Find the area of the figure bounded by the parabola
(y-2)^{2} = x-1, the tangent to it at the point with ordinate 3,
and the x-axis.

The
equation (y – 2)^{2} = x – 1 represents a parabola having vertex at
A(1, 2) and axis as y =2 . The point P on this parabola with ordinate 3
has abscissa.

x = 1+ (y-2)^{2} = 2, then P º (2, 3)

(y - 2)^{2} = x -1

&⇒
2(y-2)= 1

&⇒

Thus equation of tangent at P is given by

y - 3 =

&⇒x
– 2y +4 = 0

The region, of which
area is to be found is shown by shaded portion. We have to find the
Area(EOBAPCE). C is the point of intersection of x-2y +4 =0 and y-axis

&⇒ C º (0, 2). Obviously .

D º (0, 3).

The required area

= area(OBAPDO) – area (DPCD) + area(DEOC)

=

= sq. units

Let f(x) = maximum {x^{2}, (1-x)^{2},
2x(1-x)}. Determine the area of the region bounded by the curve y = f(x),
x-axis , x = 0 and x=1.

First we draw the graph of y = x^{2},
y = (1 - x)^{2} and y = 2x(1 - x) for

0 £ x £ 1. Obviously all these are parabolas.

Now graph of y = f(x) is
shown by dark curves. Hence C is the point of intersection of

y = x^{2}_{ }and y = (1-x)^{2} . So

C º
(1/2, ¼). B is the point of intersection of y =(1 – x)^{2},

y =2x(1-x). So B º(1/3, 4/9).

Similarly D º (2/3, 4/9)

The required area = Area (OPBDQAO)

=

=

= sq. units

Note: If f(x) = minimum {x^{2},
(1-x)^{2}, 2x(1-x)}. , then the required area will be OCAO. .

Find the area between the curves y = x^{2} + x
–2 and y = 2x, for which |x^{2} + x –2| + | 2x | = |x^{2}
+ 3x –2| is satisfied.

y = x^{2} + x –2, y = 2x

| x^{2} + x –2 | + | 2x | = | x^{2} + 3x
–2 |

This means we have to find area in the region where x^{2}
+ x –2 and 2x have same sign.

Thus, the required region is OABO and ECD.

\ x^{2} +x –2 = 2x

&⇒ x^{2} –x –2 = 0

&⇒ x = -1, x = 2

\ required area is,

A=Area of OABO+ Area of ECD

\

=

=

=

=

= =

Let f(x) = min{e^{x}, 3/2, 1+ e^{-x}
| 0 £ x £ 1
}. Find the area bounded by

y = f(x) , x-axis, y-axis and the line x = 1 . .

f(x) =

Let A be the required area as shown in fig.

A = =

= = 2+ .

Find a function ‘f’,
(x^{4} – 4x^{2} £
f(x) £ 2x^{2} – x^{3})
such that the area bounded by y = f(x), y =x^{4} –4x^{2},
the y-axis and the line x=t, (0 £
t £ 2) is k times the
area bounded by y = f(x), y = 2x^{2 }– x^{3}, y-axis and line
x =t ( 0£ t £ 2) . .

According to the given condition

Differentiate both sides w.r.t t , we get f(t) –(t^{4}
– 4t^{2}) = k(2t^{2} –t^{3} –f(t)).

&⇒
(1+k) f(t) = k 2t^{2} – kt^{3} + t^{4} – 4t^{2 }

&⇒ f(t) =

Hence, the required function ‘f’ is given by

f(x) = (x^{4} – kx^{3} +2(k – 2)x^{2})
.