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**Download NCERT solutions for Applications of Integrals as PDF.**

## NCERT Solutions class 12 Maths Applications of Integrals

**1. Find the area of the region bounded by the curve **** and the lines **** and the **** axis.**

**Ans. **Equation of the curve (rightward parabola) is

……….(i)

Required area (shaded region)

= [From eq. (i)]

= = =

= = = sq. units

### NCERT Solutions class 12 Maths Exercise 8.1

**2. Find the area of the region bounded by **** and the ****axis in the first quadrant.**

**Ans. **Equation of the curve (rightward parabola) is

……….(i)

Required area (shaded region) bounded by curve (vertical lines ) and axis in first quadrant.

= [From eq. (i)]

= =

= =

= = sq. units

### NCERT Solutions class 12 Maths Exercise 8.1

**3. Find the area of the region bounded by **** and the **** axis in the first quadrant.**

**Ans. **Equation of curve (parabola) is ……….(i)

Required (shaded) area bounded by curve (Horizontal lines ) and axis in first quadrant.

=

= = = sq. units

### NCERT Solutions class 12 Maths Exercise 8.1

**4. Find the area of the region bounded by the ellipse **** **

**Ans. **Equation of ellipse is ……….(i)

Here

From eq. (i),

……….(ii)

for arc of ellipse in first quadrant.

Ellipse (i) ia symmetrical about axis,

( On changing in eq. (i), it remains unchanged)

Ellipse (i) ia symmetrical about axis,

( On changing in eq. (i), it remains unchanged)

**Intersections of ellipse (i) with axis **

Putting in eq. (i),

Therefore, Intersections of ellipse (i) with axis are (0, 4) and .

**Intersections of ellipse (i) with axis **

Putting in eq. (i),

Therefore, Intersections of ellipse (i) with axis are (0, 3) and .

Now Area of region bounded by ellipse (i) = Total shaded area

= 4 x Area OAB of ellipse in first quadrant

= [ At end B of arc AB of ellipse; and at end A of arc AB ; ]

= =

=

= =

= sq. units

### NCERT Solutions class 12 Maths Exercise 8.1

**5. Find the area of the region bounded by the ellipse **

**Ans. **Equation of ellipse is

Here

From eq. (i),

……….(ii)

for arc of ellipse in first quadrant.

Ellipse (i) ia symmetrical about axis,

( On changing in eq. (i), it remains unchanged)

Ellipse (i) ia symmetrical about axis,

( On changing in eq. (i), it remains unchanged)

**Intersections of ellipse (i) with axis **

Putting in eq. (i),

Therefore, Intersections of ellipse (i) with axis are (0, 2) and .

**Intersections of ellipse (i) with axis **

Putting in eq. (i),

Therefore, Intersections of ellipse (i) with axis are (0, 3) and .

Now Area of region bounded by ellipse (i) = Total shaded area

= 4 x Area OAB of ellipse in first quadrant

= [ At end B of arc AB of ellipse; and at end A of arc AB ; ]

= =

=

=

= sq. units

### NCERT Solutions class 12 Maths Exercise 8.1

**6. Find the area of the region in the first quadrant enclosed by ****axis, line **** and the circle **** **

**Ans. ****Step I**. To draw the graphs and shade the region whose area we are to find.

Equation of the circle is ……….(i)

We know that eq. (i) represents a circle whose centre is (0, 0) and radius is 2.

Equation of the given line is

……….(ii)

We know that eq. (ii) being of the form where represents a straight line passing through the origin and making angle of with axis.

**Step II**. To find values of and Putting from eq. (ii) in eq. (i),

Putting in , and

Therefore, the two points of intersections of circle (i) and line (ii) are A and D.

**Step III**. Now shaded area OAM between segment OA of line (ii) and axis

=

= = = = = sq. units……….(iii)

**Step IV**. Now shaded area AMB between are AB of circle and axis.

= =

= From eq. (ii),

= =

= = = sq. units……….(iv)

**Step V**. Required shaded area OAB = Area of OAM + Area of AMB

= sq. units

### NCERT Solutions class 12 Maths Exercise 8.1

**7. Find the area of the smaller part of the circle **** cut off by the line **** **

**Ans. **Equation of the circle is ……….(i)

……….(ii)

Here Area of smaller part of the circle cut off by the line = Area of ABMC = 2 x Area of ABM

= = [From eq. (ii)]

=

=

=

=

=

=

=

= sq. units

### NCERT Solutions class 12 Maths Exercise 8.1

**8. The area between **** and **** is divided into two equal parts by the line **** find the value of **** **

**Ans. **Equation of the curve (parabola) is ……(i)

Now area bounded by parabola (i) and vertical line is divided into two equal parts by the vertical line

Area OAMB = Area AMBDNC

=

=

=

### NCERT Solutions class 12 Maths Exercise 8.1

**9. Find the area of the region bounded by the parabola **** and **** **

**Ans. **The required area is the area included between the parabola and the modulus function

To find: Area between the parabola and the ray for

Here, Limits of integration

Now, for , table of values,

if if

0 | 1 | 2 | |

0 | 1 | 2 |

0 | |||

0 | 1 | 2 |

Now, Area between parabola and axis between limits and

= = = ………..(i)

And Area of ray and axis,

= = = ………..(ii)

Required shaded area in first quadrant = Area between ray for and axis –

Area between parabola and axis in first quadrant

= Area given by eq. (ii) – Area given by eq. (i) = sq. units

### NCERT Solutions class 12 Maths Exercise 8.1

**10. Find the area bounded by the curve **** and the line **** **

**Ans. ****Step I. **Graphs and region of

Integration

Equation of the given curve is

……..(i)

Equation of the given line is

……….(ii)

0 | 1 | ||

0 | 0 |

**Step II**. Putting from eq. (i) in eq. (ii),

or

For ,from eq. (i), (2, 1)

For from eq. (i),

Therefore, the two points of intersection of parabola (i) and line (ii) are C and D (2, 1).

**Step III**. Area CMOEDN between parabola (i) and axis = =

= = = sq. units……..(iii)

**Step IV**. Area of trapezium CMND between line (ii) and axis = =

= = = =

= = sq. units……….(iv)

Required shaded area = Area given by eq. (iv) – Area given by eq. (iii)

= sq. units

### NCERT Solutions class 12 Maths Exercise 8.1

**11. Find the area of the region bounded by the curve **** and the line **** **

**Ans. **Equation of the (parabola) curve is

………(i)

………(ii)

Here required shaded area OAMB = 2 x Area OAM

= = =

= = = sq. units

**12. Choose the correct answer:**

**Area lying in the first quadrant and bounded by the circle and the lines and is**

**(A) **

**(B) **

**(C) **

**(D) **

**Ans. **Equation of the circle is ……….(i)

……….(ii)

Required area = =

=

=

= = sq. units

Therefore, option (A) is correct.

**13. Choose the correct answer:**

**Area of the region bounded by the curve axis and the line is:**

**(A) 2**

**(B) **

**(C) **

**(D) **

**Ans. **Equation of the curve (parabola) is

Required area = Area OAM = =

= = sq. units

Therefore, option (B) is correct.

## NCERT Solutions class 12 Maths Exercise 8.1

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