Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
Area of the region (in square units) bounded by the curve y=x2+4 and the line y=5 x-2 is
Q. Area of the region (in square units) bounded by the curve
y
=
x
2
+
4
and the line
y
=
5
x
−
2
is
1466
195
TS EAMCET 2019
Report Error
A
2
1
B
12
1
C
2
D
6
1
Solution:
We have,
y
=
x
2
+
4
,
y
=
5
x
−
2
Solving equation, we get
(
2
,
8
)
and
(
3
,
13
)
Area of shaded region
2
∫
3
(
(
5
x
−
2
)
−
(
x
2
+
4
)
d
x
=
2
∫
3
(
5
x
−
x
2
−
6
)
d
x
=
[
2
5
x
2
−
3
x
3
−
6
x
]
2
1
=
[
(
2
45
−
9
−
18
)
−
(
10
−
3
8
−
12
)
]
=
6
1