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All questions of Chapter 8 - Application of Integrals for JEE Exam
The area shaded in the given figure can be calculated by which of the following definite integral?
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The area shaded in the given figure can be calculated by which of the following definite integral?
a)
b)
c)
d)
|
|
Leelu Bhai answered |
The required area is calculated by the difference of area of upper curve and lower curve.Now equation of upper curve
The area of the region bounded by y = x2 – 2x and y = 4 – x2 is.- a)7 sq. units
- b)8 sq. units
- c)9 sq. units
- d)10 sq. units
Correct answer is option 'C'. Can you explain this answer?
The area of the region bounded by y = x2 – 2x and y = 4 – x2 is.
a)
7 sq. units
b)
8 sq. units
c)
9 sq. units
d)
10 sq. units
|
Vikas Kapoor answered |
To find area between y = x2 – 2x and y = 4 – x2,
We need to find POI.
x2 – 2x = 4 – x2
2x2 - 2x – 4 = 0
x2 – x – 2 = 0
(x-2)(x+1) = 0
x = -1, 2
= [(-2x3)/3 + x2 + 4x]x=2 - [(-2x3)/3 + x2 + 4x]x= -1
= -16/3 + 4 + 8 – (2/3 + 1 - 4)
= 9 sq units
We need to find POI.
x2 – 2x = 4 – x2
2x2 - 2x – 4 = 0
x2 – x – 2 = 0
(x-2)(x+1) = 0
x = -1, 2
= [(-2x3)/3 + x2 + 4x]x=2 - [(-2x3)/3 + x2 + 4x]x= -1
= -16/3 + 4 + 8 – (2/3 + 1 - 4)
= 9 sq units
Let f(x) = x + sin x. The area bounded by y = f-1 (x), y = x, x ∈ [0, π] is - a)1
- b)2
- c)3
- d)can not be found because f-1(x) cannot be determined
Correct answer is option 'B'. Can you explain this answer?
Let f(x) = x + sin x. The area bounded by y = f-1 (x), y = x, x ∈ [0, π] is
a)
1
b)
2
c)
3
d)
can not be found because f-1(x) cannot be determined
|
Geetika Shah answered |
The curves given by y = x + sin x and y = f-1(x) are images of each other in the line y = x.
Hence required area
Hence required area
The area enclosed between the lines x = 2 and x = 7 is- a)Infinite
- b)7 units
- c)5 units
- d)2 units
Correct answer is option 'A'. Can you explain this answer?
The area enclosed between the lines x = 2 and x = 7 is
a)
Infinite
b)
7 units
c)
5 units
d)
2 units
|
Manas Khantal answered |
See from graph of x=2 & x=7 it's simple
Direction: Read the following text and answer the following questions on the basis of the same:A farmer has a square plot of land. Three of its boundaries are x = 0, y = 0 and y = 4. He wants to divide this land among his three sons A, B and C as shown in figure.
Q. Area received by son B is _______ sq. units.- a)4
- b)16
- c)16/3
- d)8/3
Correct answer is option 'C'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
A farmer has a square plot of land. Three of its boundaries are x = 0, y = 0 and y = 4. He wants to divide this land among his three sons A, B and C as shown in figure.
Q. Area received by son B is _______ sq. units.
a)
4
b)
16
c)
16/3
d)
8/3
|
Vp Classes answered |
Let f : [0, ∞)→ R be a continuous and strictly increasing function such that 
The area enclosed by y = f (x), the x-axis and the ordinate at x = 3, is- a)1
- b)3/2
- c)2
- d)3
Correct answer is option 'B'. Can you explain this answer?
