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Calculus Level 3

∫ 0 1 ∣ x ∣ ∣ 1 − x ∣ d x \int _{ 0 }^{ 1 }{ \left| x \right| \left| 1-x \right| dx}

can be expressed as the form a b \frac { a }{ b } where a a and b b are co-prime integers what is the the value of a + b a+b .


The answer is 7.

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Joel Yip by Joel Yip , 21, Singapore

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1 solution

Joel Yip
Mar 14, 2015

∫ ∣ x ∣ ∣ 1 − x ∣ d x = 1 24 ( 3 ( 3 − 2 x ) x 2 s g n ( x ) + s g n ( 1 − x ) ( 3 ( 3 − 2 x ) x 2 s g n ( x ) − 6 x 3 + 9 x 2 − 4 ) + 6 x 3 − 9 x 2 + 4 ) \int { \left| x \right| \left| 1-x \right| dx } =\frac { 1 }{ 24 } \left( 3\left( 3-2x \right) { x }^{ 2 }sgn\left( x \right) +sgn\left( 1-x \right) \left( 3\left( 3-2x \right) { x }^{ 2 }sgn\left( x \right) -6{ x }^{ 3 }+9{ x }^{ 2 }-4 \right) +6{ x }^{ 3 }-9{ x }^{ 2 }+4 \right) put 1 and 0 so [ 1 24 ( 3 ( 3 − 2 x ) x 2 s g n ( x ) + s g n ( 1 − x ) ( 3 ( 3 − 2 x ) x 2 s g n ( x ) − 6 x 3 + 9 x 2 − 4 ) + 6 x 3 − 9 x 2 + 4 ) − 1 24 ( 3 ( 3 − 2 x ) x 2 s g n ( x ) + s g n ( 1 − x ) ( 3 ( 3 − 2 x ) x 2 s g n ( x ) − 6 x 3 + 9 x 2 − 4 ) + 6 x 3 − 9 x 2 + 4 ) ] 0 1 = 1 24 ( 3 + 6 − 9 + 4 ) − 1 24 ( − 4 + 4 ) { \left[ \frac { 1 }{ 24 } \left( 3\left( 3-2x \right) { x }^{ 2 }sgn\left( x \right) +sgn\left( 1-x \right) \left( 3\left( 3-2x \right) { x }^{ 2 }sgn\left( x \right) -6{ x }^{ 3 }+9{ x }^{ 2 }-4 \right) +6{ x }^{ 3 }-9{ x }^{ 2 }+4 \right) -\frac { 1 }{ 24 } \left( 3\left( 3-2x \right) { x }^{ 2 }sgn\left( x \right) +sgn\left( 1-x \right) \left( 3\left( 3-2x \right) { x }^{ 2 }sgn\left( x \right) -6{ x }^{ 3 }+9{ x }^{ 2 }-4 \right) +6{ x }^{ 3 }-9{ x }^{ 2 }+4 \right) \right] }_{ 0 }^{ 1 }=\frac { 1 }{ 24 } \left( 3+6-9+4 \right) -\frac { 1 }{ 24 } \left( -4+4 \right) \\ and it is 1 6 \frac { 1 }{ 6 }

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