Trapezium Rule

WMA12/01/IAL Edexcel IAL/Jan 2020/Q1 (The Trapezium Rule)

The table below shows corresponding values of x and y for​​ y =log2⁡2x​​ 

The values of y are given to 2 decimal places as appropriate.​​ 

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

• obtain an estimate for​​ ∫214log2⁡2xdx​​ giving your answer to one decimal place.​​ 

(3)

 ​​ ​​ ​​ ​​ ​​ ​​​​ Using your answer to part (a) and making your method clear, estimate​​ 

• ∫214log2⁡4x25dx

• ∫214log2⁡2xdx

(4)

SOLUTION​​ 

a- Using Trapezium Rule to estimate​​ ∫214log2⁡2xdx.

(The number of trapezium strips are 4 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=14-24=3

∫214log2⁡2xdx =322+4.81+23.32+4+4.46

=9111200

=45.555

=45.6

b- i-

∫214log2⁡4x25dx=15∫214log2⁡2x2dx

Applying the power rule of logarithms to bring​​ power down as the coefficient.

∫214log2⁡4x25dx=15∫2142log2⁡2xdx

∫214log2⁡4x25dx=25∫214log2⁡2xdx

Where,​​ ∫214log2⁡2xdx=45.6

∫214log2⁡4x25dx=2xx 45.6

∫214log2⁡4x25dx=18.2

b- ii- Multiplying and dividing the expression by 2.​​ 

∫214log2⁡2xdx=∫214log2⁡2x×22dx

=∫214log2⁡4xdx

= ∫214log2⁡4 -log2⁡2xdx

= ∫214log2⁡22 -log2⁡2xdx

= ∫2142log2⁡2 -log2⁡2xdx

Remember,​​ loga⁡a=1; thus,​​ log2⁡2=1

= ∫2142×1 -log2⁡2xdx

=∫2142-log2⁡2xdx

= ∫2142dx- ∫214log2⁡2xdx

=2x214-45.6

=24-45.6

∫214log2⁡2xdx=-21.6