Let f : [0, ∞)→ R be a continuous and strictly increasing function such that The area enclosed by y = f (x), the x-axis and the ordinate at x = 3, is
The area enclosed by y = f (x), the x-axis and the ordinate at x = 3, isa)
1
b)
3/2
c)
2
d)
3
|
Om Desai answered |
Given, f3(x) = ∫(0 to x)tf(t)dt
Differentiating throughout w.r.t x,
3f2(x) f′(x) = xf2(x)
⇒ f′(x)= x/3
So, f(x)=∫(0 to x) x/3 dx = (x2)/6
Area = ∫f(x)dx=∫(0 to 3) (x2)/6 dx
= {(x3)/18}(0 to 3)
= 3/2
Differentiating throughout w.r.t x,
3f2(x) f′(x) = xf2(x)
⇒ f′(x)= x/3
So, f(x)=∫(0 to x) x/3 dx = (x2)/6
Area = ∫f(x)dx=∫(0 to 3) (x2)/6 dx
= {(x3)/18}(0 to 3)
= 3/2
The area bounded by the curve y2 = 16x , x = 1, y = 3 and X-axis is: - a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The area bounded by the curve y2 = 16x , x = 1, y = 3 and X-axis is:
a)
b)
c)
d)
|
Moksh Sharma answered |
Plese give detailed solution of this question
such that min {PA, PB, PC} = 1. The area formed by the curve traced by P is ....... sq. units
The area between the curves y = x2 and y = x3 is:
- a) 1/12 sq.units
- b) 1/8 sq.units
- c) 1/10 sq.unit
- d) 1/6 sq.units
Correct answer is option 'A'. Can you explain this answer?
The area between the curves y = x2 and y = x3 is:
a)
1/12 sq.unitsb)
1/8 sq.unitsc)
1/10 sq.unitd)
1/6 sq.units
|
Arun Khanna answered |
First we note that the curves intersect at the points (0,0) and (1,1). Then we see that
x^3 < x^2
in this interval. Hence the area is given by
= 1/3 - 1/4 = 1/12.
The area enclosed between the curves y = √x , x = 2y+3and the x-axis is- a)9
- b)18
- c)27
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The area enclosed between the curves y = √x , x = 2y+3and the x-axis is
a)
9
b)
18
c)
27
d)
none of these
|
Manisha Mehta answered |
The two curves meet where;
Therefore, the two curves meet where x = 9.
Therefore,required Area:
Therefore, the two curves meet where x = 9.
Therefore,required Area:
is : 
Can you explain the answer of this question below:Let f(x) be a continuous function such that the area bounded by the curve y = f(x), the xaxis and the two ordinates x = 0 and 
- A:
1/2
- B:
- C:
- D:
The answer is a.
Let f(x) be a continuous function such that the area bounded by the curve y = f(x), the xaxis and the two ordinates x = 0 and
1/2
|
Prisha Yadav answered |
Differentiating both sides w.r.t. a, we get
The area bounded by the curve y = x[x], the x – axis and the ordinates x = 1 and x = -1 is given by- a)0
- b)1/2
- c)2/3
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The area bounded by the curve y = x[x], the x – axis and the ordinates x = 1 and x = -1 is given by
a)
0
b)
1/2
c)
2/3
d)
none of these
|
Shruti Sharma answered |
Understanding the Problem
To find the area bounded by the curve y = x[x], the x-axis, and the vertical lines x = -1 and x = 1, we first need to determine the behavior of the function y = x[x].
Analyzing the Function
- The function y = x[x] involves the greatest integer function [x], which takes the largest integer less than or equal to x.
- For x in the interval [-1, 0), [x] = -1, resulting in y = x * (-1) = -x.
- For x in the interval [0, 1), [x] = 0, resulting in y = x * 0 = 0.
This means we can break the area calculation into two parts:
Calculating Areas
1. For x in [-1, 0):
- The function simplifies to y = -x.
- The area under this curve from x = -1 to x = 0 is given by the integral:
Area = ∫[-1 to 0] (-x) dx = [-(x^2)/2] from -1 to 0 = [0 - (-1/2)] = 1/2.
2. For x in [0, 1):
- The function simplifies to y = 0.
- The area under this curve from x = 0 to x = 1 is 0.
Total Area Calculation
- The total area bounded by the curve, the x-axis, and the ordinates is simply the area calculated from the interval [-1, 0], which is:
Total Area = Area from [-1, 0] + Area from [0, 1] = 1/2 + 0 = 1/2.
Conclusion
Thus, the area bounded by the curve y = x[x], the x-axis, and the ordinates x = -1 and x = 1 is indeed 1/2, confirming that the correct answer is option 'B'.
To find the area bounded by the curve y = x[x], the x-axis, and the vertical lines x = -1 and x = 1, we first need to determine the behavior of the function y = x[x].
Analyzing the Function
- The function y = x[x] involves the greatest integer function [x], which takes the largest integer less than or equal to x.
