NCERT Exemplar problems are important for students in their exam preparations. They will get advanced knowledge of the subjects, which is important from an examination point of view. The NCERT Exemplar Solutions have been developed to solve the exercise problems in the chapters. Also, these solutions are according to the latest CBSE guidelines.

The 8^{th} chapter of NCERT Exemplar Solutions for Class 12 Maths is the Application of Integrals. In this chapter, students will learn the application of integrals with the method of finding the area under simple curves, the area of the region bounded by a curve and a line and area between two curves. Students who want to clear their doubts and understand how to solve the exercise problems in a step-by-step manner can access the solutions PDF of NCERT Exemplar Solutions for Class 12 Maths Chapter 8 Application of Integrals, available in the link below.

## Download the PDF of NCERT Exemplar Solutions for Class 12 Maths Chapter 8 Application of Integrals

### Access Answers to NCERT Exemplar Class 12 Maths Chapter 8 Application of Integrals

Exercise 8.3 Page No: 176

**1. Find the area of the region bounded by the curves y^{2} = 9x, y = 3x.**

**Solution: **

Given curves are *y*^{2} = 9*x* and *y *= 3*x*

Now, solving the two equations we have

(3x)^{2} = 9x

9x^{2} = 9x

9x^{2} â€“ 9x = 0 â‡’ 9x(x – 1) = 0

Thus, x = 0, 1

So, the area of the shaded region is given by

= ar(region OAB) â€“ ar (Î” OAB)

Therefore, the required area is Â½ sq.units.

**2. Find the area of the region bounded by the parabola y^{2} = 2px, x^{2} = 2py.**

**Solution: **

Given parabolas are *y*^{2} = 2*px *â€¦. (i) and *x*^{2} = 2*py *â€¦. (ii)

Now, from equation (ii) we have

y = x^{2}/2p

Putting the value of y in equation (i), we have

(x^{2}/2p)^{2} = 2*px*

x^{4}/4p^{2} = 2px

x^{4} = 8p^{3}x

x^{4} â€“ 8p^{3}x = 0

x(x^{3} â€“ 8p^{3}) = 0

So, x = 0 or x^{3} â€“ 8p^{3} = 0 â‡’ x = 2p

Now, the required area is

Therefore, the required area is 4/3 p^{2} sq. units.

**3. Find the area of the region bounded by the curve y = x^{3} and y = x + 6 and x = 0.**

**Solution: **

Given curves are *y *= *x*^{3}, *y *= *x *+ 6 and *x *= 0

On solving *y *= *x*^{3} and *y *= *x *+ 6, we have

x^{3} = x + 6

x^{3} â€“ x â€“ 6 = 0

x^{2}(x – 2) + 2x(x – 2) + 3(x – 2) = 0

(x – 2) (x^{2 }+ 2x + 3) = 0

Itâ€™s seen that x^{2 }+ 2x + 3 = 0 has no real roots

So, x = 2 is the only root for the above equation.

Now, the required area of the shaded region is given by

**4. Find the area of the region bounded by the curve y^{2} = 4x and x^{2} = 4y.**

**Solution: **

Given curves are *y*^{2} = 4*x â€¦ *(i) and *x*^{2} = 4*y â€¦ *(ii)

On solving the equations, we get

From (ii),

y = x^{2}/4

Putting value of y in (i), we have

(x^{2}/4)^{2} = 4x

x^{4}/16 = 4x

x^{4} = 64x

x^{4} â€“ 64x = 0

So, x = 0, 4

Now, the required area is the shaded region

**5. Find the area of the region included between y^{2} = 9x and y = x**

**Solution: **

Given curves are *y*^{2} = 9*x *and *y *= *x*

Solving the above equations, we have

x^{2} = 9x â‡’ x^{2} â€“ 9x = 0

x(x – 9) = 0

So, x = 0, 9

Now, the required area is

Therefore, the required area is 27/2 sq. units.

**6. Find the area of the region enclosed by the parabola x^{2} = y and the line y = x + 2**

**Solution: **

Given, equation of parabola *x*^{2} = *y *and line *y *= *x *+ 2

Solving the above equations, we get

x^{2} = x + 2

x^{2} â€“ x â€“ 2 = 0

x^{2} â€“ 2x + x â€“ 2 = 0

x(x – 2) + 1(x – 2) = 0

(x + 1) (x – 2) = 0

So, x = -1, 2

Now,

Therefore, the area of the required region is 9/2 sq. units.

