4PM1/02 Jan 2023, Question 11: Definite Integrals and Areas
Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 14 marks
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The question, in short
For the curve y = 4 − e^(2x) meeting the axes at A and B, determine A's y value, verify B's x value, then find the normal at B and the area of region R bounded by the curve, normal and y-axis.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (a) · 3 marks
State the y coordinate where the curve meets the y-axis, and demonstrate that the x intercept equals ln 2.
- 1x = 0: y = 4 − 1 = 3
- 2y = 0: e^(2x) = 4
- 32x = ln 4
- 4x = ½ ln 4 = ln 2
Answer(i) y = 3 (ii) x = ln 2 shown
Part (b) · 4 marks
Obtain the equation of the normal line to the curve at B, expressed as y = mx + c.
- 1dy/dx = −2e^(2x)
- 2At x = ln 2: dy/dx = −2 × 4 = −8
- 3Normal gradient = 1/8
- 4y − 0 = (1/8)(x − ln 2)
- 5y = (1/8)x − (1/8) ln 2
Answery = (1/8)x − (1/8) ln 2
Part (c) · 7 marks
Use integration to calculate the area of the region R enclosed by the curve, the normal and the y-axis, to 1 decimal place.
- 1Area under C = ∫ from 0 to ln 2 of (4 − e^(2x)) dx
- 2= [4x − ½e^(2x)] from 0 to ln 2 = (4 ln 2 − 2) − (−½) = 4 ln 2 − 3/2
- 3l meets the y-axis at (0, −(1/8) ln 2), below the x-axis
- 4Triangle area = ½ × ln 2 × (1/8) ln 2 = (ln 2)²/16
- 5Area of R = 4 ln 2 − 3/2 + (ln 2)²/16 ≈ 1.3
Answer1.3
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