Important For Questions Class 12 Maths Chapter 8 Application of Integrals are provided at BYJUâ€™S along with the detailed explanations. These questions are given as per the new guidelines ofÂ **CBSE**Â for standard 12th board examination. Students can also practice all the problems ofÂ **NCERT**Â book along with these questions ofÂ Application of Integrals for class 12.

Application of Integrals is the chapter for which students can score good marks after getting thorough with the formulas. We have provided formulas of Class 12 Maths syllabus to help the students revise during their examination. Students can also refer to theÂ important questions for all chapters of Class 12 mathsÂ here to prepare well for the CBSE final exam.

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## Important Questions & Answers For Class 12 Maths Chapter 8 Application of Integrals

**Q. No.1: Find the area enclosed by the ellipse x ^{2}/a^{2} + y^{2}/b^{2} =1.**

**Solution:**

Given,

We know that,

Ellipse is symmetrical about both x-axis and y-axis.

Area of ellipse = 4 Ã— Area of AOB

Substituting the positive value of y in the above expression since OAB lies in the first quadrant.

= 2ab Ã— sin^{-1}(1)

= 2ab Ã— Ï€/2

= Ï€ab

Hence, the required area is Ï€ab sq.units.

**Q. No. 2: Find the area of the region bounded by y ^{2} = 9x, x = 2, x = 4 and the x-axis in the first quadrant.**

**Solution:**

We can draw the figure of y^{2} = 9x; x = 2, x = 4 and the x-axis in the first curve as below.

y^{2} = 9x

y = Â±âˆš(9x)

y = Â±3âˆšx

We can consider the positive value of y since the required area is in the first quadrant.

The required area is the shaded region enclosed by ABCD.

= 2 [(2)^{3} – (âˆš2)^{3}]

= 2[8 – 2âˆš2]

= 16 – 4âˆš2

Hence, the required area is 16 – 4âˆš2 sq.units.

**Q. No. 3: Find the area of the curve y = sin x between 0 and Ï€.**

**Solution:**

Given,

y = sin x

Area of OAB

= – [cos Ï€ – cos 0]

= -(-1 -1)

= 2 sq. units

**Q. No. 4: Find the area of the region bounded by the two parabolas y = x ^{2} and y^{2} = x.**

**Solution:**

Given two parabolas are y = x^{2} and y^{2} = x.

The point of intersection of these two parabolas is O (0, 0) and A (1, 1) as shown in the below figure.

Now,

y^{2} = x

y = âˆšx = f(x)

y = x^{2} = g(x), where, f (x) â‰¥ g (x) in [0, 1].

Area of the shaded region

= (â…”) – (â…“)

= â…“

Hence, the required area is â…“ sq.units.

**Q. No. 5: Smaller area enclosed by the circle x ^{2}+ y^{2} = 4 and the line x + y = 2 is**

**(A) 2 (Ï€ â€“ 2) **

**(B) Ï€ â€“ 2 **

**(C) 2Ï€ â€“ 1 **

**(D) 2 (Ï€ + 2)**

**Solution:**

Option (B) is the correct answer.

Explanation:

Given,

Equation of circle is x^{2}+ y^{2} = 4â€¦â€¦â€¦.(i)

x^{2}+ y^{2} = 2^{2}

y = âˆš(2^{2} – x^{2}) â€¦â€¦â€¦â€¦(ii)

Equation of a lines is x + y = 2 â€¦â€¦â€¦(iii)

y = 2 – x

x | 0 | 2 |

y | 2 | 0 |

Therefore, the graph of equation (iii) is the straight line joining the points (0, 2) and (2, 0).

From the graph of a circle (i) and straight-line (iii), it is clear that points of intersections of circle

(i) and the straight line (iii) is A (2, 0) and B (0.2).

Area of OACB, bounded by the circle and the coordinate axes is

= [ 1 Ã— âˆš0 + 2 sin^{-1}(1) – 0âˆš4 – 2 Ã— 0]

= 2 sin^{-1}(1)

= 2 Ã— Ï€/2

= Ï€ sq. units

Area of triangle OAB, bounded by the straight line and the coordinate axes is

= 4 – 2 – 0 + 0

= 2 sq.units

Hence, the required area = Area of OACB – Area of triangle OAB

= (Ï€ – 2) sq.units

### Practice Questions For Class 12 Maths Chapter 8 Application of Integrals

- Find the area of the region bounded by y =âˆša and y = a.
- Find the area bounded by the curve y = 4 â€“ x
^{2}and the lines y = 0 and y = 3. - Find the area including between the parabolas y
^{2}= 4ax and x^{2}= 4by. - What is the area bounded by the parabola y
^{2}= 8x and x^{2}= 8y? - Find the area bounded by the parabola y
^{2}= 4ax, latus rectum and the x-axis. - Using integration, find the area of the region bounded by the line 2y = 5x + 7, x- axis and the lines x = 2 and x = 8.
- Find the area of the smaller part of the circle x
^{2}+ y^{2}= a^{2}cut off by the line x = a/âˆš2.