Area Solution 5

Processing math: 100%

tex2html_wrap_inline125

SOLUTION 5:

$\ \$ If

$y= \displaystyle{ { x^4 \over 4 } + { 1 \over 8x^2 } = (1/4)x^4 + (1/8)x^{-2} }$ for

$2 \le x \le 4$ , then

$\displaystyle{ { dy \over dx} = (1/4)4x^3 + (1/8)(-2)x^{-3} }$

$= \displaystyle{ x^3 - { 1 \over 4x^3 } }$

$= \displaystyle{ x^3 { 4x^3 \over 4x^3} - { 1 \over 4x^3 } }$

$= \displaystyle{ { { 4x^6 - 1 }\over 4x^3} }$ so that

$ARC = \displaystyle{ \int_{2}^{4} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx }$

$= \displaystyle{ \int_{2}^{4} \sqrt{ 1 + \Big({ { 4x^6 - 1 }\over 4x^3} \Big)^2 } \ dx }$

$= \displaystyle{ \int_{2}^{4} \sqrt{ { 16x^6 \over 16x^6} + { {16x^{12} - 8x^6 + 1 }\over 16x^6} } \ dx }$

$= \displaystyle{ \int_{2}^{4} \sqrt{ { {16x^{12} + 8x^6 + 1 } \over 16x^6} } \ dx }$

$= \displaystyle{ \int_{2}^{4} { \sqrt{ 16x^{12} + 8x^6 + 1 } \over \sqrt{ 16x^6} } \ dx }$

$= \displaystyle{ \int_{2}^{4} { \sqrt{ (4x^{6} + 1)^2 } \over \sqrt{ (4x^3)^2} } \ dx }$

$= \displaystyle{ \int_{2}^{4} { { 4x^{6} + 1 } \over 4x^3 } \ dx }$

$= \displaystyle{ \int_{2}^{4} { \Big( { 4x^{6} \over 4x^3 } + { 1 \over 4x^3} \Big) } \ dx }$

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

$= \displaystyle{ { \Big( {x^{4} \over 4} + (1/4) {x^{-2} \over -2 } } \Big) \ \Big\vert_{2}^{4} }$

$= \displaystyle{ { \Big( {x^{4} \over 4} - { 1 \over 8x^2 } } \Big) \ \Big\vert_{2}^{4} }$

$= \displaystyle{ { \Big( {(4)^{4} \over 4} - { 1 \over 8(4)^2 } } \Big) } - \displaystyle{ { \Big( {(2)^{4} \over 4} - { 1 \over 8(2)^2 } } \Big) }$

$= \displaystyle{ 64 - { 1 \over 128 } -4 + { 1 \over 32 } }$

$= \displaystyle{ 60 - { 1 \over 128 } + { 4 \over 128 } }$

$= \displaystyle{ { 7680 \over 128 } + { 3 \over 128 } }$

$= \displaystyle{ 7683 \over 128 }$

Click HERE to return to the list of problems.