If ∫ f ( x ) d x = F ( x ) \displaystyle\large\int { f\left( x \right) dx=F\left( x \right) } ∫ f ( x ) d x = F ( x ) . Then ∫ x 3 f ( x 2 ) d x \displaystyle \large \int { { x }^{ 3 } } f\left( { x }^{ 2 } \right) dx ∫ x 3 f ( x 2 ) d x is equal to :
A) 1 2 [ x 2 F ( x 2 ) − ∫ F ( x 2 ) d ( x 2 ) ] \displaystyle \cfrac { 1 }{ 2 } \left[ { x }^{ 2 }{ F\left( { x }^{ 2 } \right) }-\int { F\left( { x }^{ 2 } \right) d\left( { x }^{ 2 } \right) } \right] 2 1 [ x 2 F ( x 2 ) − ∫ F ( x 2 ) d ( x 2 ) ]
B) 1 2 [ x 2 { F ( x ) } 2 − ∫ { F ( x ) } 2 d x ] \displaystyle \cfrac { 1 }{ 2 } \left[ { x }^{ 2 }{ \left\{ F\left( x \right) \right\} }^{ 2 }-\int { { \left\{ F\left( x \right) \right\} }^{ 2 }dx } \right] 2 1 [ x 2 { F ( x ) } 2 − ∫ { F ( x ) } 2 d x ]
C) 1 2 [ x 2 F ( x ) − 1 2 ∫ { F ( x ) } 2 d x ] \displaystyle \cfrac { 1 }{ 2 } \left[ { x }^{ 2 }F\left( x \right) -\cfrac { 1 }{ 2 } \int { { \left\{ F\left( x \right) \right\} }^{ 2 }dx } \right] 2 1 [ x 2 F ( x ) − 2 1 ∫ { F ( x ) } 2 d x ]
D) None Of These
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Using partial integration:
∫ u d v = u v − ∫ v d u \displaystyle \int {u dv} = uv - \int {v du} ∫ u d v = u v − ∫ v d u
And let:
{ u = x 2 2 ⇒ d u = d ( x 2 ) 2 d v = 2 x f ( x 2 ) ⇒ v = F ( x 2 ) \begin{cases} u = \dfrac {x^2}{2} & \Rightarrow du = \dfrac{d(x^2)}{2} \\ dv = 2xf(x^2) & \Rightarrow v = F(x^2) \end{cases} ⎩ ⎨ ⎧ u = 2 x 2 d v = 2 x f ( x 2 ) ⇒ d u = 2 d ( x 2 ) ⇒ v = F ( x 2 )
∫ x 3 f ( x 2 ) d x = ∫ x 2 2 ˙ 2 x f ( x 2 ) d x = x 2 2 ˙ F ( x 2 ) − ∫ F ( x 2 ) ˙ d ( x 2 ) 2 = 1 2 ( x 2 F ( x 2 ) − ∫ F ( x 2 ) d ( x 2 ) ) \begin{aligned} \displaystyle \int {x^3f(x^2)dx} & = \int {\dfrac {x^2}{2}\dot{} 2xf(x^2)dx} \\ & = \dfrac {x^2}{2} \dot{} F(x^2) - \int {F(x^2) \dot{} \dfrac {d(x^2)}{2}} \\ & = \boxed {\displaystyle \frac {1}{2} \left( x^2F(x^2)- \int {F(x^2) d(x^2)} \right)} \end{aligned} ∫ x 3 f ( x 2 ) d x = ∫ 2 x 2 ˙ 2 x f ( x 2 ) d x = 2 x 2 ˙ F ( x 2 ) − ∫ F ( x 2 ) ˙ 2 d ( x 2 ) = 2 1 ( x 2 F ( x 2 ) − ∫ F ( x 2 ) d ( x 2 ) )
Yes ! Really cute :D
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Using partial integration:
∫ u d v = u v − ∫ v d u
And let:
⎩ ⎨ ⎧ u = 2 x 2 d v = 2 x f ( x 2 ) ⇒ d u = 2 d ( x 2 ) ⇒ v = F ( x 2 )
∫ x 3 f ( x 2 ) d x = ∫ 2 x 2 ˙ 2 x f ( x 2 ) d x = 2 x 2 ˙ F ( x 2 ) − ∫ F ( x 2 ) ˙ 2 d ( x 2 ) = 2 1 ( x 2 F ( x 2 ) − ∫ F ( x 2 ) d ( x 2 ) )