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# Mathematics Internal Assessment: Finding area under a curve

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Introduction

Mathematics Internal Assessment: Finding area under a curve

IBDP- 11-A

Introduction:

In this IA, I will attempt to find the area present below the curve f(x) = +3 within the domain [0, 1]. It is generally considered not possible to find the area under a curve due to it not being in the form of a conventional polygon. However, a close approximation of the area is possible. This is done by dividing the curve into a number of polygons.  To achieve this, I will first plot the graph of the function +3 within the stated domain and then divide it into a certain number of trapeziums. I shall then calculate the area of each trapezium using technology and add them to come up with the approximate total area of the curve. The process will be repeated with increasing number of trapeziums to increase accuracy to come up with a general solution for any equation f(x) from x=a to x=b using “n” number of trapeziums.

Investigation:

To find the area under the curve f(x) = +3 with domain [0,1], I must first plot the function. When graphed, the function looks like this: Case1:  n=2

Middle 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

Conclusion

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:

1. 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

1. 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

1. 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.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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