Maths · Calculus · chapter 8 of 13
Class 12 Maths Chapter 8: Application of Integrals case study question with answers
One case-based question of the kind the board asks: a passage and a figure, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
Case study · 4 marks
A carpenter in Jodhpur is making a wooden gate for an arched opening in a haveli wall. Taking the left foot of the opening as the origin and measuring in metres, the floor is the x-axis and the top edge of the opening follows y = 3 sin πx6 for 0 ≤ x ≤ 6, as in the figure. The opening is the region bounded by this curve and the x-axis. To estimate the timber, he treats the opening as a large number of thin vertical strips and adds their areas.
Part 1 · 1 mark
The width of the opening at a height of 1.5 m, in metres, is
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Answer: (c) 4
3 sin πx6 = 1.5 gives sin πx6 = 12, so πx6 = π6 or 5π6, that is x = 1 or x = 5. The width at that height is 5 − 1 = 4 m. 6 is the width at the floor; 3 is the greatest height.
Part 2 · 1 mark
The elementary area of a vertical strip of width dx at a distance x metres from the origin is
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Answer: (b) 3 sin πx6 dx
A vertical strip has height y = 3 sin πx6 and width dx, so dA = y dx = 3 sin πx6 dx. x dy is a horizontal strip.
Part 3 · 2 marks
Find the area of the opening.
Answer and marking scheme
A = ∫06 3 sin πx6 dx = 3 · 6π[−cos πx6]06 = 18π(1 + 1) = 36π ≈ 11.46 m2.
- sets up ∫_{0}^{6} 3 sin \frac{πx}{6} dx and integrates to −\frac{18}{π} cos \frac{πx}{6} (1)
- area \frac{36}{π} m^{2} (about 11.46 m^{2}) (1)
Or
The part of the opening between x = 0 and x = 2 is to be fitted with glass. Find the area of that part.
Answer and marking scheme
A = ∫02 3 sin πx6 dx = 18π[−cos πx6]02 = 18π(1 − cos π3) = 18π · 12 = 9π ≈ 2.86 m2.
- sets up ∫_{0}^{2} 3 sin \frac{πx}{6} dx and integrates (1)
- area \frac{9}{π} m^{2} (about 2.86 m^{2}) (1)