Area Solution 2

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SOLUTION 2:

$\ \$ If

$y = x^{3/2}$ for

$0 \le x \le 4$ , then

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

$ARC = \displaystyle{ \int_{0}^{4} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx }$

$= \displaystyle{ \int_{0}^{4} \sqrt{ 1 + ((3/2)x^{1/2})^2 } \ dx }$

$= \displaystyle{ \int_{0}^{4} \sqrt{ 1 + ((3/2)^2(x^{1/2})^2 } \ dx }$

$= \displaystyle{ \int_{0}^{4} \sqrt{ 1 + (9/4)x } \ dx }$ (Now integrate using the method of u-substitution. Let

$u= 1 + (9/4)x \ \longrightarrow \ du= (9/4)dx \ \longrightarrow \ (4/9)du = dx \.$ )

$= \displaystyle{ (4/9) \int_{x=0}^{x=4} \sqrt{ u } \ du }$

$= \displaystyle{ (4/9)(2/3) u^{3/2} \ \Big\vert_{x=0}^{x=4} }$

$= \displaystyle{ (8/27) (1 + (9/4)x)^{3/2} \ \Big\vert_{x=0}^{x=4} }$

$= \displaystyle{ (8/27) (1 + (9/4)(4))^{3/2} - (8/27) (1 + (9/4)(0))^{3/2} }$

$= \displaystyle{ (8/27) (10)^{3/2} - (8/27) (1)^{3/2} }$

$= \displaystyle{ (8/27) 10^{3/2} - 8/27 }$

$= \displaystyle{ (8/27) ( 10^{3/2} - 1 ) }$

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