The figure, once graphed in totality:
Here, the two trapeziums present are ADEB and BCFE. The area of trapezium ADEB is 1.56 cm2 and the area of trapezium BCFE is 1.81 cm2. Therefore, one can conclude that the approximate area present underneath the curve f(x) = x2+3 [0, 1] is the combined areas of the trapeziums the curve has been divided into.
Therefore, area under curve f(x) = x2+3 [0, 1] is,
Area (ADEB) +Area (BCFE) = (1.56+1.81) cm2 or 3.37cm2
Case 2: n=3
To gain a more approximate area of the space present underneath the same curve f(x) =x2+3 [0, 1], I shall divide the curve into 3 trapeziums instead of the two I have done in the example above. By dividing a curve in to more number of trapeziums, I reduce the amount of area left uncalculated, thereby increasing the accuracy of the area.
When divided into 3 trapeziums, the curve looks like this:
In this figure, the curve f(x) = x2+3 [0, 1] is divided in to 3 trapeziums, namely trap. (ABEF) whose area is 1.02cm2 trap.(BCGE) whose area is 1.09cm2 and trap.(CDHG) whose area is 1.24cm2 . The trapeziums have been drawn at 3.33 cm intervals. As done above, the coordinates of the divided parts have been joined to create the trapeziums. Using the same mathematical process as before, I can estimate the area of the curve by adding the individual areas of the curves.
Therefore, approx. area of the curve f(x) = x2+3 [0, 1]=
Area(BCGE)+ area(CDHG)+Area(ABEF)= 1.02+ 1.09+1.24= 3.35cm2 .
Case3: n=4
Next, I shall divide the curve f(x) = x2+3 [0, 1] into 4 trapeziums, drawing perpendicular lines to the x axis at intervals of 2.5cm. As done above, I shall join the coordinates of the formed trapeziums and find there area. Once graphed, the function looks like this:
The trapeziums formed here are trap.(ABFG) with area 0.76cm2, trap. (BCHF) with area 0.79cm2, trap. (CDIH) with area 0.85cm2 and trap. (DEJI) with area 0.95cm2.
By increasing the number of trapeziums, I have reduced the area between the side of the trapezium and the curve f(x) = x2+3 [0, 1], thereby increasing the approximation of the area. Therefore, the approximate area present under the curve f(x) = x2+3 [0, 1] when it is divided into 4 trapeziums=
Area(ABFG)+ Area(BCHF)+ Area(CDIH)+ Area(DEJI)=
(0.76+0.79+0.85+0.95)cm2 = 3.35cm2
Case 4: n=5
Finally, I shall divide the curve f(x) = x2+3 [0, 1] into 5 triangles to obtain the maximum approximation of area.
Approximate area of curve f(x) = x2+3 [0, 1] when it is divided into 5 trapeziums is:
Area( ABGH+ BCIG+CDJI+DEKJ+EFLK) = 0.6+0.62+0.65+0.7+0.76= 3.33cm2
General Statement:
To find a general statement for the area present beneath the curve is relatively simple. The area present beneath a curve is the sum total of the trapeziums the curve has been divided into.
The area of a trapezium=
h(a+b) where ‘h” is the height and “a” and “b” are the bases of the trapezium.
Therefore, when a curve is divided into equal intervals by the x axis, the area of the space present beneath the curve would be the sum total of the areas of all the trapeziums it has been divided into.
Therefore,
Area of space beneath a curve when divided into “n” trapeziums=
h(f(a)+f(a+m))+
h(f(a+m)+f(a+m+m))+…………..+
h(f(b-m)+f(b)), where m= the length of the “x” interval used to divide the curve into trapeziums and the curve has a domain [a,b] This process must be followed till ”n+1” values are obtained for accurate area.
Further simplifying,
Area of the space=
h(f(a)+2(f(0+m)+…….+2(f(b-m))+ f(b)). When the curve has a domain [a,b].
This formula is applicable for all curves above the “x” axis where trapeziums are made at equal intervals. When trapeziums are drawn at unequal intervals, the area of each trapezium should be found individually and then added to get the total approximate area and the formula cannot be generalized as the value of “m” is different in each case.
Verification:
-
When n=2, area was found to be 3.37cm2 . Using the obtained formula,
When n=2,
Area=
×0.5(f(0)+2(f(0.5)+f(1))=
0.25(3+6.5+4)=3.37cm2
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When n=4, area was found to be 3.35cm2 . Using the obtained formula,
When n=4,
Area=
×0.25(f(0)+2(f(0+0.25)+2(f(0+2(0.25))+2(f(0+3(0.25)+ f(1))
= 3.35cm2
-
When n=5, area was found to be 3.33cm2. using obtained formula,
When n=5,
Area=
×0.2(f(0)+2(f(0.2+0)+2(f(0+2(0.2))+2(f(0)+3(0.2))+2(0+4(0.2))+f(1))
=3.33cm2
Hence proved.
by
priya95 (student)
