Trapezium Rule

Edexcel IAL WMA12/01/P2/June 2022/Q2 (Trapezium Rule, Integration)

Figure 1 shows the graph of​​ 

y = 1 -log10⁡sin⁡x    0 < x < π 

where x is in radians.​​ 

The table below shows some values of x and y for this graph, with values of y given to 3 decimal places.

(a) Complete the table above, giving values of y to 3 decimal places.​​ 

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ (2)

(b) Use the trapezium rule with all the y values in the completed table to find, to 2 decimal places, an estimate for​​ 

∫0.531-log10⁡(sin⁡x)dx

(3)

(c) Use your answer to part (b) to find an estimate for​​ 

∫0.533+log10⁡(sin⁡x)dx

(3)

SOLUTION​​ 

a-​​ 

y = 1 -log10⁡sin⁡x 

When​​ x=1, then find​​ y​​ by using the given equation.

y=1-log10⁡sin⁡1=1.075

When​​ x=2, then find​​ y​​ by using the given equation.

y=1-log10⁡sin⁡2=1.041

b- ​​ Using Trapezium Rule, to estimate the value of​​ ∫0.531-log10⁡(sin⁡x)dx.

(First, divide the area under the curve into 5 equal trapezium strips as we are given 6 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=3-0.55=2.55=12=0.5

A=0.52 1.319+21.075+1.001+1.041+1.223+1.850

A=2.06225

A=2.96 sq. unit

c-​​ 

(Since the question has asked us to use the answer of part b, so first let us rewrite the question so that we inculcate the answer we got in part b. We may simply write​​ (4-1)​​ instead of 3, and on splitting the integrals we get​​ ∫0.534dx​​ &​​ ∫0.531-log10⁡(sin⁡x)dx​​ with a minus in between. And now, we may substitute the answer of part b in place of​​ ∫0.531-log10⁡(sin⁡x)dx​​ which was found through trapezium rule.)

∫0.533+log10⁡(sin⁡x)dx=∫0.534-1+log10⁡(sin⁡x)dx

∫0.533+log10⁡(sin⁡x)dx=∫0.534-(1-log10⁡(sin⁡x))dx

Splitting the integrals on the RHS.

∫0.533+log10⁡(sin⁡x)dx=∫0.534dx-∫0.531-log10⁡(sin⁡x)dx

Now, Integrating​​ ∫0.534dx, whereas substituting the value​​ ∫0.531-log10⁡(sin⁡x)dx​​ as​​ 2.96225​​ we got in part b.​​ 

∫0.533+log10⁡(sin⁡x)dx=4x0.53-2.96225

Applying the limits.

∫0.533+log10⁡(sin⁡x)dx=[43-40.5]-2.96225

∫0.533+log10⁡x)dx=(12-2)-2.96225

∫0.533+log10⁡x)dx=7.03775

∫0.533+log10⁡x)dx=7.04