NCERT Solutions for Class 12 Maths Chapter 8: Application of Integrals

Calculus helps students connect mathematics with real-life applications, and Chapter 8: Application of Integrals is one of the most important and scoring chapters in Class 12 Maths. This chapter focuses on using definite integrals to find areas under curves and between two curves, making it essential for understanding the practical side of integration.

NCERT Solutions for Class 12 Maths Chapter 8 – Application of Integrals are prepared strictly as per the CBSE syllabus and exam pattern. These solutions include important CBSE board questions asked between 2018 and 2025 and are explained in a clear, step-by-step manner using simple language and proper mathematical presentation. This helps students build strong conceptual clarity, practise effectively, and score well in board examinations.

NCERT Solutions for Class 12 Maths Chapter 8: Application of Integrals

Application of Integrals
Difficult
Q.
Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
Application of Integrals
Easy
Q.
Find the area enclosed between the parabola y2 = 4ax and the line y = mx.
Application of Integrals
Easy
Q.
The area of the circle x2+y2 =16 exterior to  the parabola y2 = 6x is(A)43(4π-3)         (B)43(4π + 3)(C)  43(8π-3)      (D)43(8π + 3)
Application of Integrals
Easy
Q.
The area bounded by the curve y = x |x|, x-axis and the ordinates x = –1 and x = 1 is given by
(A) 0     (B) 1/3   (C) 2/3    (D) 4/3
Application of Integrals
Easy
Q.
Area bounded by the curve y = x3, the x-axis and the ordinates x=  –2 and x = 1 is
(A) – 9 (B) – 15/4    (C) 15/4   (D) 17/4
Application of Integrals
Easy
Q.
  1
Application of Integrals
Easy
Q.
Using method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
Application of Integrals
Easy
Q.
  Find the area bounded by the curves {(x,y):y≥x2 and y=∣x∣}.\{ \left(\right. x , y \left.\right) : y \geq x^{2} {\ and\ } y = \left|\right. x \left|\right. \} . ​​
Application of Integrals
Easy
Q.
Find the area of the smaller region bounded by the ellipse
x29+y24=1 \frac{x^2}{9}+\frac{y^2}{4}=1 and line x3+y2=1.\frac{x}{3} + \frac{y}{2} = 1 . ​​
Application of Integrals
Medium
Q.
Findtheareaboundedbythecurvey=sinxbetweenx=0andx=2π.
Application of Integrals
Difficult
Q.
Find the area of the region bounded by y2= 9x,   x = 2, x = 4 and the x-axis in the first quadrant.
Application of Integrals
Easy
Q.
Sketch the graph of y=| x+3 | and evaluate  ∫ −6 0 | x+3 |​dx. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaWHtbGaaC4AaiaahwgacaWH0bGaaC4yaiaahIgacaqGGaGaaCiDaiaahIgacaWHLbGaaeiiaiaahEgacaWHYbGaaCyyaiaahchacaWHObGaaeiiaiaah+gacaWHMbGaaeiiaiaahMhacqGH9aqpdaabdaqaaiaahIhacqGHRaWkcaWHZaaacaGLhWUaayjcSdGaaeiiaiaahggacaWHUbGaaCizaiaabccacaWHLbGaaCODaiaahggacaWHSbGaaCyDaiaahggacaWH0bGaaCyzaiaaykW7daWdXaqaamaaemaabaacbeGaa8hEaiaa=TcacaWFZaaacaGLhWUaayjcSdGaaGzaVlaa=rgacaWF4bGaaiOlaaWcbaGaeyOeI0IaaGOnaaqaaiaaicdaa0Gaey4kIipaaaa@6D98@
Application of Integrals
Difficult
Q.
Draw the rough sketch of the region enclosed by the curves y = x and y = x2, and find its area.
Application of Integrals
Difficult
Q.
Area lying between the curves y2 = 4x and y = 2x is
(A) 2/3    (B) 1/3    (C) 1/4  (D) 3/4
Application of Integrals
Difficult
Q.
Choose the correct answer
Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is
(a)  π     (b)   π 2      (c)  π 3     (d)   π 4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaGGOaGaamyyaiaacMcacaaMe8UaaGjbVlabec8aWjaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaiikaiaadkgacaGGPaGaaGjbVlaaysW7daWcaaqaaiabec8aWbqaaiaaikdaaaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caGGOaGaam4yaiaacMcacaaMe8+aaSaaaeaaiiaacqWFapaCaeaacaaIZaaaaiaaysW7caaMe8UaaGjbVlaaysW7caGGOaGaamizaiaacMcacaaMe8UaaGjbVpaalaaabaGaeqiWdahabaGaaGinaaaaaaa@6BAF@
Application of Integrals
Difficult
Q.
Findtheareaoftheregioninthefirst  quadrantenclosedbyx-axis,linex=3yandthecirclex2+y2=4.
Application of Integrals
Difficult
Q.
Find the area of the region bounded by the ellipse (x2/4) + (y2/9) = 1.
Application of Integrals
Difficult
Q.
Find the area of the region bounded by the ellipse (x2/16) + (y2/9) = 1.
Application of Integrals
Difficult
Q.
Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Application of Integrals
Medium
Q.
The area bounded by the y-axis, y = cosx and y = sinx  when 0≤x≤ π 2  is  ( A ) 2( 2 −1 )     ( B )  2 −1   ( C ) 2 +1    ( D ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGHbGaaeOCaiaabwga caqGHbGaaeiiaiaabkgacaqGVbGaaeyDaiaab6gacaqGKbGaaeyzai aabsgacaqGGaGaaeOyaiaabMhacaqGGaGaaeiDaiaabIgacaqGLbGa aeiiaiaabMhacaqGTaGaaeyyaiaabIhacaqGPbGaae4CaiaabYcaca qGGaGaaeyEaiaab2dacaqGJbGaae4BaiaabohacaqG4bGaaeiiaiaa bggacaqGUbGaaeizaiaabccacaqG5bGaaeypaiaabccacaqGZbGaae yAaiaab6gacaqG4bGaaeiiaaqaaiaabEhacaqGObGaaeyzaiaab6ga caqGGaGaaeimaiabgsMiJkaabIhacqGHKjYOdaWcaaqaaGGaaiab=b 8aWbqaaiaabkdaaaGaaeiiaiaabMgacaqGZbaabaGaaeiOamaabmaa baGaaeyqaaGaayjkaiaawMcaaGqabiaa+bcacaqGYaWaaeWaaeaada GcaaqaaiaabkdaaSqabaGccqGHsislcaqGXaaacaGLOaGaayzkaaGa a4hOaiaa+bkacaGFGcGaa4hOaiaaxMaadaqadaqaaiaabkeaaiaawI cacaGLPaaacaqGGaWaaOaaaeaacaqGYaaaleqaaOGaeyOeI0Iaaeym aiaa+bkacaGFGcGaaCzcamaabmaabaGaae4qaaGaayjkaiaawMcaai aa+bcadaGcaaqaaiaabkdaaSqabaGccaqGRaGaaeymaiaabckacaGF GcGaa4hOaiaaxMaadaqadaqaaiaabseaaiaawIcacaGLPaaacaGFGa WaaOaaaeaacaqGYaaaleqaaaaaaa@9635@

