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Question: \(\int _ { 0 } ^ { \mathrm { n } ^ { 2 } } [ \sqrt { \mathrm { x } } ]\) dx is equal to -...

∫0n2[x]\int _ { 0 } ^ { \mathrm { n } ^ { 2 } } [ \sqrt { \mathrm { x } } ] dx is equal to -

A

B

C

n(n−1)(4n−1)6\frac { \mathrm { n } ( \mathrm { n } - 1 ) ( 4 \mathrm { n } - 1 ) } { 6 }

D

None of these

Answer

Explanation

Solution

dx = dx + dx + dx + ….dx

= + + + …. + ∫(n−1)2n2(n−1)\int _ { ( \mathrm { n } - 1 ) ^ { 2 } } ^ { \mathrm { n } ^ { 2 } } ( \mathrm { n } - 1 ) dx

= 1(4 – 1) + 2(9 – 4) + ….. + (n – 1) [n2 – (n – 1)2]

= – (12 + 22 + 32 + …. + n2) + n3

= n3 – n(n+1)(2n+1)6\frac { \mathrm { n } ( \mathrm { n } + 1 ) ( 2 \mathrm { n } + 1 ) } { 6 }= n(n−1)(4n+1)6\frac { \mathrm { n } ( \mathrm { n } - 1 ) ( 4 \mathrm { n } + 1 ) } { 6 }.