# `y = x^2/2 , [0,4]` Find the arc length of the curve over the given interval.

## Expert Answers

To find the arc length of a curve, we follow the formula:

`S = int_a^b sqrt(1+((dy)/(dx))^2)`  if `y=f(x)` , `alt=xlt=b` or ` [a,b]` .

For the given problem: `y =x^2/2` on interval `[0,4]` , we have boundary values: `a= 0` and `b=4` .

Apply Power Rule for differentiation:` d/(dx) x^n = n * x^(n-1) * dx` .

`(dy)/(dx) = d/(dx) (x^2/2)`

` = 1/2d/(dx) (x^2)`

` = 1/2 * [ 2 *x^(2-1) * 1 ]`

` =1/2 * [ 2x]`

` = (2x)/2 `

` = x`

Plug-n `a=0` , `b = 4` , and `(dy)/(dx)= x` on the formula `S = int_a^b sqrt(1+((dy)/(dx))^2)` , we get:

`S = int_0^4 sqrt(1+x^2) dx`

From indefinite integral table, the problem resembles the formula for integral with roots:

`int sqrt(u^2+-a^2) dx=1/2usqrt(a^2+-u^2)+-1/2a^2ln|u+sqrt(u^2+-a^2)|` .

Take note we have "`+` " sign inside the radical part then we follow formula as:

`int sqrt(u^2+a^2)dx=1/2*usqrt(a^2+u^2)+1/2*a^2ln|u+sqrt(u^2+a^2)|` .

Applying the formula, we get

`S = int_0^4 sqrt(1+(x)^2)`

` =[1/2*xsqrt(1^2+x^2)+1/2*1^2ln|x+sqrt(x^2+1^2)|]|_0^4`

`=[1/2*xsqrt(1+x^2)+1/2*ln|x+sqrt(x^2+1)|]|_0^4`

` =[(xsqrt(1+x^2))/2+(ln|x+sqrt(x^2+1)|)/2]|_0^4`

Apply definite integration formula: `F(x)|_a^b= F(b)-F(a)` .

`S =[(4sqrt(1+4^2))/2+(ln|4+sqrt(4^2+1)|)/2]-[(0sqrt(1+0^2))/2+(ln|0+sqrt(0^2+1)|)/2]`

` =[(4sqrt(1+16))/2+(ln|4+sqrt(16+1)|)/2]-[(0sqrt(1+0))/2+(ln|0+sqrt(0+1)|)/2]`

` =[(4sqrt(17))/2+(ln|4+sqrt(17)|)/2]-[0/2+(ln|0+sqrt(1)|)/2]`

` =[ 2sqrt(17)+(ln|4+sqrt(17)|)/2]-[0+(ln|1|)/2]`

` =[ 2sqrt(17)+(ln|4+sqrt(17)|)/2]-[0+0/2]`

` =[ 2sqrt(17)+(ln|4+sqrt(17)|)/2]-[0]`

` =2sqrt(17)+(ln|4+sqrt(17)|)/2`  or `9.29` (approximated value)

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