Area of Shaded Region

Edexcel IAL WMA12/01/P2/Jan 2023/Q9 (Integration Area)

In this question you must show all stages Of your working.

Solutions based entirely on calculator technology are not acceptable.

Figure 3 shows​​ 

• the curve C with equation​​ y = x2— 4x + 5​​ 

• the line I with equation y = 2​​ 

The curve C intersects the yaxis at the point D.​​ 

• Write down the coordinates of D.​​ 

(1)

The curve C intersects the line I at the points E and F, as shown in Figure 3.​​ 

• Find the x coordinate of E and the x coordinate of F.​​ 

(2)

Shown shaded in Figure 3 is​​ 

• the region R1​​ which is bounded by C, I and the y-axis​​ 

• the region R2​​ which is bounded by C and the line segments EF and DF​​ 

Given that​​ area of R1area of R2​​ = k, where k is a constant,​​ 

• use algebraic integration to find the exact value of k, giving your answer as a simplified fraction.​​ 

(5)

SOLUTION

a-​​ 

Point D is the point where the curve cuts the y-axis. This means it is the y-intercept. You could either find it directly through the equation by comparing it to the general equation of astright line or else you may find the value of y coordinate​​ of D by substituting the value of x as 0 in the equation of the curve.​​ 

x=0

y=x2-4x+5

y=02-40+5=

y=5

Hence, coordinate of point D is​​ 0,5. 

b- ​​ Solve both equations (of curve and line) simultaneously to get the x cordinates of point E and F. ​​ 

(Point E and F are the points where the curve and line cuts each other and to find there point of intersection we have to solve both of them simultaneously.)

Substituting the value of y from equation of line​​ (y=2)​​ in equation of curve.​​ 

y=x2-4x+ 5

2=x2-4x+5

x2-4x+3=0

x-3x-1=0

x=3 ,  x=1

Hence, to choose which x value reoresents the x cordinate of which point, see which point is closer to the origin; therefore, the x-cordinate of point E ​​ and F is​​ 1 & 3.​​ 

E,   x-coordinate=1

F, x-coordinate=3

c- Finding area​​ R1​​ by integrating the equation of the curve within the limits 0 and 1 and subtracting from it the area of rectangle enclosed between the line​​ L​​ and x-axis.​​ 

By integrating the equation of the curve we get the Area enclosed between the curve and x-axis. However, area of region​​ R1​​ is area bounded between the curve and line (y=2) which corresponds to the shape of rectangle. Therefore, we may say simply that​​ Area of R1=∫01y.dx-Area of Rectangle. ​​ For better understanding have a look at the figure given.​​ 

Area of R1=∫01x2-4x1+5 dx-2

Area of R1=x33-4x22+5x01-2 

=133-212+51-2

=13-2+5-2=1+13=43

Area of R1=43 sqr unit

For Area of​​ R2, consider the region 1 and 2, they collective correspond to a right angle triangle. So, we will find the area of this triangle and subtract from it the area of​​ R1.​​ 

Area of Δ=12 x 3 x 3=92 

∴Area of R2=Area of ∆-Area of R1 

Area of R2=92-43=276=86=196cm2

Area of R1Area of R2=k

43÷196=k

k=43 x 619

k=819 cm2