Integration (8)

Calculus Level 3

0 1 4 x 3 [ d 2 d x 2 ( 1 x 2 ) 5 ] d x = ? \large \int_0^1 4x^3 \left [ \dfrac{d^2}{dx^2} (1-x^2)^5 \right ] \, dx = \, ?


The answer is 2.

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1 solution

4 x 3 { d 2 d x 2 [ 1 x 2 ] 5 } d x \int { 4{ x }^{ 3 } } \left\{ \frac { { d }^{ 2 } }{ d{ x }^{ 2 } } \left[ 1-{ x }^{ 2 } \right] ^{ 5 } \right\} dx

= [ 4 x 3 d d x [ 1 x 2 ] 5 ] 1 0 0 1 [ d d x ( 4 x 3 ) ] . [ d d x ( 1 x 2 ) 5 ] . d x \left[ 4{ x }^{ 3 }\frac { d }{ dx } { \left[ 1-{ x }^{ 2 } \right] }^{ 5 } \right] \begin{matrix} 1 \\ 0 \end{matrix}\quad -\int _{ 0 }^{ 1 }{ \left[ \frac { d }{ dx } \left( 4{ x }^{ 3 } \right) \right] } .\left[ \frac { d }{ dx } { \left( 1-{ x }^{ 2 } \right) }^{ 5 } \right] .dx

= 4 x 3 ( 10 x ) ( 1 x 2 ) 5 1 0 0 1 12 x 2 ( 1 x 2 ) 4 ( 10 x ) d x 4{ x }^{ 3 }\left( -10x \right) { \left( 1-{ x }^{ 2 } \right) }^{ 5 }\begin{matrix} 1 \\ 0 \end{matrix}\quad -\quad \int _{ 0 }^{ 1 }{ 12{ x }^{ 2 } } { \left( 1-{ x }^{ 2 } \right) }^{ 4 }\left( -10x \right) dx

= ( 0 0 ) + 120 0 1 x 3 ( 1 x 2 ) 4 d x (0-0)\quad +\quad 120\int _{ 0 }^{ 1 }{ { x }^{ 3 } } { \left( 1-{ x }^{ 2 } \right) }^{ 4 }dx

n o w p u t 1 x 2 = t 2 x d x = d t x d x = 0.5 d t x 2 = 1 t now\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad put\quad 1-{ x }^{ 2 }=t\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -2xdx\quad =dt\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad xdx=-0.5dt\\ { \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x }^{ 2 }=1-t

= 120 1 0 ( 1 t ) ( t 4 ) ( 0.5 ) d t = ( 0.5 ) 120 1 0 ( 1 t ) ( t 4 ) d t = 60 1 0 ( 1 t ) ( t 4 ) d t = 60 0 1 ( 1 t ) ( t 4 ) d t =120\int _{ 1 }^{ 0 }{ \left( 1-t \right) } \left( { t }^{ 4 } \right) \left( -0.5 \right) dt\\ =\left( -0.5 \right) 120\int _{ 1 }^{ 0 }{ \left( 1-t \right) } \left( { t }^{ 4 } \right) dt\\ =-60\int _{ 1 }^{ 0 }{ \left( 1-t \right) } \left( { t }^{ 4 } \right) dt\\ =60\int _{ 0 }^{ 1 }{ \left( 1-t \right) } \left( { t }^{ 4 } \right) dt

= 60 0 1 ( t 4 t 5 ) d t =60\int _{ 0 }^{ 1 }{ \left( { t }^{ 4 }-{ t }^{ 5 } \right) } dt

= 60 [ t 5 5 t 6 6 ] 1 0 =60\left[ \frac { { t }^{ 5 } }{ 5 } -\frac { { t }^{ 6 } }{ 6 } \right] \begin{matrix} 1 \\ 0 \end{matrix}

= 60 { [ 1 5 5 1 6 6 ] [ 0 5 5 0 6 6 ] } = 60 { 1 5 1 6 } = 60 { 6 5 5 6 } = 60 { 1 30 } = 2 =60\left\{ \left[ \frac { { 1 }^{ 5 } }{ 5 } -\frac { { 1 }^{ 6 } }{ 6 } \right] -\left[ \frac { { 0 }^{ 5 } }{ 5 } -\frac { { 0 }^{ 6 } }{ 6 } \right] \right\} \\ =60\left\{ \frac { 1 }{ 5 } -\frac { 1 }{ 6 } \right\} \\ =60\left\{ \frac { 6-5 }{ 5*6 } \right\} \\ =60\left\{ \frac { 1 }{ 30 } \right\} =2

Moderator note:

Nice usage of integration by parts so that we don't have to evaluate d 2 d x 2 ( 1 x 2 ) 5 \frac{d^2}{dx^2} ( 1 - x^2)^5 .

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