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All questions of 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
and let f –1 be the inverse function of f. Then (f –1)' (2) is equal to
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
Then F(e) equals- a)1
- b)2
- c)1/2
- d)0
Correct answer is option 'C'. Can you explain this answer?
Then F(e) equalsa)
1
b)
2
c)
1/2
d)
0
|
Vikash Yadav answered |
Here assume that F(x) = g(x)
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
|
Vikash Yadav answered |
https://www.youtube.com/watch?v=QylwvR8WCQM
watch this for solution
The area enclosed between the curves y2 = x and y = | x | is- a)1/6
- b)1/3
- c)2/3
- d)1
Correct answer is option 'A'. Can you explain this answer?
The area enclosed between the curves y2 = x and y = | x | is
a)
1/6
b)
1/3
c)
2/3
d)
1
|
Lavanya Menon answered |
The area enclosed between the curves y2 = x and y = | x |
From the figure, area lies between y2 = x and y = x
From the figure, area lies between y2 = x and y = x
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.
Then which one of the following is true?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Then which one of the following is true?a)
b)
c)
d)
|
|
Anand Kumar answered |
1 1
I = { [ (sinx)/√x ] dx J = { [ (cosx)/√x ] dx
0 0
replace √x by "y"
1 1
I = { 2(siny)dy J = { 2(cosy)dy
0 0
1 1
I = -2 cosy| J = 2sinx|
0 0
I = -2cos1 + 2 J = 2sin1
we know,
sin1 < sin60="" but="" cos1="" /> cos60
J < 2="" and="" i="" /><2 note:-="" 1="" radian="56.67" degree="" note:-="" 1="" radian="56.67">
I = { [ (sinx)/√x ] dx J = { [ (cosx)/√x ] dx
0 0
replace √x by "y"
1 1
I = { 2(siny)dy J = { 2(cosy)dy
0 0
1 1
I = -2 cosy| J = 2sinx|
0 0
I = -2cos1 + 2 J = 2sin1
we know,
sin1 < sin60="" but="" cos1="" /> cos60
J < 2="" and="" i="" /><2 note:-="" 1="" radian="56.67" degree="" note:-="" 1="" radian="56.67">
then the expression for l(m, n) in terms of l(m + 1, n – 1) is
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 enclosed between the curve y = loge (x +e) and the coordinate axes is- a)1
- b)2
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
The area enclosed between the curve y = loge (x +e) and the coordinate axes is
a)
1
b)
2
c)
3
d)
4
|
Dhruv Kulkarni answered |
The graph of the curve y = loge (x+e) is as shown in the fig.
e -e - 0 + 1=1
Hence the required area is 1 square unit.
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
Let f : [– 1, 2] → [0, ∞) be a continuous function such that f (x) = f (1 – x) for all x ∈ [–1, 2]
and R2 be the area of the region bounded by y = f (x), x = –1, x = 2, and the x-axis.
Then- a)R1 = 2R2
- b)R1 = 3R2
- c)2R1 = R2
- d)3R1 = R2
Correct answer is option 'C'. Can you explain this answer?
Let f : [– 1, 2] → [0, ∞) be a continuous function such that f (x) = f (1 – x) for all x ∈ [–1, 2]
and R2 be the area of the region bounded by y = f (x), x = –1, x = 2, and the x-axis.Then
a)
R1 = 2R2
b)
R1 = 3R2
c)
2R1 = R2
d)
3R1 = R2
|
|
Ankit Singh answered |
increases in
If f (x) is differentiable and 
equals- a)2/5
- b)–5/2
- c)1
- d)5/2
Correct answer is option 'A'. Can you explain this answer?
If f (x) is differentiable and equals
equalsa)
2/5
b)
–5/2
c)
1
d)
5/2
|
|
Harsh Singhal answered |
Differentiate & put t=2/5
Then the real roots of the equation x2 – f '(x) = 0 are 
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'.
The area of the region bounded by the parabola (y – 2)2 = x –1, the tangent of the parabola at the point (2, 3) and the x-axis is:- a)6
- b)9
- c)12
- d)3
Correct answer is option 'B'. Can you explain this answer?
The area of the region bounded by the parabola (y – 2)2 = x –1, the tangent of the parabola at the point (2, 3) and the x-axis is:
a)
6
b)
9
c)
12
d)
3
|
Prisha Das answered |
The given parabola is (y – 2)2 = x – 1 Vertex (1, 2) and it meets x–axis at (5, 0) Also it gives y2 – 4y – x + 5 = 0
So, that equation of tangent to the parabola at (2, 3) is
So, that equation of tangent to the parabola at (2, 3) is
which meets x-axis at (– 4, 0).
In the figure shaded area is the required area.
Let us draw PD perpendicular to y – axis.
In the figure shaded area is the required area.
Let us draw PD perpendicular to y – axis.
Then required area = Ar ΔBOA + Ar (OCPD) – Ar (ΔAPD)
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
then g (x + π) equals- a)
- b)g (x) + g (π)
- c)g (x) – g (π)
- d)g (x) . g (π)
Correct answer is option 'B,C'. Can you explain this answer?
then g (x + π) equalsa)
b)
g (x) + g (π)
c)
g (x) – g (π)
d)
g (x) . g (π)
|
|
Rahul Kumar answered |
Integral of cos4t = (sin4t)/4 after limit process we got:-
(sin4x)/4 - (sin0)/4 = (sin4x)/4
hence:-
g(x) = (sin4x)/4
therefore:-
g(x + π) = (sin(4π + 4x))/4 = (sin4x)/4
from option (a):-
((sin4x)/4)/((sin4π)/4) = infinity
from option (b):-
((sin4x)/4) + ((sin4π)/4) = ((sin4x)/4) = g(x + π)
from option (c):-
((sin4x)/4) - ((sin4π)/4) = ( (sin4x)/4) = g( x +π)
from option (d):-
((sin4x)/4).((sin4π)/4) = 0
THUS WE CAN SAY THAT THE OPTION (b) AND (C) IS CORRECT
(sin4x)/4 - (sin0)/4 = (sin4x)/4
hence:-
g(x) = (sin4x)/4
therefore:-
g(x + π) = (sin(4π + 4x))/4 = (sin4x)/4
from option (a):-
((sin4x)/4)/((sin4π)/4) = infinity
from option (b):-
((sin4x)/4) + ((sin4π)/4) = ((sin4x)/4) = g(x + π)
from option (c):-
((sin4x)/4) - ((sin4π)/4) = ( (sin4x)/4) = g( x +π)
from option (d):-
((sin4x)/4).((sin4π)/4) = 0
THUS WE CAN SAY THAT THE OPTION (b) AND (C) IS CORRECT
The solution for x of the equation 
- a)
- b)
- c)2
- d)None
Correct answer is option 'D'. Can you explain this answer?
The solution for x of the equation
a)
b)
c)
2
d)
None
|
|
Anand Kumar answered |
Output of the integration will be sec inverse t i.e, sec^(-1)t
sec^(-1) x - sec^(-1)√2 = π/2
sec^(-1)x - π/4 = π/2
sec^(-1)x = π/2 + π/4 = 3π/4
x = sec(3π/4) = 1/cos(3π/4) = 1/cos(π - π/4)
x = - 1/cos(π/4) = -1/(1/√2) = - √2.
sec^(-1) x - sec^(-1)√2 = π/2
sec^(-1)x - π/4 = π/2
sec^(-1)x = π/2 + π/4 = 3π/4
x = sec(3π/4) = 1/cos(3π/4) = 1/cos(π - π/4)
x = - 1/cos(π/4) = -1/(1/√2) = - √2.
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 :
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