Question 1 (Level: Difficult)

Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.

Solution

Answer: Area = ∫14 √x dx = (2/3)(8 − 1) = 14/3 sq. units.

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Question 2 (Level: Easy)

Find the area of the smaller region bounded by the ellipse and line

Solution

Answer: Required area = (Area of ellipse part) − (Area under line inside ellipse) (compute by integration).

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Question 3 (Level: Easy)

Find the area bounded by the curves

Solution

Answer: Area = 2∫01 (x − x2) dx = 1/3 sq. units.

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Question 4 (Level: Difficult)

Draw the rough sketch of the region enclosed by the curves y = x and y = x2, and find its area.

Solution

Answer: Area = ∫01 (x − x2) dx = 1/6 sq. units.

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Question 5 (Level: Easy)

Using method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0.

Solution

Answer: Required area = area of triangle formed by intersection points of the three lines.

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Question 6 (Level: Easy)

Find the area of the region satisfying: y2 ≥ 4x and 4x2 + 4y2 ≤ 9.

Solution

Answer: Required area = common region of parabola y2=4x and circle x2+y2=9/4.

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Question 7 (Level: Difficult)

Find the area of the region in the first quadrant enclosed by x-axis, line x = √3 y and the circle x2 + y2 = 4.

Solution

Answer: Area = area of sector of circle (radius 2) between angles 0 and π/6 = (1/2)·r2·θ = π/3.

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Question 8 (Level: Difficult)

Choose the correct answer:

Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is:

(a) π    (b) π/2    (c) π/3    (d) π/4

Solution

Answer: Correct option: (b) π/2.

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Question 9 (Level: Difficult)

Area lying between the curves y2 = 4x and y = 2x is:

(A) 2/3    (B) 1/3    (C) 1/4    (D) 3/4

Solution

Answer: Correct option: (A) 2/3.

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Question 10 (Level: Medium)

Find the area bounded by the curve y = sin x between x = 0 and x = 2π.

Solution

Answer: Area = ∫02π |sin x| dx = 4 sq. units.

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FAQs: NCERT Solutions for Class 12 Maths Chapter 8 – Application of Integrals

Q1. Why is Application of Integrals important for Class 12 Maths?

This chapter explains the practical use of integration and carries good weightage in CBSE board exams, especially in long-answer questions.

Q2. How many marks are usually asked from Chapter 8 in board exams?

Typically, 6 to 8 marks worth of questions are asked from Application of Integrals.

Q3. Are NCERT Solutions enough to score well in this chapter?

Yes. NCERT textbook questions and NCERT Solutions are sufficient for CBSE board exams, as most questions are directly based on NCERT examples and exercises.

Q4. What are the most important topics in this chapter?

The most important topics include:

  • Area under curves

  • Area between two curves

  • Determining correct limits of integration

Q5. Why do students find Application of Integrals difficult?

Many students struggle with:

  • Drawing correct graphs

  • Choosing correct limits

  • Identifying the required region

Regular practice and clear understanding of graphs help overcome these difficulties.

Q6. Is Application of Integrals useful for competitive exams?

Yes. Concepts from this chapter are frequently used in JEE and other competitive exams, especially in questions related to area and calculus-based applications.