Find the area bounded by the
curve
x2 = 4y and the X= 3
Answers
Step-by-step explanation:
The area bounded by the curve, x
2
=4y, and line, x=4y−2, is represented by the shaded area OBAO.
Let A and B be the points of intersection of the line and parabola.
Coordinates of point A are(−1,
4
1
)
Coordinates of point B are (2,1).
We draw AL and BM perpendicular to x-axis.
It can be observed that,
Area OBAO=Area OBCO+Area OACO ........... (1)
Then, Area OBCO=Area OMBC−Area OMBO
=∫
0
2
4
x+2
dx−∫
0
2
4
x
2
dx
=
4
1
[
2
x
2
+2x]
0
2
−
4
1
[
3
x
3
]
0
2
=
4
1
[2+4]−
4
1
[
3
8
]
=
2
3
−
3
2
=
6
5
Similarly, Area OACO = Area OLAC - Area OLAO
=∫
−1
0
4
x+2
dx−∫
−1
0
4
x
2
dx
=
4
1
[
2
x
2
+2x]
−1
0
−
4
1
[
3
x
3
]
−1
0
=−
4
1
[
2
(−1)
2
+2(−1)]−[−
4
1
(
3
(−1)
3
)]=−
4
1
[
2
1
−2]−
12
1
=
2
1
−
8
1
−
12
1
=
24
7
Therefore, required area =(
6
5
+
24
7
)=
8
9
Answer:
The area bounded by the curve, x2=4y, and line, x=4y−2, is represented by the shaded area OBAO.
Let A and B be the points of intersection of the line and parabola.
Coordinates of point A are(−1,41)
Coordinates of point B are (2,1).
We draw AL and BM perpendicular to x-axis.
It can be observed that,
Area OBAO=Area OBCO+Area OACO........... (1)
Then, Area OBCO=Area OMBC−Area OMBO
=∫024x+2dx−∫024x2dx
=41[2x2+2x]02−41[3x3]02
=41[2+4]−41[38]
=2