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Byju's Answer
Standard XIII
Mathematics
Definite Integral as Limit of Sum
If I = ∫ 168 ...
Question
If
I
=
16
∫
8
(
√
x
+
√
32
x
−
256
+
√
x
−
√
32
x
−
256
)
d
x
then
(
I
16
)
2
is
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Solution
I
=
16
∫
8
(
|
√
x
−
8
)
+
2
√
2
|
+
|
√
x
−
8
−
2
√
2
|
d
x
⇒
I
=
16
∫
8
√
x
−
8
+
2
√
2
+
2
√
2
−
√
x
−
8
d
x
⇒
I
=
8
×
4
√
2
=
32
√
2
⇒
(
I
16
)
2
=
(
32
√
2
16
)
2
=
8
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1
Similar questions
Q.
If
I
=
16
∫
8
(
√
x
+
√
32
x
−
256
+
√
x
−
√
32
x
−
256
)
d
x
then
(
I
16
)
2
is
Q.
If
3
2
x
=
200
, then
x
is between
Q.
Factorise :
(i)
x
3
−
2
x
2
−
x
+
2
(ii)
x
3
+
13
x
2
+
32
x
+
20
Q.
If
16
2
=
256
, then which of the following are perfect squares?
Q.
Solve the equation
x
4
−
8
x
3
+
24
x
2
−
32
x
+
20
=
0
if
3
+
i
is a root of it.
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