- For x in the interval [-1, 0), [x] = -1, resulting in y = x * (-1) = -x.
- For x in the interval [0, 1), [x] = 0, resulting in y = x * 0 = 0.
This means we can break the area calculation into two parts:
Calculating Areas
1. For x in [-1, 0):
- The function simplifies to y = -x.
- The area under this curve from x = -1 to x = 0 is given by the integral:
Area = ∫[-1 to 0] (-x) dx = [-(x^2)/2] from -1 to 0 = [0 - (-1/2)] = 1/2.
2. For x in [0, 1):
- The function simplifies to y = 0.
- The area under this curve from x = 0 to x = 1 is 0.
Total Area Calculation
- The total area bounded by the curve, the x-axis, and the ordinates is simply the area calculated from the interval [-1, 0], which is:
Total Area = Area from [-1, 0] + Area from [0, 1] = 1/2 + 0 = 1/2.
Conclusion
Thus, the area bounded by the curve y = x[x], the x-axis, and the ordinates x = -1 and x = 1 is indeed 1/2, confirming that the correct answer is option 'B'.
Can you explain the answer of this question below:Area of region bounded by x2 + y2 < 4 and (|x| + |y|) < 2 is ____ square units.
- A:
4
- B:
6
- C:
8
- D:
10
The answer is c.
Area of region bounded by x2 + y2 < 4 and (|x| + |y|) < 2 is ____ square units.
4
6
8
10
|
Akshita Nair answered |
Using circles representation, required area = 8 sq. units
Direction: Read the following text and answer the following questions on the basis of the same:A farmer has a square plot of land. Three of its boundaries are x = 0, y = 0 and y = 4. He wants to divide this land among his three sons A, B and C as shown in figure.
Q. Area received by son A is _______ sq. units.- a)4
- b)16
- c)16/3
- d)8/3
Correct answer is option 'C'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
A farmer has a square plot of land. Three of its boundaries are x = 0, y = 0 and y = 4. He wants to divide this land among his three sons A, B and C as shown in figure.
Q. Area received by son A is _______ sq. units.
a)
4
b)
16
c)
16/3
d)
8/3
|
|
Anjali Sharma answered |
Direction: Read the following text and answer the following questions on the basis of the same:
The bridge connects two hills 100 feet apart. The arch on the bridge is in a parabolic form. The highest point on the bridge is 10 feet above the road at the middle of the bridge as seen in the figure.Q. The value of the integral
- a)1000/3
- b)250/3
- c)1200
- d)0
Correct answer is option 'A'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
The bridge connects two hills 100 feet apart. The arch on the bridge is in a parabolic form. The highest point on the bridge is 10 feet above the road at the middle of the bridge as seen in the figure.
Q. The value of the integral
a)
1000/3
b)
250/3
c)
1200
d)
0
|
|
Anjali Sharma answered |
The area bounded by the angle bisectors of the lines x2−y2+2y = 1 and x+y = 3 is- a)6 sq. units
- b)2 sq. units
- c)3 sq.units
- d)4 sq. units
Correct answer is option 'B'. Can you explain this answer?
The area bounded by the angle bisectors of the lines x2−y2+2y = 1 and x+y = 3 is
a)
6 sq. units
b)
2 sq. units
c)
3 sq.units
d)
4 sq. units
|
Jatin Dasgupta answered |
The angle bisectors of the line given by x2−− y2+2y = 1 are x = 0 , y = 1. Required area = 1/2 .2.2 = 2 sq. units
The area bounded by the curve y = 2x - x2 and the line x + y = 0 is- a)9/2 sq. units
- b)35/6 sq. units
- c)19/6 sq. units
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The area bounded by the curve y = 2x - x2 and the line x + y = 0 is
a)
9/2 sq. units
b)
35/6 sq. units
c)
19/6 sq. units
d)
none of these
|
Pranav Pillai answered |
The equation y = 2x − x2 i.e. y – 1 = - (x - 1)2 represents a downward parabola with vertex at (1, 1) which meets x – axis where y = 0 .i .e . where x = 0 , 2. Also , the line y = - x meets this parabola where – x = 2x − x2 i.e. where x = 0 , 3.
