definite integral
But when the graph line is below the ‘x’ axis,
the definite integral is negative
And as you can see definite integral is not automatically area;
it is rather NET accumulation of the function over interval [a, b].
If 𝑦 = 𝑓(𝑥) is nonnegative and integrable over a closed interval [𝑎, 𝑏] then the area under the curve
𝑦 = 𝑓(𝑥) from 𝑎 to 𝑏 is the definite integral of 𝑓 from 𝑎 to 𝑏.
𝑏
𝐴 = ∫ 𝑓(𝑥)𝑑𝑥
𝑎
If 𝑦 = 𝑓(𝑥) is negative and integrable over a closed interval [𝑎, 𝑏], then the area under the curve 𝑦 =
𝑓(𝑥) from 𝒂 to 𝒃 is the OPPOSITE of the definite integral of 𝑓 from 𝑎 to 𝑏.
𝑏
𝐴 = − ∫ 𝑓(𝑥)𝑑𝑥
𝑎
AREA IS ALWAYS POSITIVE IN MATH
𝑏
If 𝑦 = 𝑓(𝑥) is both positive and negative on closed interval[𝑎, 𝑏], then ∫𝑎 𝑓(𝑥)𝑑𝑥 will NOT give us the
area.
𝑏
In general, ∫𝑎 𝑓(𝑥)𝑑𝑥 does NOT give us the area but rather the NET accumulation over the interval
from 𝑥 = 𝑎 to 𝑥 = 𝑏.
In velocity – time graph, net accumulation of integral is displacement, the area
(absolute value) is distance !!!!!!
If we are integrating by hand, we must decide if and where the graph crosses the x-axis, then split up
our interval, manually making negative regions positive. The following property will help accomplish
this:
𝑏
𝑐
𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
𝑎
𝑐
4
if the calculator is permitted, then evaluate
𝑏
∫|𝑓(𝑥)|𝑑𝑥
𝑎
If you use the absolute value function, you don’t need to find the roots.
Difference between the Value of a Definite Integral and Total Area
Find the mean (average) value of the function
𝑓(𝑥) = 𝑠𝑖𝑛𝑥 over the interval [0, 2𝜋]
Value of the Definite Integral
2𝜋
∫0 𝑠𝑖𝑛𝑥 𝑑𝑥 = − cos 𝑥|2𝜋
0
=
−𝑐𝑜𝑠(2𝜋) − (−𝑐𝑜𝑠(0)) =
−(1) − (−1) = 0
Total Area
2𝜋
𝜋
2𝜋
∫0 𝑠𝑖𝑛𝑥 𝑑𝑥 = ∫0 𝑠𝑖𝑛𝑥 𝑑𝑥 − ∫𝜋 𝑠𝑖𝑛𝑥 𝑑𝑥
− cos 𝑥|𝜋0 + cos 𝑥|2𝜋
𝜋
= −[(−1) − (−1)] + [1 − (−1)] = 4
−𝑐𝑜𝑠(2𝜋) − (−𝑐𝑜𝑠(0)) =
Example
−(1) −1:(−1) = 0
The graph of 𝑦 = 𝑓(𝑥) is shown below. If A1 and A2 are positive numbers that represent the
areas of the shaded regions, then shaded area A1 and A2 is: