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Exercise 8.1 : Application of Integrals NCERT Solutions | Mathematics (Maths) Class 12 - JEE PDF Download

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 Page 1


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
 
Exercise 8.1                                                                Page No: 365 
 
1. Find the area of the region bounded by the curve  and the lines x=1, x = 4  and 
the x- axis in the first quadrant. 
Solution: Equation of the curve (rightward parabola) is  
 
 
 
  ……….(1) 
Required area is shaded region: 
=  [From equation (1)] 
=   
=   
=  
Page 2


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
 
Exercise 8.1                                                                Page No: 365 
 
1. Find the area of the region bounded by the curve  and the lines x=1, x = 4  and 
the x- axis in the first quadrant. 
Solution: Equation of the curve (rightward parabola) is  
 
 
 
  ……….(1) 
Required area is shaded region: 
=  [From equation (1)] 
=   
=   
=  
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
=  =  =  sq. units 
2. Find the area of the region bounded by  and the x-axis in the first 
quadrant. 
Solution: Equation of the curve (rightward parabola) is  
  ……….(1) 
 
 
Required area is shaded region, which is bounded by curve   and vertical lines x=2, 
x=4 and x-axis in first quadrant. 
=  [From equation (1)] 
=  =  
=  =  
Page 3


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
 
Exercise 8.1                                                                Page No: 365 
 
1. Find the area of the region bounded by the curve  and the lines x=1, x = 4  and 
the x- axis in the first quadrant. 
Solution: Equation of the curve (rightward parabola) is  
 
 
 
  ……….(1) 
Required area is shaded region: 
=  [From equation (1)] 
=   
=   
=  
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
=  =  =  sq. units 
2. Find the area of the region bounded by  and the x-axis in the first 
quadrant. 
Solution: Equation of the curve (rightward parabola) is  
  ……….(1) 
 
 
Required area is shaded region, which is bounded by curve   and vertical lines x=2, 
x=4 and x-axis in first quadrant. 
=  [From equation (1)] 
=  =  
=  =  
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
=  =  sq. units 
 
3. Find the area of the region bounded by  and the y- axis in the first 
quadrant. 
Solution: Equation of curve (parabola) is   
or  ……….(1) 
Required region is shaded, that is area bounded by curve  and Horizontal 
lines  and y-axis in first quadrant. 
 
=  
=  =  =  sq. units 
4. Find the area of the region bounded by the ellipse  
Solution: Equation of ellipse is  ……….(1) 
Page 4


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
 
Exercise 8.1                                                                Page No: 365 
 
1. Find the area of the region bounded by the curve  and the lines x=1, x = 4  and 
the x- axis in the first quadrant. 
Solution: Equation of the curve (rightward parabola) is  
 
 
 
  ……….(1) 
Required area is shaded region: 
=  [From equation (1)] 
=   
=   
=  
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
=  =  =  sq. units 
2. Find the area of the region bounded by  and the x-axis in the first 
quadrant. 
Solution: Equation of the curve (rightward parabola) is  
  ……….(1) 
 
 
Required area is shaded region, which is bounded by curve   and vertical lines x=2, 
x=4 and x-axis in first quadrant. 
=  [From equation (1)] 
=  =  
=  =  
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
=  =  sq. units 
 
3. Find the area of the region bounded by  and the y- axis in the first 
quadrant. 
Solution: Equation of curve (parabola) is   
or  ……….(1) 
Required region is shaded, that is area bounded by curve  and Horizontal 
lines  and y-axis in first quadrant. 
 
=  
=  =  =  sq. units 
4. Find the area of the region bounded by the ellipse  
Solution: Equation of ellipse is  ……….(1) 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
 
 
 
 
 
Here  
From equation (1), 
  
 ……….(2) 
for arc of ellipse in first quadrant. 
Ellipse (1) is symmetrical about x-axis and about y-axis (if we change y to –y or x to –x, 
equation remain same). 
Intersections of ellipse (1) with x-axis (y=0) 
Put y=0 in equation (1), we have 
    
Therefore, Intersections of ellipse (1) with x-axis are (0, 4) and (0, -4). 
 
Now again, 
Intersections of ellipse (1) with y-axis (x=0) 
 
Page 5


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
 
Exercise 8.1                                                                Page No: 365 
 
1. Find the area of the region bounded by the curve  and the lines x=1, x = 4  and 
the x- axis in the first quadrant. 
Solution: Equation of the curve (rightward parabola) is  
 
 
 
  ……….(1) 
Required area is shaded region: 
=  [From equation (1)] 
=   
=   
=  
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
=  =  =  sq. units 
2. Find the area of the region bounded by  and the x-axis in the first 
quadrant. 
Solution: Equation of the curve (rightward parabola) is  
  ……….(1) 
 
 
Required area is shaded region, which is bounded by curve   and vertical lines x=2, 
x=4 and x-axis in first quadrant. 
=  [From equation (1)] 
=  =  
=  =  
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
=  =  sq. units 
 
3. Find the area of the region bounded by  and the y- axis in the first 
quadrant. 
Solution: Equation of curve (parabola) is   
or  ……….(1) 
Required region is shaded, that is area bounded by curve  and Horizontal 
lines  and y-axis in first quadrant. 
 
=  
=  =  =  sq. units 
4. Find the area of the region bounded by the ellipse  
Solution: Equation of ellipse is  ……….(1) 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
 
 
 
 
 
Here  
From equation (1), 
  
 ……….(2) 
for arc of ellipse in first quadrant. 
Ellipse (1) is symmetrical about x-axis and about y-axis (if we change y to –y or x to –x, 
equation remain same). 
Intersections of ellipse (1) with x-axis (y=0) 
Put y=0 in equation (1), we have 
    
Therefore, Intersections of ellipse (1) with x-axis are (0, 4) and (0, -4). 
 
Now again, 
Intersections of ellipse (1) with y-axis (x=0) 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
Integrals 
 
 
Putting x=0 in equation (1),    
Therefore, Intersections of ellipse (1) with y-axis are (0, 3) and (0,-3). 
 
Now , 
Area of region bounded by ellipse (1) = 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 
5. Find the area of the region bounded by the ellipse  
Solution: Equation of ellipse is  
 
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FAQs on Exercise 8.1 : Application of Integrals NCERT Solutions - Mathematics (Maths) Class 12 - JEE

1. How to find the area under a curve using integration?
Ans. To find the area under a curve using integration, you need to integrate the given function over the specified interval. This involves finding the antiderivative of the function and evaluating it at the upper and lower limits of the interval, then subtracting the two values.
2. What is the significance of the definite integral in real-life applications?
Ans. The definite integral is significant in real-life applications as it helps in finding quantities such as area, volume, work done, and displacement. It plays a crucial role in various fields like physics, engineering, economics, and biology.
3. How can integration be used to calculate the total distance traveled by an object?
Ans. Integration can be used to calculate the total distance traveled by an object by integrating the absolute value of the velocity function over the given time interval. The result gives the total distance covered by the object irrespective of the direction of motion.
4. Why is it important to understand the concept of integration in calculus?
Ans. Understanding the concept of integration in calculus is essential as it allows us to solve a wide range of problems related to area, volume, motion, and accumulation. It provides a powerful tool for analyzing and modeling real-world phenomena.
5. How can the definite integral be interpreted geometrically?
Ans. Geometrically, the definite integral represents the signed area between the curve and the x-axis over a specified interval. If the function is above the x-axis, the integral gives the area above the x-axis, and if the function is below the x-axis, the integral gives the area below the x-axis (considered negative).
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