Therefore , required area is :
Therefore , required area is :
The area bounded by [x] +[y] = 8 such that x, y > 0 is .... sq. unitsWhere [.] is G.I.F.- a)3
- b)6
- c)9
- d)12
Correct answer is option 'C'. Can you explain this answer?
The area bounded by [x] +[y] = 8 such that x, y > 0 is .... sq. units
Where [.] is G.I.F.
a)
3
b)
6
c)
9
d)
12
|
Jaya Das answered |
Given Equation:
The equation of the line is given as [x] + [y] = 8. Here, [.] denotes the greatest integer function.
Finding the Area:
To find the area bounded by this equation, we need to visualize the region formed by the line [x] + [y] = 8 in the first quadrant where x and y are greater than or equal to 0.
Understanding the Region:
The line [x] + [y] = 8 passes through the points (8,0) and (0,8). This line divides the first quadrant into two regions.
One region is a triangle with vertices at (0,0), (8,0), and (0,8). The area of this triangle is 1/2 * base * height = 1/2 * 8 * 8 = 32 sq. units.
The other region is a square with side length 8 units. The area of this square is 8 * 8 = 64 sq. units.
Calculating the Bounded Area:
The area bounded by the line [x] + [y] = 8 in the first quadrant is the sum of the areas of the triangle and the square, which is 32 + 64 = 96 sq. units.
Therefore, the area bounded by [x] + [y] = 8 such that x, y are greater than or equal to 0 is 96 sq. units.
Conclusion:
The correct option is (C) 9 sq. units.
The equation of the line is given as [x] + [y] = 8. Here, [.] denotes the greatest integer function.
Finding the Area:
To find the area bounded by this equation, we need to visualize the region formed by the line [x] + [y] = 8 in the first quadrant where x and y are greater than or equal to 0.
Understanding the Region:
The line [x] + [y] = 8 passes through the points (8,0) and (0,8). This line divides the first quadrant into two regions.
One region is a triangle with vertices at (0,0), (8,0), and (0,8). The area of this triangle is 1/2 * base * height = 1/2 * 8 * 8 = 32 sq. units.
The other region is a square with side length 8 units. The area of this square is 8 * 8 = 64 sq. units.
Calculating the Bounded Area:
The area bounded by the line [x] + [y] = 8 in the first quadrant is the sum of the areas of the triangle and the square, which is 32 + 64 = 96 sq. units.
Therefore, the area bounded by [x] + [y] = 8 such that x, y are greater than or equal to 0 is 96 sq. units.
Conclusion:
The correct option is (C) 9 sq. units.
The area bounded by the curves y2 = 4x and y = x is equal to- a)8/3
- b)35/6
- c)1/3
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The area bounded by the curves y2 = 4x and y = x is equal to
a)
8/3
b)
35/6
c)
1/3
d)
none of these
|
|
Pranav Pillai answered |
The two curves y2 = 4x and y = x meet where x2 = 4x i.e ..where x = 0 or x = 4 . Moreover , the parabola lies above the line y = x between x = 0 and x = 4 . Hence , the required are is :
Area of the region bounded by the curves y = ex, x = a , x = b and the x- axis is given by- a)eb−ea
- b)eb−a
- c)eb+ea
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Area of the region bounded by the curves y = ex, x = a , x = b and the x- axis is given by
a)
eb−ea
b)
eb−a
c)
eb+ea
d)
none of these
|
Nisha Sen answered |
To find the area of the region bounded by the curves y = ex, x = a, x = b, and the x-axis, we can use integration.
Step 1: Determine the limits of integration
The region is bounded by the vertical lines x = a and x = b, so the limits of integration for x will be from a to b.
Step 2: Set up the integral
The area of the region can be found by integrating the function y = ex with respect to x over the given limits of integration.
The integral for the area is given by:
∫(ex) dx from a to b
Step 3: Evaluate the integral
Integrating the function ex with respect to x gives us ex.
So, the integral becomes:
∫(ex) dx = ex
Evaluating this integral from a to b gives us:
ex evaluated from a to b = eb - ea
Step 4: Subtracting the areas below the x-axis
Since the region is bounded by the x-axis, we need to consider the areas below the x-axis as negative.
If there are any values of x within the interval (a, b) where ex is negative, we need to subtract the absolute value of the integral of ex in that range. However, since the function ex is always positive, we don't need to consider this step.
