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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
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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 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
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
|
Anjali Sharma answered |
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
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.
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 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 : 
The area bounded by the curves y = x2, y = [x+1], x < 1 and the y - axis, where[.] denotes the greatest integer not exceeding x, is- a)2/3
- b)1/3
- c)1
- d)2
Correct answer is option 'A'. Can you explain this answer?
The area bounded by the curves y = x2, y = [x+1], x < 1 and the y - axis, where[.] denotes the greatest integer not exceeding x, is
a)
2/3
b)
1/3
c)
1
d)
2
|
Snehal Mehta answered |
To find the area bounded by the given curves, we need to determine the points of intersection between the curves and the boundaries of the region.
1. Finding the point of intersection:
Let's first find the point of intersection between the curves y = x^2 and y = [x].
For y = x^2, we have:
x^2 = [x]
To find the intersection points, we need to consider two cases:
Case 1: When x is an integer
In this case, [x] will be equal to x. So, the equation becomes:
x^2 = x
x^2 - x = x(x - 1) = 0
This gives us two solutions: x = 0 and x = 1.
Case 2: When x is not an integer
In this case, [x] will be equal to the greatest integer less than or equal to x. So, the equation becomes:
x^2 = x - 1
x^2 - x + 1 = 0
Using the quadratic formula, we can solve for x, but we can see that the discriminant (b^2 - 4ac) is negative, which means there are no real solutions.
Therefore, the only points of intersection are (0, 0) and (1, 1).
2. Determining the boundaries of the region:
The given boundaries are y = [x], y = 0, and x = 1.
Since y = [x] represents the greatest integer less than or equal to x, it means that y will be an integer for all values of x between two consecutive integers. Therefore, the boundary y = [x] is equivalent to y = 0 for x < 1="" and="" y="1" for="" 1="" ≤="" x="" />< />
3. Calculating the area:
The area bounded by the curves can be divided into two regions:
Region 1: Bounded by y = x^2, y = 0, and x = 1
This region is a right-angled triangle with base 1 and height 1 (the point of intersection (1, 1)). Therefore, the area of this region is 1/2 * base * height = 1/2 * 1 * 1 = 1/2.
Region 2: Bounded by y = x^2, y = 1, and x = 2
This region is a trapezium with bases of lengths 2 and 1 (the points of intersection (0, 0) and (1, 1)). The height of the trapezium is 1 (the difference between the y-coordinates of the points of intersection). Therefore, the area of this region is 1/2 * (base1 + base2) * height = 1/2 * (2 + 1) * 1 = 3/2.
Total area = Area of Region 1 + Area of Region 2 = 1/2 + 3/2 = 2/3.
Therefore, the correct answer is option A, 2/3.
1. Finding the point of intersection:
Let's first find the point of intersection between the curves y = x^2 and y = [x].
For y = x^2, we have:
x^2 = [x]
To find the intersection points, we need to consider two cases:
Case 1: When x is an integer
In this case, [x] will be equal to x. So, the equation becomes:
x^2 = x
x^2 - x = x(x - 1) = 0
This gives us two solutions: x = 0 and x = 1.
Case 2: When x is not an integer
In this case, [x] will be equal to the greatest integer less than or equal to x. So, the equation becomes:
x^2 = x - 1
x^2 - x + 1 = 0
Using the quadratic formula, we can solve for x, but we can see that the discriminant (b^2 - 4ac) is negative, which means there are no real solutions.
Therefore, the only points of intersection are (0, 0) and (1, 1).
2. Determining the boundaries of the region:
The given boundaries are y = [x], y = 0, and x = 1.
Since y = [x] represents the greatest integer less than or equal to x, it means that y will be an integer for all values of x between two consecutive integers. Therefore, the boundary y = [x] is equivalent to y = 0 for x < 1="" and="" y="1" for="" 1="" ≤="" x="" />< />
3. Calculating the area:
The area bounded by the curves can be divided into two regions:
Region 1: Bounded by y = x^2, y = 0, and x = 1
This region is a right-angled triangle with base 1 and height 1 (the point of intersection (1, 1)). Therefore, the area of this region is 1/2 * base * height = 1/2 * 1 * 1 = 1/2.