**7. Find the area of region bounded by the line x = 2 and the parabola y^{2} = 8x**

**Solution: **

Given, equation of line x = 2 and parabola y^{2} = 8x

Putting value of x in the other equation, we have

y^{2} = 8(2)

y^{2} = 16

So, y = Â± 4

Now, the required area is

Therefore, the area of the region = 32/3 sq. units

**8. Sketch the region {( x, 0) : y = âˆš(4 â€“ x^{2})} and x-axis. Find the area of the region using integration.**

**Solution: **

Given, {(*x*, 0) : *y *= âˆš(4 â€“ *x*^{2})}

So, y^{2} = 4 â€“ x^{2}

x^{2} + y^{2} = 4 which is a circle.

Now, the required area

**9. Calculate the area under the curve y = 2 âˆšx included between the lines x = 0 and x = 1.**

**Solution: **

**10. Using integration, find the area of the region bounded by the line 2 y = 5x + 7, x-axis and the lines x = 2 and x = 8.**

**Solution: **

Given, 2*y *= 5*x *+ 7, *x-*axis, *x *= 2 and *x *= 8

Letâ€™s draw the graph of 2*y *= 5*x *+ 7 â‡’ y = (5x + 7)/2

x | 1 | -1 |

y | 6 | 1 |

Now, letâ€™s plot the straight line on a graph with other lines.

The area of the required region is

Therefore, the required area the region = 96 sq. units

**11. Draw a rough sketch of the curve y = âˆš(x â€“ 1) in the interval [1, 5]. Find the area under the curve and between the lines x = 1 and x = 5.**

**Solution: **

Given curve *y *= âˆš(*x *â€“ 1)

â‡’ y^{2} = x â€“ 1

Plotting the curve and finding the area of the shaded region between the lines *x *= 1 and *x *= 5, we have

Therefore, the area of the required region = 16/3 sq.units

**12. Determine the area under the curve y = âˆš(a^{2} â€“ x^{2}) included between the lines x = 0 and x = a.**

**Solution: **

Given curve *y *= âˆš(*a*^{2} â€“ *x*^{2}) and lines *x *= 0 and *x *= *a*

*y *= âˆš(*a*^{2} â€“ *x*^{2}) â‡’ *y ^{2} *=

*a*

^{2}â€“

*x*

^{2}

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

*a*

^{2}which is equation of a circle.

Now, the required region is found by plotting the curve and lines.

So, the area of the shaded region is

Therefore, the required area = Ï€a^{2}/4

**13. Find the area of the region bounded by y = âˆšx and y = x.**

**Solution: **

Given equations of curve *y *= âˆš*x *and line y = x

Solving the equations *y *= âˆš*x *â‡’ y^{2} = x and y = x, we get

x^{2} = x

x^{2} â€“ x = 0

x(x – 1) = 0

So, x = 0, 1

Now, the required area of the shaded region

Therefore, the required area = 1/6 sq.units.

**14. Find the area enclosed by the curve y = â€“x^{2} and the straight-line x + y + 2 = 0.**

**Solution: **

Given curve *y *= â€“*x*^{2} or *x*^{2} = –*y *and the line *x *+ *y *+ 2 = 0

Solving the two equation, we get

x â€“ x^{2} + 2 = 0

x^{2} â€“ x â€“ 2 = 0

x^{2} â€“ 2x + x â€“ 2 = 0

x(x – 2) + 1 (x – 2) = 0

(x – 2) (x + 1) = 0

So, x = -1, 2

Now,

The area of the required shaded region

Therefore, the required area = 9/2 sq.units

**15. Find the area bounded by the curve y = âˆšx , x = 2y + 3 in the first quadrant and x-axis.**

**Solution: **

Given curve *y *= âˆš*x *and line *x *= 2*y *+ 3, first quadrant and x-axis.

Solving *y *= âˆš*x *and *x *= 2*y *+ 3, we get

**Long Answer (L.A.)**

**16. Find the area of the region bounded by the curve y^{2} = 2x and x^{2} + y^{2} = 4x.**

**Solution: **

Given equation of curves are *y*^{2} = 2*x *and *x*^{2} + *y*^{2} = 4*x*

Solving the equations, we have

x^{2 }â€“ 4x + y^{2 }= 0

x^{2 }â€“ 4x + 4 â€“ 4 + y^{2 }= 0

(x – 2)^{2} + y^{2} = 4

Itâ€™s clearly seen that the equation of the circle having its centre (2, 0) and radius 2.

Solving x^{2} + y^{2} = 4x and y^{2} = 2x

x^{2} + 2x = 4x

x^{2} + 2x â€“ 4x = 0

x^{2} â€“ 2x = 0

x(x – 2) = 0

So, x = 0, 2

Now, the area of the required region is given as

**17. Find the area bounded by the curve y = sin x between x = 0 and x = 2p.**

**Solution: **

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