Area Solution 8

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:browse confirm e Area Solution 8

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SOLUTION 8:

$\ \$ If

$y = (2/3)(x^2+1)^{3/2}$ for

$0 \le x \le 2$ , then

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

$ARC = \displaystyle{ \int_{0}^{2} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx }$

$= \displaystyle{ \int_{0}^{2} \sqrt{ 1 + ( 2x(x^2+1)^{1/2} )^2 } \ dx }$

$= \displaystyle{ \int_{0}^{2} \sqrt{ 1 + 2^2x^2(x^2+1) } \ dx }$

$= \displaystyle{ \int_{0}^{2} \sqrt{ 1 + 4x^4+4x^2 } \ dx }$

$= \displaystyle{ \int_{0}^{2} \sqrt{ (2x^2)^2+2(2x^2) + 1 } \ dx }$

$= \displaystyle{ \int_{0}^{2} \sqrt{ (2x^2+1)^2 } \ dx }$

$= \displaystyle{ \int_{0}^{2} (2x^2+1) \ dx }$

$= \displaystyle{ \Big( 2 {x^3 \over 3} +x \Big) \ \Big\vert_{0}^{2} }$

$= \displaystyle{ \Big( 2 {(2)^3 \over 3} + 2 \Big) - \Big( 2 {(0)^3 \over 3} + 0 \Big) }$

$= \displaystyle{ {16 \over 3} + { 6 \over 3 } }$

$= \displaystyle{ {22 \over 3} }$

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