Region 2: Bounded by y = x^2, y = 1, and x = 2
This region is a trapezium with bases of lengths 2 and 1 (the points of intersection (0, 0) and (1, 1)). The height of the trapezium is 1 (the difference between the y-coordinates of the points of intersection). Therefore, the area of this region is 1/2 * (base1 + base2) * height = 1/2 * (2 + 1) * 1 = 3/2.
Total area = Area of Region 1 + Area of Region 2 = 1/2 + 3/2 = 2/3.
Therefore, the correct answer is option A, 2/3.
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:
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 |
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 |
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 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 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 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
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
|
Nandini Kumar answered |
To find the area of the part of the circle above the line, we first need to determine the points of intersection between the circle and the line.
1. Finding the points of intersection:
We have two equations:
- Circle equation: x^2 + y^2 - 2x - 4y - 1 = 0
- Line equation: 2x - y = 0
To find the points of intersection, we can substitute the value of y from the line equation into the circle equation:
x^2 + (2x)^2 - 2x - 4(2x) - 1 = 0
x^2 + 4x^2 - 2x - 8x - 1 = 0
5x^2 - 10x - 1 = 0
Using the quadratic formula, we can solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
where a = 5, b = -10, and c = -1.
Calculating the values of x using the quadratic formula, we get:
x = (-(-10) ± √((-10)^2 - 4(5)(-1))) / (2(5))
x = (10 ± √(100 + 20)) / 10
x = (10 ± √120) / 10
x = (10 ± 2√30) / 10
x = 1 ± 0.632
So, the two x-values of intersection are x = 1.632 and x = 0.368.
Now, substitute these x-values back into the line equation to find the corresponding y-values:
For x = 1.632: 2(1.632) - y = 0
y = 3.264
For x = 0.368: 2(0.368) - y = 0
y = 0.736
2. Calculating the area:
The area of the part of the circle above the line can be calculated by finding the area of the sector formed by the two intersection points and subtracting the area of the triangle formed by the two intersection points and the origin (0,0).
- Area of the sector:
The angle formed by the two intersection points and the origin is the angle at the center of the circle. This angle can be calculated using trigonometry:
θ = 2 * arctan(y/x)
For x = 1.632 and y = 3.264:
θ = 2 * arctan(3.264/1.632)
θ ≈ 2.312 radians
The area of the sector is given by:
A_sector = (θ/2π) * πr^2
where r is the radius of the circle.
Since the equation of the circle is x^2 + y^2 - 2x - 4y - 1 = 0, we can complete the square to find the radius:
(x^2 - 2x + 1) + (y^2 - 4y + 4) = 6
(x - 1)^2 + (y - 2)^2 = 6
Comparing this to the standard equation of a circle (x - h)^2 + (y - k)^
1. Finding the points of intersection:
We have two equations:
- Circle equation: x^2 + y^2 - 2x - 4y - 1 = 0
- Line equation: 2x - y = 0
To find the points of intersection, we can substitute the value of y from the line equation into the circle equation:
x^2 + (2x)^2 - 2x - 4(2x) - 1 = 0
x^2 + 4x^2 - 2x - 8x - 1 = 0
5x^2 - 10x - 1 = 0
Using the quadratic formula, we can solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
where a = 5, b = -10, and c = -1.
Calculating the values of x using the quadratic formula, we get:
x = (-(-10) ± √((-10)^2 - 4(5)(-1))) / (2(5))
x = (10 ± √(100 + 20)) / 10
x = (10 ± √120) / 10
x = (10 ± 2√30) / 10
x = 1 ± 0.632
So, the two x-values of intersection are x = 1.632 and x = 0.368.