Step 5: Finalize the answer
The area of the region bounded by the curves y = ex, x = a, x = b, and the x-axis is given by:
Area = eb - ea
Therefore, the correct answer is option A) eb - ea.
Step 1: Determine the limits of integration
The region is bounded by the vertical lines x = a and x = b, so the limits of integration for x will be from a to b.
Step 2: Set up the integral
The area of the region can be found by integrating the function y = ex with respect to x over the given limits of integration.
The integral for the area is given by:
∫(ex) dx from a to b
Step 3: Evaluate the integral
Integrating the function ex with respect to x gives us ex.
So, the integral becomes:
∫(ex) dx = ex
Evaluating this integral from a to b gives us:
ex evaluated from a to b = eb - ea
Step 4: Subtracting the areas below the x-axis
Since the region is bounded by the x-axis, we need to consider the areas below the x-axis as negative.
If there are any values of x within the interval (a, b) where ex is negative, we need to subtract the absolute value of the integral of ex in that range. However, since the function ex is always positive, we don't need to consider this step.
Step 5: Finalize the answer
The area of the region bounded by the curves y = ex, x = a, x = b, and the x-axis is given by:
Area = eb - ea
Therefore, the correct answer is option A) eb - ea.
The area of the part of circle x2 + y2 - 2x - 4y -1 = 0 above 2x - y = 0 is ...... sq. units- a) 3π
- b)2π
- c)6π
- d)π
Correct answer is option 'A'. Can you explain this answer?
The area of the part of circle x2 + y2 - 2x - 4y -1 = 0 above 2x - y = 0 is ...... sq. units
a)
3π
b)
2π
c)
6π
d)
π
|
Sarthak Khanna answered |
Required area = Area of semi-circle
Since given line is diameter of circle
Since given line is diameter of circle
Direction: Read the following text and answer the following questions on the basis of the same:
The bridge connects two hills 100 feet apart. The arch on the bridge is in a parabolic form. The highest point on the bridge is 10 feet above the road at the middle of the bridge as seen in the figure.Q. The area formed by the curve x2 = 250y, x-axis, y = 0 and y = 10 is- a)1000√2/3
- b)4/3
- c)1000/3
- d)0
Correct answer is option 'C'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
The bridge connects two hills 100 feet apart. The arch on the bridge is in a parabolic form. The highest point on the bridge is 10 feet above the road at the middle of the bridge as seen in the figure.
Q. The area formed by the curve x2 = 250y, x-axis, y = 0 and y = 10 is
a)
1000√2/3
b)
4/3
c)
1000/3
d)
0
|
Mohit Rajpoot answered |
X2 = 250 y
at = y = 0 x = 0
at y = 10 x = 50, -50
∴ Area formed by curve
The area of the region bounded by the parabola (y−2)2 = x−1,the tangent to yhe parabola at the point (2 , 3) and the x – axis is equal to- a)12 sq. units
- b)6 sq. units
- c)9 sq. units
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
The area of the region bounded by the parabola (y−2)2 = x−1,the tangent to yhe parabola at the point (2 , 3) and the x – axis is equal to
a)
12 sq. units
b)
6 sq. units
c)
9 sq. units
d)
none of these
|
Mohit Patel answered |
Therefore, tangent at (2, 3) is y – 3 = ½ (x – 2). i.e. x – 2y +4 = 0 . therefore required area is
Let y be the function which passes through (1 , 2) having slope (2x + 1) . The area bounded between the curve and the x – axis is- a)1/6 sq. units
- b)6 sq. units
- c)5/6 sq. units
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Let y be the function which passes through (1 , 2) having slope (2x + 1) . The area bounded between the curve and the x – axis is
a)
1/6 sq. units
b)
6 sq. units
c)
5/6 sq. units
d)
none of these
|
Rhea Joshi answered |
Given slope of the curve is 2x + 1.
Also , it passes through (1, 2).
∴ 2 = 1+1+c ⇒ c = 0
Equation of curve is : y = x2 + x. Therefore , points of intersection of y = x (x +1) and the x – axis are x = 0 , x = - 1.
Required area :
Also , it passes through (1, 2).
∴ 2 = 1+1+c ⇒ c = 0
Equation of curve is : y = x2 + x. Therefore , points of intersection of y = x (x +1) and the x – axis are x = 0 , x = - 1.
Required area :
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