WMA12 Jan 2023

Edexcel IAL WMA12/01/P2/Jan 2023/Q1 (Trapezium Rule, Integration)

Figure 1 shows a sketch of part of the curve with equation​​ y = f (x)

The table below shows some corresponding values of x and y for this curve.

The values of y are given to 3 decimal places.

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

(a) obtain an estimate for

∫-11fxdx

giving your answer to 2 decimal places.        (3)

(b) Use your answer to part (a) to estimate​​ 

• ∫-11fx-2dx

• ∫13fx-2dx

SOLUTION​​ 

a- ​​ Using Trapezium Rule to estimate the value of​​ ∫-11fxdx.

(First, divide the area under the curve into 4 equal trapezium strips as we are given 5 values of x and y in the table and we know that the number of strip is always 1 less than the total number of values of x and y given. Then, find h or width of each trapezium strip.)

h=b-an=1-(-1)4=24=12=0.5

A=0.522.287+2.834+24.470+6.719+7.291

A=10.52 1025

A=10.52 square 

b- ​​ 

i- To estimate​​ ∫-11fx-2dx, simply integrate it.​​ 

(Remember when you are integrating a polynomial function you can integrate one term at a time.

∫abfx+gx.dx=∫abfx.dx+∫abgxdx

So, we are going to first split it and integrate each term separately. And don’t forget to substiute the answer of part a in place of​​ ∫-11fx.dx.​​ Additionally, here the transformation rule​​ may also be applied that is when​​ subtracting a constant inside a function, the graph ​​ translates vertically downwards. In this way the area enclosed between the curve and x-axis is reduced. And this is what happens when the ​​ given polynomial function is integrated.)

∫-11fx-2dx=∫-11fxdx-∫-112 dx

=10.52-2x-11

=10.52-21-2-1=0.52-4

∫-11fx-2dx= 6.52 square

ii- To estimate​​ ∫13fx-2dx, use the concept of translation along the x-axis.​​ 

(Recall, the rule of transformation learnt in pure maths 1 that when adding or subtracting a constant inside the function, the graph ​​ translates horizontally. So the translation is given as​​ 20.​​ Hence, the translated graph when sketched looks like the following. Now, since the difference in the limit​​ (3-1)​​ of the given integration is 2 and of the previous was also 2 when the limits of the integral were​​ -1 and 1, so the area under the curve would remain the same. Don’t get confused with translation because with translation along the x –axis, area under the curve within the same difference of limits remains the same.)

∫13fx-2dx=10.52