Consider the two functions h(x) = x² and g(x) = Vx on the interval [0, 1]. (a) Plot the two functions on the interval. (b) Based on the plots, do you think the arc-lengths of the two are equal? (c) Recall the formula for arc-length V1+ f'(x)²dx. Apply this to get the arclength of h and g as definite integrals. Call these numbers (definite integral values) In, Ig respectively.

Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered.

Plot the graph blue for $y=\sqrt{x}$ and red for $y=x^2$. 

It looks like that the two arcs are equal in length. 

To find arc length of $y=h\left(x\right)$: 

$$\begin{gathered} I_h={\int }_0^1\sqrt{1+h\text{'}{\left(x\right)}^2}dx \\ ={\int }_0^1\sqrt{1+{\left(\frac{d}{dx}{x}^2\right)}^2}dx \\ ={\int }_0^1\sqrt{1+{\left(2x\right)}^2}dx \\ ={\int }_0^1\sqrt{1+4x^2}dx \\ =2{\int }_0^1\sqrt{\frac{1}{4}+x^2}dx \\ =2{\left(\frac{x}{2}\sqrt{\frac{1}{4}+x^2}+\frac{1}{8}\ln \left(x+\sqrt{\frac{1}{4}+x^2}\right)\right)}_0^1 \\ =\sqrt{\frac{1}{4}+1^2}+\frac{1}{4}\ln \left(1+\sqrt{\frac{1}{4}+1^2}\right)-\frac{1}{4}\ln \left(\sqrt{\frac{1}{4}}\right) \\ =\sqrt{\frac{5}{4}}+\frac{1}{4}\ln \left(1+\sqrt{\frac{5}{4}}\right)+\frac{1}{4}\ln \left(2\right) \end{gathered}$$