2010 Paper 2 Q7 - area under the curve

The diagram shows part of the curve y = √(2x + 5)  passing through the point P and meeting the x-axis at the point Q. The line  x = 2 passes through P and intersects the x-axis at the point S. Lines from Q meet x = 2 at the points R and T such that QR is parallel to the tangent to the curve at P, and RS = ST. Find

(i) the equation of QR,

(ii) the area of the shaded region

(i) To find the equation QR, we will need the gradient of the tangent of the curve at the point where x = 2. To get the gradient, we will have to differentiate the equation of the curve.

When x = 2, gradient is 1/√(4+5) = 1/3

Therefore, the equation of QR is  y = (1/3)x + c

At the point Q, y = 0. Substituting this into the equation of the curve, we get

0 =  √(2x + 5)

2x = -5

x = -5/2

Substituting the coordinates of the point Q(-5/2 , 0) into the equation of QR, we get

0 = -5/6 + c

c = 5/6

Hence, the equation of QR is y = (¹/₃)x + ⁵/₆

(ii) For the area of the shaded region, we will have to find the area under the curve as well as the area of triangle QST. Let’s find the length of ST first. This is the same as SR.  Subsituting x = 2 into the equation of QR, we get  y = ²/₃ + ⁵/₆ = ⁹/₆ = ³/₂

Hence, the area of QST = ½(⁹/₂)(³/₂) = ²⁷/₈ units²

For area under the curve,

hence, the area of the shaded region is 30⅜ units².