Area of Shaded Region

Edexcel IAL WMA12/01/P2/Jan 2022/Q9 (Integration, Area under Curves)

Figure 2 shows​​ 

• the curve C with equation​​ y = x – x2​​ 

• the line l with equation y = mx, where m is a constant and 0 < m < 1​​ 

The line and the curve intersect at the origin O and at the point P.​​ 

• Find, in terms of m, the coordinates of P.​​ 

(2)​​ 

The region R1, shown shaded in Figure 2, is bounded by C and l.​​ 

• Show that the area of R1​​ is​​ 

1-m36

(5)​​ 

The region R2, also shown shaded in Figure 2, is bounded by C, the x‑axis and l.​​ 

Given that the area of R1​​ is equal to the area of R2​​ 

• find the exact value of m.​​ 

(3)

SOLUTION

a- ​​ Solving both equations simultaneously to find the point of intersection of line and curve.​​ 

y=x-x2      -1

y=mx            -2

Using the substitution method.

mx=x-x2

x2+mx-x=0

xx+m-1=0

x=0                              x+m-1=0  

                                 x=1-m

To find the y cordinate when​​ x=1-m, put the value of x in either equation of line or curve.​​ 

y=mx

y=m(1-m)

Hence, the cordinates of​​ P 1-m,   m1-m.​​ 

b- To find the area​​ R1, subtract the area under the line from the area under the curve.

Area of R1=area under the curve-area under line

Area of R1=∫ab( ycurve-yline )dx

We will be considering upper limit as 1-m as the extreme right point of​​ R1​​ is P, which has the x cordinate​​ (m-1).

Area of R1=∫01-m[x-x2)-(mx] dx

=∫01-mx-mx-x2dx  

=∫01-mx1-m-x2dx

Now, integrating the expression.​​ 

=x21-m2-x3301-m

=1-m1-m22-1-m33 

=1-m32-1-m33

=31-m36-21-m36

Area of R1=1-m36

c- ​​ since​​ R1=R2,​​ 

R1+R2=R1+R1=2R1  R1=R2

R1=1-m36 

Now, the total shaded region must be equal to​​ 

Total shaded Area=R1+R2

Total shaded Area=2R1

Total shaded Area=2×1-m36

Total shaded Area=1-m33

Another expression of total shaded region is equal to area under the curve enclosed between its two solutions of x. so first, we will find the solutions of equation of a curve so that we may have the limts and then we will find the total shaded area by integrating the equation of curve. ​​ 

The equation of curve is

y=x-x2= 0

x1-x=0

x=0      x=1

Hence, the upper limit is 1 and lower limit is 0.

Total shaded area=∫01x-x2dx=x22-x3301 

On applyting limits.​​ 

Total shaded area=12-13-0

Total shaded area=16

Equating it with the expression we found earlier while solving this part of total shaded area.​​ 

1-m33=16

1-m3=12

1-m=123 

1-123=m

m= 1-123