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(ii)The required area is represented by the shaded area ADCBA as
The points of intersection of the curves,
y
=
x
and
y
=
x
2
, is A (1, 1).
We draw AC perpendicular to
x
-axis.
∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)
Question 3. Find the area of the region lying in the first quadrant and bounded by
y
= 4
x
2
,
x
= 0,
y
= 1 and
y
= 4
Solution :
The area in the first quadrant bounded by
y
= 4
x
2
,
x
= 0,
y
= 1, and
y
= 4 is represented by the shaded area ABCDA as
Question 4. Sketch the graph of y = |x + 3| and evaluate
Solution :
The given equation is y = |x + 3|
The corresponding values of
x
and
y
are given in the following table.
On plotting these points, we obtain the graph of y = |x + 3| as follows.
Question 5. Find the area bounded by the curve
y
= sin
x
between
x
= 0 and
x
= 2π
Solution :
The graph of
y
= sin
x
can be drawn as
∴ Required area = Area OABO + Area BCDB
Question 6. Find the area enclosed between the parabola
y
2
= 4
ax
and the line
y
=
mx
Solution :
The area enclosed between the parabola,
y
2
= 4
ax
, and the line,
y
=
mx
, is represented by the shaded area OABO as
The points of intersection of both the curves are (0, 0) and
.
We draw AC perpendicular to
x
-axis.
∴ Area OABO = Area OCABO – Area (ΔOCA)
Question 7. Find the area enclosed by the parabola 4
y
= 3
x
2
and the line 2
y
= 3
x
+ 12
Solution :
The area enclosed between the parabola, 4
y
= 3
x
2
, and the line, 2
y
= 3
x
+ 12, is represented by the shaded area OBAO as
The points of intersection of the given curves are A (–2, 3) and (4, 12).
We draw AC and BD perpendicular to
x-
axis.
∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)
Question 8. Find the area of the smaller region bounded by the ellipse
and the line
Solution :
The area of the smaller region bounded by the ellipse,
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Question 9. Find the area of the smaller region bounded by the ellipse
and the line
Solution :
The area of the smaller region bounded by the ellipse,
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Question 10. Find the area of the region enclosed by the parabola
x
2
=
y
, the line
y
=
x
+ 2 and
x
-axis
Solution :
The area of the region enclosed by the parabola,
x
2
=
y
, the line,
y
=
x
+ 2, and
x
-axis is represented by the shaded region OACO as
The point of intersection of the parabola,
x
2
=
y
, and the line,
y
=
x
+ 2, is A (–1, 1) and C(2, 4).
Question 11.Using the method of integration, find the area enclosed by the curve |x| + |y| = 1
The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0).
It can be observed that the given curve is symmetrical about
x
-axis and
y
-axis.
∴ Area ADCB = 4 × Area OBAO
Question 12. Find the area bounded by curves {(x, y) : y ≥ x
2
and y = |x|}.
Solution :
The area bounded by the curves, {(x, y) : y ≥ x
2
and y = |x|}
.
, is represented by the shaded region as
It can be observed that the required area is symmetrical about
y
-axis.
Question 13.Using the method of integration, find the area of the triangle whose vertices are A (2, 0), B (4, 5) and C (6, 3).
Solution :
Vertices of the given triangle are A (2, 0), B (4, 5) and C (6, 3).
Equation of side AB is
Equation of side BC
Equation of side CA is
Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)
Question 14.Using the method of integration, find the area of the region bounded by the lines: 2
x
+
y
= 4, 3
x
– 2
y
= 6 and
x
– 3
y
+ 5 = 0
Solution :
The given equations of lines are
2
x
+
y
= 4 … (1)
3
x
– 2
y
= 6 … (2)
And,
x
– 3
y
+ 5 = 0 … (3)
The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on
x
-axis.
Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)
Question 15. Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}.
Solution :
The area bounded by the curves, {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}, is represented as
The points of intersection of both the curves are (1/2,√2) and (1/2, -√2).
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about
x
-axis.
∴ Area OABCO = 2 × Area OBC
Area OBCO = Area OMC + Area MBC
Therefore, the required area is
units.
Question 16.Choose the correct answer:
Therefore, option (D) is correct.
Question 17.Choose the correct answer:
The area bounded by the curve y = x|x|, axis and the ordinates
x
= –1 and
x
= 1 is given by:
Therefore, option (C) is correct.
Question 18.Choose the correct answer:
Solution :
The given equations are
x
2
+
y
2
= 16 … (1)
y
2
= 6
x
… (2)
Area bounded by the circle and parabola
Thus, the correct answer is C.
Question 19.Choose the correct answer:
Required area = Area (ABLA) + area (OBLO)
Required area = Area (AB
Therefore, option (B) is correct.
