Trapezium Rule

Edexcel IAL WMA12/01/P2/June 2021/Q5 (The Trapezium Rule, Integration)

Figure 2 shows a sketch of part of the graph of the curves C1​​ and C2​​ 

The curves intersect when x = 2.5 and​​ when x = 4​​ 

A table of values for some points on the curve C1​​ is shown below, with y values given to 3 decimal places as appropriate.

Using the trapezium rule with all the values of y​​ in the table,​​ 

• find, to 2 decimal places, an estimate for the area bounded by the curve C1, the line with equation x = 2.5, the x-axis and the line with equation x = 4​​ 

(4)​​ 

The curve C2​​ has equation​​ 

y=x12-3x+9        x>0

• Find​​ ∫x12-3x+9dx

(3)

The region R, shown shaded in Figure 2, is​​ bounded by the curves C1​​ and C2​​ 

• Use the answers to part (a) and part (b) to find, to one decimal place, an estimate for the area of the region R.​​ 

(3)

SOLUTION

a- Using Trapezium Rule to estimate the area bounded by the curve​​ C1.

(The number of trapezium strips are 6 as we know that the number of strip is always 1 less than the total number of values of x and y given.)

h=b-an=4-2.56=14=0.25

A=0.2525.453+5+27.764+9.375+9.964+9.367+7.626

Area=12.330625 

=12.33 sq.units

b- To find​​ ∫x12-3x+9dx, lets integrate it simply.​​ 

∫x12-3x+9dx=x5252-3x22+9x+c

∫x12-3x+9dx=2x522-3x22+9x+c

∫x12-3x+9dx=2x5-3x22+9x+c 

c- To find the area​​ R, subtract the area under​​ C1from the area under​​ C2.

Area of R=Area under C1-Area under C2

We already​​ calculated the area under​​ C1​​ via trapezium rule as​​ 12.33.

Area of R=12.33-∫2.54x32-3x+9dx

Area of R=12.33-2x55-3x22+9x2.54

Applying limits.

Area of R=12.33-2455-3422+94-22.552-32.522+92.5

Area of R=4.60847  

Area of R=4.6 sq.units