NCERT Solutions · Class 12 · Chapter 8: Application of Integrals
NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1
Exercise 8.1: Area under simple curves. Area between \(y = f(x)\), the x-axis and \(x = a\), \(x = b\) is \(\int_a^b|y|\,dx\); between \(x = g(y)\), the y-axis and \(y = c\), \(y = d\) it is \(\int_c^d|x|\,dy\). Sketch first and use symmetry (an ellipse is four equal quarters). \(\int_0^a\sqrt{a^2 - x^2}\,dx = \tfrac{\pi a^2}{4}\).
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Exercise 8.1 questions and solutions
Exercise 8.1, Question 1
Find the area enclosed by the ellipse \(\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1\).
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In the first quadrant \(y = \tfrac34\sqrt{16 - x^2}\), \(0 \le x \le 4\); the ellipse is symmetric about both axes.
The area in the first quadrant bounded by \(x^2 + y^2 = 4\), \(x = 0\) and \(x = 2\) is: (A) \(\pi\) (B) \(\tfrac{\pi}{2}\) (C) \(\tfrac{\pi}{3}\) (D) \(\tfrac{\pi}{4}\)
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A quarter of a circle of radius 2: \[\displaystyle\int_0^2\sqrt{4 - x^2}\,dx = \tfrac14\pi(2)^2\]
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Application of Integrals has 39 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.
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