NCERT Solutions · Class 12 · Chapter 8: Application of Integrals

NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise

The Miscellaneous Exercise on Application of Integrals. Where the curve dips below the x-axis, integrate each part separately and add the absolute values; odd and even symmetry halves the work.

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Miscellaneous Exercise questions and solutions

Miscellaneous Exercise, Question 1

Find the area under the curve between the given lines and the x-axis.

(i) \(y = x^2\), \(x = 1\), \(x = 2\)

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\[\begin{aligned}\displaystyle\int_1^2 x^2\,dx &= \left[\dfrac{x^3}{3}\right]_1^2 \\ &= \dfrac{8 - 1}{3}\end{aligned}\]

Answer: \(\tfrac73\) square units

(ii) \(y = x^4\), \(x = 1\), \(x = 5\)

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\[\begin{aligned}\displaystyle\int_1^5 x^4\,dx &= \left[\dfrac{x^5}{5}\right]_1^5 \\ &= \dfrac{3125 - 1}{5}\end{aligned}\]

Answer: \(\tfrac{3124}{5}\) square units

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Miscellaneous Exercise, Question 2

Sketch \(y = |x + 3|\) and evaluate \(\displaystyle\int_{-6}^{0}|x + 3|\,dx\).

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The graph is a V with its vertex at \((-3, 0)\): \(y = -(x + 3)\) for \(x < -3\), \(y = x + 3\) for \(x \ge -3\).
\[\displaystyle\int_{-6}^{-3}-(x + 3)\,dx + \int_{-3}^{0}(x + 3)\,dx = \tfrac92 + \tfrac92\] (two triangles of base 3, height 3).

Answer: 9

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Miscellaneous Exercise, Question 3

Find the area bounded by \(y = \sin x\) between \(x = 0\) and \(x = 2\pi\).

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\(\sin x \ge 0\) on \([0, \pi]\) and \(\le 0\) on \([\pi, 2\pi]\).
\[\displaystyle\int_0^{\pi}\sin x\,dx + \left|\int_{\pi}^{2\pi}\sin x\,dx\right| = 2 + |-2|\]

Answer: 4 square units

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Miscellaneous Exercise, Question 4

The area bounded by \(y = x^3\), the x-axis and \(x = -2\), \(x = 1\) is: (A) \(-9\) (B) \(-\tfrac{15}{4}\) (C) \(\tfrac{15}{4}\) (D) \(\tfrac{17}{4}\)

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\[\left|\displaystyle\int_{-2}^{0}x^3\,dx\right| + \int_0^1 x^3\,dx = |-4| + \tfrac14\]

Answer: (D) \(\tfrac{17}{4}\)

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Miscellaneous Exercise, Question 5

The area bounded by \(y = x|x|\), the x-axis and \(x = -1\), \(x = 1\) is: (A) 0 (B) \(\tfrac13\) (C) \(\tfrac23\) (D) \(\tfrac43\)

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\(y = x^2\) for \(x > 0\) and \(-x^2\) for \(x < 0\); the two pieces are mirror images.
Area \[= 2\displaystyle\int_0^1 x^2\,dx = \tfrac23\] (the signed integral would be 0).

Answer: (C) \(\tfrac23\)

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Done the NCERT exercises? The board paper asks more

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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Application of Integrals in our sample papers: Sample paper 1 (question 33) · Sample paper 2 (question 29) · Sample paper 3 (question 33) · Sample paper 4 (question 33).

Also useful: free MCQs and case studies for Application of Integrals · formulas for this chapter (Class 12 formula sheet, free PDF) · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan

Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.