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Byju's Answer
Standard XII
Mathematics
Definite Integral as Limit of Sum
Evaluate, ∫...
Question
Evaluate,
∫
2
1
(
3
x
2
−
1
)
d
x
as the limit of a sum
Open in App
Solution
Consider the problem
Given
∫
2
1
(
3
x
2
−
1
)
d
x
Consider
x
−
1
+
r
h
where,
h
=
2
−
1
n
=
1
n
As,
x
→
1
,
r
→
0
And
x
→
2
,
r
→
1
n
∫
2
1
(
3
x
2
−
1
)
d
x
=
lim
n
→
∞
∑
n
r
=
0
h
[
3
(
1
+
r
h
)
2
−
1
]
=
lim
n
→
∞
∑
n
r
=
1
1
n
[
3
(
1
+
r
n
)
2
−
1
]
=
lim
n
→
∞
∑
n
r
0
1
n
(
(
3
+
3
r
2
n
2
+
6
r
n
)
−
1
)
=
lim
n
→
∞
∑
n
r
1
n
(
2
+
3
r
2
n
2
+
6
r
n
)
=
lim
n
→
∞
1
n
(
∑
n
r
=
0
(
2
)
+
∑
n
r
=
0
(
3
r
2
n
2
)
+
∑
n
r
=
0
(
6
r
2
)
)
=
lim
n
→
∞
1
n
(
∑
n
r
=
0
(
2
)
+
3
n
2
∑
n
r
=
0
(
r
2
)
+
6
n
∑
n
r
=
0
(
r
)
)
=
lim
n
→
∞
1
n
(
2
n
+
3
n
2
(
n
(
n
+
1
)
(
2
n
+
1
)
6
)
+
6
n
(
n
(
n
+
1
)
2
)
)
=
lim
n
→
∞
(
2
n
n
+
3
n
3
(
n
(
n
+
1
)
(
2
n
+
1
)
6
)
+
6
n
2
(
n
(
n
+
1
)
2
)
)
=
lim
n
→
∞
(
2
+
1
n
3
(
n
(
n
+
1
)
(
2
n
+
1
)
2
)
+
3
n
2
(
n
(
n
+
1
)
1
)
)
=
lim
n
→
∞
(
2
+
1
2
(
1
)
(
1
+
1
n
)
(
2
+
1
n
)
+
3
(
1
)
(
1
+
1
n
)
)
=
2
+
1
2
lim
n
→
∞
(
(
1
+
1
n
)
(
2
+
1
n
)
+
3
lim
n
→
∞
(
1
+
1
n
)
)
=
2
+
1
2
×
2
+
3
×
1
=
2
+
1
+
3
=
6
Hence,
∫
2
1
(
3
x
2
−
1
)
d
x
=
6
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