Solving Equations Using Logarithms

Coach Name: Sir Muhammad Abdullah Shah

Edexcel IAL WMA12/01/P2/Oct 2021/Q3 (Logarithms,​​ The Trapezium Rule)

A graph of a function

Description automatically generated

Figure 1 shows a sketch of part of the curve with equation​​ y =log10x​​ 

The region R, shown shaded in Figure 1, is bounded by​​ the curve, the line with equation x = 2, the x-axis and the line with equation x = 14​​ 

Using the trapezium rule with four strips of equal width,​​ 

  • show that the area of R is approximately 10.10​​ 

(3)​​ 

  • Explain how the trapezium rule could be used to obtain a​​ more accurate estimate for the area of R.​​ 

(1)​​ 

  • Using the answer to part (a) and making your method clear, estimate the value of​​ 

  • 214log10xdx

  • 214log10100x3dx

(4)

SOLUTION

a-​​ Using Trapezium Rule to estimate the shaded area below the curve​​ y =log10x.

(First, divide the area under the​​ curve into 4 equal trapezium strips as given in the question. To know the x values, let us first find the value of h or the width of each trapezium strip.)

h=b-an=14-24=124=3

Let us now construct the table of x and y values. Picking the x values at the difference of 3 as h=3, and finding their respoective values of y which are the parallel lengths of the trapezium strips.​​ 

x

2

5

8

11

14

y =log10x

log102

log105

log108

log1011

log1014

 

Now, applying the trapezium formula from the flash card above.​​ 

A=32log2 +log14 +2 log5+log8+log11

A=10.10 1

A=10.10 sq unit

b- By dividing the area under the curve into more trapezium​​ strips.​​ 

c- i-​​ 

214log10xdx=214log10x12dx

214log10xdx=21412log10x.dx

214log10xdx=12214log10x dx

Substituting the value of​​ 214log10x dx=10.10

214log10xdx=12x 10.10

214log10xdx=5.05

c-ii-​​ 

214log10100x3dx=214log10100dx+214log10x3dx

=214log10102dx+214log10x3dx

=2142log1010dx+214log10x3dx

Remember,​​ logaa=1; thus,​​ log1010=1

=2214(1)dx+214log10x3dx

=2214dx+214log10x3dx

Applying the power rule on​​ 214log10x3dx​​ to bring 3 down as the coefficient of the​​ log10x.

=2214dx+2143log10xdx

=2214dx+3214log10xdx

Substituting the value of​​ 214log10xdx​​ as 10.10.

=2214dx+3(10.10)

Solving the rest of the integration.​​ 

=2x214+3(10.10)

Applying the limits.

=28-4+30.30

=24+30.30

214log10100x3dx=54.30