Find the area bounded by y = |x-1| and y = 3 - |x| .

Let y = f_{1}(x) =|x-1| and y = f_{2}(x)
= 3 - |x|. Now .

f_{1}(x) = and f_{2}(x) =

The graph of y = f_{1}(x)
and y = f_{2}(x) are as shown : We have to find the area ABCD. Here

A º (1, 0), Cº (0, 3). B is the point of
intersection of y = x–1 and y = 3-x . .

So B º (2, 1) and D is the point of
intersection of y = – x + 1 and y = 3 + x .

Hence D º (-1, 2) and N º (2, 0). .

The required area =

=

=

= 9 –

= 9 –

= 9 –

= 9 – 5 = 4 sq. units .

Note:

Obviously ABCD is a rectangle with AB =and AD = 2. Hence the required area = 2= 4 sq. units.

Find the smaller area enclosed by y = f(x) ,
when f(x) is a polynomial of least degree satisfying and
the circle x^{2 }+ y^{2 }=2 above the x- axis.

Since exists, = 0.

Thus f(x) = a_{4}x^{4}+a_{5}x^{5}+-------+
a_{n}x^{n}, a_{n} >
0 , n ≥ 4.

Since f(x) is of
least degree, f(x) = a_{4}x^{4}. Given that.

= e

&⇒ = e

&⇒ a_{4} = 1

&⇒ f(x) = x^{4}.

The graph of y = x^{4}
and x^{2}+ y^{2}= 2 are shown below.

The point A º(1, 1). The required
area =

=

= =sq. units

Find the area bounded by y = f(x) and the curve y =satisfying the conditions f(x) . f(y) = f(xy) " x, y ∈ R and f¢(1) =2, f(1) =1 .

f¢(x) = =

= = =

&⇒
.

Integrating both
sides, we get f(x) = cx^{2}. Since f(1) = 1

&⇒ c = 1.

So, f(x) = x^{2}
. Now

&⇒ x^{4} + x^{2} –2 = 0

&⇒ x^{2} =1

&⇒ x = ±1

Required area = 2

= 2= . sq. units.

Find the area of the
region bounded by y^{2} + 4ax = 0 and (y^{2} + 4a^{2})x
+ 8a^{3} = 0, a > 0.

y^{2}
= – 4ax is the parabola opening towards left and vertex at origin.

x = -is a curve existing on the negative side of x – axis only and symmetrical w.r.t. x – axis. The lowest value of x occurs when y = 0 (then x = -2a). The curve tends to approach the y – axis as x approaches 0.

Required area = Area (OABCO) = 2 Area(OBCO)

= 2[Area (OBCDO) – Area(OCDO)]

Point D can be found by solving the two curve equations for y.

i.e. -= -

&⇒ (y^{2})^{2}
+ 4a^{2}y^{2} – 32a^{4} = 0

&⇒ y^{2} =

= (y^{2}
cannot be negative)

= 4a^{2 }

&⇒ y
= ±2a

So, D is (0, 2a).

\ Required Area

=

= 2

= 2= sq. units.

Find the area
enclosed by the region formed by x^{2} +y^{2} –6x –4y +12 £ 0, y £ x and x £
.

The area enclosed by the curve is as shown

x^{2} +y^{2}
– 6x – 4y +12 = 0

&⇒ (y –2)^{2}
+( x – 3)^{2} =1

&⇒ y – 2 = ±

&⇒ y –2 = – [
since y £ 5/2 ]

&⇒ y = 2 –

Required Area = Area of trapezium ABCDE- Area ABCE . .

=

=

=

= = .