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<[∵  sin x is periodic with period π and sin x > 0 if 0 < x π]
If y = f(x) makes +ve intercept of 2 and 0 unit on x and y axes and encloses an area of 3/4 square unit with the axes then
The area bounded by the curves y = lnx, y = ln x,y= ln x  and y =  ln x  is
First we draw each curve as separate graph
NOTE : Graph of y =  f(x)  can be obtained from the graph of the curve y = f(x) by drawing the mirror image of the portion of the graph below xaxis, with respect to xaxis.
Clearly the bounded area is as shown in the following figure.
The area of the region bounded by the curves y = x  1 and y = 3  x is
is equal to
Let f(x) be a function satisfying f '(x) = f(x) with f(0) = 1 and g(x) be a function that satisfies f(x) + g(x) = x^{2} . Then the value of the integral
The value of the integral
[Using definite integrals as limit of sum]
and then the value of is
The area of the region boun ded by the curves y =  x  2 , x = 1,x = 3 and the xaxis is
The required area is shown by shaded region
The area enclosed between the curve y = log_{e} (x +e) and the coordinate axes is
The graph of the curve y = log_{e} (x+e) is as shown in the fig.
e e  0 + 1=1
Hence the required area is 1 square unit.
The parabolas y^{2} = 4x and x^{2} = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If S_{1} , S_{2} , S_{3} are respectively the areas of these parts numbered from top to bottom; then S_{1} : S_{2} : S_{3} is
Intersection points of x^{2} = 4y and y^{2} = 4x are (0, 0) and (4, 4). The graph is as shown in the figure.
∴ S_{1} : S_{2} :S_{3} = 1 : 1:1
Let f (x) be a non – negative continuous function such that the area bounded by the curve y = f (x), x  axis and the
ordinates
Differentiating w. r . t β
Adding equations (1) and (2) we get
Adding equation (1) and (2)
is equal to
[ using the property of even and odd function]
The value of denotes the greatest integer not exceeding x is
Let a = k + h where k is an integer such that [a] =k and 0 < h < 1
{f (2) – f (1)} + 2{f (3) – f (2)} + 3{f (4) –f (3)} + ........ + (k – 1) {f(k) – f(k – 1)} + k{f(k + h) – f(k)}
= – f (1) – f (2) – f (3) ......... – f (k) + kf (k + h)
= [a] f (a)  {f (1) + f (2) + f (3)+ .......... f ([a])}
Then F(e) equals
and limit for t = 1 ⇒ z = 1 and for t = 1/e ⇒ z = e
[∴ log1 = 0]
Equation (A) becomes
Let log t = x
[for limit t = 1, x = 0 and t = e, x = log e = 1]
The solution for x of the equation
The area enclosed between the curves y^{2} = x and y =  x  is
The area enclosed between the curves y^{2} = x and y =  x 
From the figure, area lies between y^{2} = x and y = x
Then which one of the following is true?
for x∈(0, 1)
The area of the plane region bounded by the curves x + 2y^{2} = 0 and x + 3y^{2} = 1is equal to
[Left handed parabola with vertex at (0, 0)]
[Left handed parabola with vertex at (1, 0)] Solving the two equations we get the points of intersection as (–2, 1), (–2, –1)
The required area is ACBDA, given by
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 xaxis is:
The given parabola is (y – 2)^{2} = x – 1 Vertex (1, 2) and it meets x–axis at (5, 0) Also it gives y^{2} – 4y – x + 5 = 0
So, that equation of tangent to the parabola at (2, 3) is
which meets xaxis at (– 4, 0).
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)
denotes the greatest integer function, is equal to :
Adding two values of I in eq^{n} s (1) & (2), We get
The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and
Area above xaxis = Area below xaxis
Let p(x) be a function defined on R such that p '(x) = p'(1 – x), for all x ∈ [0, 1], p (0) = 1 and p (1) = 41. Then
p '( x) = p '(1x) ⇒ p( x) =  p(1  x)+c
at x = 0
p(0) = – p(1) + c ⇒ 42 = c
Now, p(x) =  p(1 x)+ 42 ⇒ p (x) + p (1 x) = 42
on adding (i) and (ii),
The value of
The area of the region enclosed by the curves y = x, x = e, y = 1/x and the positive xaxis is
Area of required region AOBC
The area between the parabolas and the straight line y = 2 is :
Given curves are the parabolas whose equations can be written as y = 4x^{2} and
Also, given y = 2.
Now, shaded portion shows the required area which is symmetric.
then g (x + π) equals
Putting t = π + y in second integral, we get
Statement1 : The value of the integral
Also, Given, I
…(2)
By adding (1) and (2), we get
It is fundamental property.
The area (in square units) bounded by the curves and lying in the first quadrant is :
Given curves are
…(1)
and 2y – x + 3 = 0 …(2)
On solving both we get y = –1, 3
The integral
The area of the region described by
Given curves are x^{2} + y^{2} = 1 and y^{2} = 1 –x. Intersecting points are x = 0, 1
Area of shaded portion is the required area.
So, Required Area = Area of semicircle + Area bounded by parabola
(∵ radius of circle = 1)
The area (in sq. units) of the region described by {(x, y) : y^{2} < 2x and y > 4x – 1} is
Required area
= Area of ABCD – ar (ABOCD)
The integral
`..(i) ...(ii)
Adding (1) and (2)
The area (in sq. units) of the region {(x, y) : y^{2} > 2x and x^{2} + y^{2} < 4x, x > 0, y > 0} is :
Points of intersection of the two curves are (0, 0), (2, 2) and (2, –2)
Area = Area (OAB) – area under parabola (0 to 2)
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