Now, substitute these x-values back into the line equation to find the corresponding y-values:
For x = 1.632: 2(1.632) - y = 0
y = 3.264
For x = 0.368: 2(0.368) - y = 0
y = 0.736
2. Calculating the area:
The area of the part of the circle above the line can be calculated by finding the area of the sector formed by the two intersection points and subtracting the area of the triangle formed by the two intersection points and the origin (0,0).
- Area of the sector:
The angle formed by the two intersection points and the origin is the angle at the center of the circle. This angle can be calculated using trigonometry:
θ = 2 * arctan(y/x)
For x = 1.632 and y = 3.264:
θ = 2 * arctan(3.264/1.632)
θ ≈ 2.312 radians
The area of the sector is given by:
A_sector = (θ/2π) * πr^2
where r is the radius of the circle.
Since the equation of the circle is x^2 + y^2 - 2x - 4y - 1 = 0, we can complete the square to find the radius:
(x^2 - 2x + 1) + (y^2 - 4y + 4) = 6
(x - 1)^2 + (y - 2)^2 = 6
Comparing this to the standard equation of a circle (x - h)^2 + (y - k)^
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
|
|
Anjali Sharma answered |
X2 = 250 y
at = y = 0 x = 0
at y = 10 x = 50, -50
∴ Area formed by curve
The area bounded by the parabolas y = 5x2and y − 9 = 2x2 is- a)12√3 sq.units
- b)6√2 sq.units
- c)12√2 sq.units
- d)4√3 sq.units
Correct answer is option 'A'. Can you explain this answer?
The area bounded by the parabolas y = 5x2and y − 9 = 2x2 is
a)
12√3 sq.units
b)
6√2 sq.units
c)
12√2 sq.units
d)
4√3 sq.units
|
Saranya Choudhary answered |
Explanation:
Finding the Points of Intersection:
To find the points of intersection of the two parabolas, set them equal to each other:
5x^2 = 9 - 2x^2
7x^2 = 9
x^2 = 9/7
x = ±√(9/7)
Calculating the Area Between the Curves:
Integrate the difference of the two equations between the points of intersection to find the area:
∫(9 - 2x^2) - 5x^2 dx from -√(9/7) to √(9/7)
= ∫(9 - 7x^2) dx from -√(9/7) to √(9/7)
= [9x - (7x^3)/3] from -√(9/7) to √(9/7)
Substitute the Values:
= [9√(9/7) - (7(9/7)^(3/2))/3] - [-9√(9/7) - (7(9/7)^(3/2))/3]
= (27/√7 - 21√7/3) - (-27/√7 - 21√7/3)
= 54/√7
Final Calculation:
To find the area, square the result:
(54/√7)^2 = 2916/7 = 123 sq. units
Therefore, the area bounded by the parabolas y = 5x^2 and y = 9 - 2x^2 is 123 sq. units.
Finding the Points of Intersection:
To find the points of intersection of the two parabolas, set them equal to each other:
5x^2 = 9 - 2x^2
7x^2 = 9
x^2 = 9/7
x = ±√(9/7)
Calculating the Area Between the Curves:
Integrate the difference of the two equations between the points of intersection to find the area:
∫(9 - 2x^2) - 5x^2 dx from -√(9/7) to √(9/7)
= ∫(9 - 7x^2) dx from -√(9/7) to √(9/7)
= [9x - (7x^3)/3] from -√(9/7) to √(9/7)
Substitute the Values:
= [9√(9/7) - (7(9/7)^(3/2))/3] - [-9√(9/7) - (7(9/7)^(3/2))/3]
= (27/√7 - 21√7/3) - (-27/√7 - 21√7/3)
= 54/√7
Final Calculation:
To find the area, square the result:
(54/√7)^2 = 2916/7 = 123 sq. units
Therefore, the area bounded by the parabolas y = 5x^2 and y = 9 - 2x^2 is 123 sq. units.
where [.] is G.I.F
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 :
Chapter doubts & questions for Chapter 8 - Application of Integrals - Mathematics (Maths) Class 12 2024 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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