Question

Find the value of dx by first principle rule –

• A. $\frac { 1 } { 3 }$(a³ – b³)
• B. $\frac { 1 } { 3 }$ (b³ – a³)
• C. $\frac { 1 } { 2 }$ (a³ – b³)
• D. None of these

Answer 1

Answer: (B)

$\frac { 1 } { 3 }$ (b³ – a³)

Answer 2

Here ƒ(x) = x², nh = b – a from the definition,

= $\operatorname { Lim } _ { \mathrm { h } \rightarrow 0 }$ h $\sum _ { r = 0 } ^ { \mathrm { n } - 1 } ( \mathrm { a } + \mathrm { rh } ) ^ { 2 }$

= h $\sum _ { r = 0 } ^ { \mathrm { n } - 1 } \left( \mathrm { a } ^ { 2 } + 2 \mathrm { ahr } + \mathrm { h } ^ { 2 } \mathrm { r } ^ { 2 } \right)$

= h

= h $\left\{ a ^ { 2 } n + 2 a h \frac { ( n - 1 ) n } { 2 } + \frac { h ^ { 2 } ( n - 1 ) n ( 2 n - 1 ) } { 6 } \right\}$

= h $\left\{ \mathrm { a } ^ { 2 } ( \mathrm { nh } ) + \mathrm { a } ( \mathrm { nh } ) ( \mathrm { nh } - \mathrm { h } ) + \frac { ( \mathrm { nh } - \mathrm { h } ) ( \mathrm { nh } ) ( 2 \mathrm { nh } - \mathrm { h } ) } { 6 } \right\}$

=

= a² (b – a) + a (b – a)² + $\frac { ( b - a ) ^ { 3 } } { 3 }$

= $\frac { ( b - a ) } { 3 }${b² + ab + a²}

= $\frac { 1 } { 3 }$ (b³ – a³).