Answer in Calculus for JaytheCreator #264742

Question 3

Determine the length of the curve = ^2 /2 for 0 ≤ ≤ 1/2 . Assume positive.

“L= int sqrt{1+(y’)^2}dx\ny=sqrt{2x}\ny’=frac{1}{sqrt{2x}}\nL= int_0^frac{1}{2}sqrt{1+frac{1}{2x}}dx\”

This can also be written as “L= int_0^frac{1}{2}sqrt{frac{1}{2x}+1}dx\”

we apply linearity

“frac{1}{sqrt{2}}int sqrt{frac{1}{x}+2}dx\”

let u = “sqrt{frac{1}{x}+2}\” , dx = -2“sqrt{frac{1}{x}+2x^2}du, dx=-2int frac{u^2}{(u^2-2)}du\”

we integrate “int frac{u^2}{(u^2-2)}du = int (frac{u^2-2}{(u^2-2)^2}+frac{2}{(u^2-2)^2})du=\”

“int (frac{1}{(u^2-2)}+frac{2}{(u^2-2)^2})du\”

we split and integrate differently.

“int (frac{1}{(u^2-2)}du implies frac{Ln(u-sqrt{2})}{2^{frac{3}{2}}}- frac{Ln(u+sqrt{2})}{2^{frac{3}{2}}}”

also

“intfrac{2}{(u^2-2)^2}du implies int (frac{1}{(u^2-2)}du implies frac{Ln(u-sqrt{2})}{2^{frac{3}{2}}}- frac{Ln(u-sqrt{2})}{2^{frac{3}{2}}} – frac{1}{4(u+sqrt{2})}- frac{1}{4(u-sqrt{2})}\”

but substituting the solution of the integrations back, we get

“frac{Ln(u-sqrt{2})}{2^{frac{3}{2}}}- frac{Ln(u+sqrt{2})}{2^{frac{3}{2}}}+ frac{Ln(u-sqrt{2})}{2^{frac{3}{2}}}- frac{Ln(u-sqrt{2})}{2^{frac{3}{2}}} – frac{1}{4(u+sqrt{2})}- frac{1}{4(u-sqrt{2})}”

recall, u= “sqrt{frac{1}{x}+2}\” then we substitute u in terms of x back and introduce the lower and upper limits

“frac{Ln(u-sqrt{2})}{2^{frac{3}{2}}}- frac{Ln(u+sqrt{2})}{2^{frac{3}{2}}}+ frac{Ln(u-sqrt{2})}{2^{frac{3}{2}}}- frac{Ln(u-sqrt{2})}{2^{frac{3}{2}}} – frac{1}{4(u+sqrt{2})}- frac{1}{4(u-sqrt{2})}|_0^frac{1}{2}”

We have.

“frac{1}{sqrt{2}}+frac{Ln(sqrt{2}+1)-Ln(sqrt{2}-1)}{4}”