Area Solution 6

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SOLUTION 6:

$\ \$ If

$y= \displaystyle{ { x^3 \over 6 } + { 1 \over 2x } = (1/6)x^3 + (1/2)x^{-1} }$ for

$1 \le x \le 3$ , then

$\displaystyle{ { dy \over dx} = (1/6)(3)x^{2} + (1/2)(-1)x^{-2} }$

$= \displaystyle{ (1/2)x^2 - { 1 \over 2x^2 } }$

$= \displaystyle{ {x^2 \over 2} { x^2 \over x^2} - { 1 \over 2x^2 } }$

$= \displaystyle{ { x^4 - 1 } \over 2x^2 }$ so that

$ARC = \displaystyle{ \int_{1}^{3} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx }$

$= \displaystyle{ \int_{1}^{3} \sqrt{ 1 + \Big({ { x^4 - 1 } \over 2x^2 } \Big)^2 } \ dx }$

$= \displaystyle{ \int_{1}^{3} \sqrt{ { 4x^4 \over 4x^4} + { {x^{8} - 2x^4 + 1 }\over 4x^4} } \ dx }$

$= \displaystyle{ \int_{1}^{3} \sqrt{ { {x^{8} + 2x^4 + 1 } \over 4x^4} } \ dx }$

$= \displaystyle{ \int_{1}^{3} { \sqrt{ x^{8} + 2x^4 + 1 } \over \sqrt{ 4x^4} } \ dx }$

$= \displaystyle{ \int_{1}^{3} { \sqrt{ (x^{4} + 1)^2 } \over \sqrt{ (2x^2)^2} } \ dx }$

$= \displaystyle{ \int_{1}^{3} { { x^{4} + 1 } \over 2x^2 } \ dx }$

$= \displaystyle{ \int_{1}^{3} { \Big( { x^{4} \over 2x^2 } + { 1 \over 2x^2} \Big) } \ dx }$

$= \displaystyle{ \int_{1}^{3} { \Big( (1/2)x^{2} + (1/2)x^{-2} \Big) } \ dx }$

$= \displaystyle{ { \Big( (1/2){x^{3} \over 3} + (1/2) {x^{-1} \over -1 } } \Big) \ \Big\vert_{1}^{3} }$

$= \displaystyle{ { \Big( {x^{3} \over 6} - { 1 \over 2x } } \Big) \ \Big\vert_{1}^{3} }$

$= \displaystyle{ { \Big( {(3)^{3} \over 6} - { 1 \over 2(3) } } \Big) } - \displaystyle{ { \Big( {(1)^{3} \over 6} - { 1 \over 2(1) } } \Big) }$

$= \displaystyle{ {27 \over 6} - { 1 \over 6 } - { 1 \over 6 } + { 1 \over 2 } }$

$= \displaystyle{ { 25 \over 6 } + { 3 \over 6 } }$

$= \displaystyle{ 28 \over 6 }$

$= \displaystyle{ 14 \over 3 }$

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