Calculus

Calculus Level 2

F i n d t h e v a l u e o f y . y = 1 6 x 2 . d x 3 4 x . d x Find\quad the\quad value\quad of\quad y.\\ y=\int _{ 1 }^{ 6 }{ { x }^{ 2 } } .dx\quad -\quad \int _{ 3 }^{ 4 }{ x } .dx

Till 2 decimals


The answer is 68.17.

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1 solution

Justin Tuazon
Nov 8, 2014

y = 1 6 x 2 d x 3 4 x d x = [ x 3 3 ] 1 6 [ x 2 2 ] 3 4 = 6 3 3 1 3 3 4 2 2 + 3 2 2 = 216 3 1 3 16 2 + 9 2 = 215 3 7 2 = 430 21 6 = 409 6 = 68.16666... y = 68.17 y=\int _{ 1 }^{ 6 }{ { x }^{ 2 }dx } -\int _{ 3 }^{ 4 }{ x } dx\\ \\ =\quad { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 1 }^{ 6 }-\quad { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 3 }^{ 4 }\\ \\ =\frac { { 6 }^{ 3 } }{ 3 } -\frac { { 1 }^{ 3 } }{ 3 } -\frac { { 4 }^{ 2 } }{ 2 } +\frac { { 3 }^{ 2 } }{ 2 } \\ \\ =\quad \frac { 216 }{ 3 } -\frac { 1 }{ 3 } -\frac { 16 }{ 2 } +\frac { 9 }{ 2 } \\ \\ =\frac { 215 }{ 3 } -\frac { 7 }{ 2 } =\frac { 430-21 }{ 6 } \\ \\ =\frac { 409 }{ 6 } =68.16666...\\ \\ \therefore \quad \boxed { y=68.17 }

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