The point A with x-coordinate 2 lies on the curve y = √4x+1. The diagram shows part of this curve and the tangent to the curve at A. Find the area of the shaded region enclosed by the curve, the tangent and the x-axis.

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

1. [B1] [When x = 2] y = 3
2. [M1] dy/dx = 1/2 (4x+1)^(-1/2) × 4
3. [M1] [gradient of tangent =] 1/2 (4(2)+1)^(-1/2) × 4 soi
4. [M1] Dep on first M1; y - their 3 = their (2/3) (x-2)
5. [A1] Must be from correct straight line equation; [x-intercept for tangent =] 5/2 soi
6. [B1] For area of triangle – allow unsimplified (1/2 × (5/2 + 2) × 3) soi
7. [B1] [x-intercept for curve =] 1/4 soi
8. [M1] M1 for integration in the form of k(4x+1)^(3/2), k ≠ 0 or 4
9. [M1] Dep on previous M1 for area under the curve; correct use of upper and lower limits: ( (4(2)+1)^(3/2) / 6 ) - ( (4(-0.25)+1)^(3/2) / 6 ) oe
10. [A1] Shaded area = 27/4 - 9/2 = 9/4 oe