I'm not getting the correct answer?

The problem asks to find the length of the curve $f(x)=x^2$ on the interval $[1,3]$. Using the definition of Arc Length = $$s=\int_1^3\left(\sqrt{1+[f'(x)]^2}\right)dx$$
Substitute $f'(x)= 2x$ $$s=\int_1^3\left(\sqrt{1+[2x]^2}\right)dx$$ Then u-substitute for $2x$.
$u=2x$ $\frac{du}{2}=dx$


From here I decided to use the Pythagorean Idenitie: $sec^2(\theta)-tan^2(\theta)=1$
if $tan(\theta)= u$,

$$s=\frac12\int_2^6\left(sec(\theta)\right)du$$ convert $du$ to $d\theta$



Then rewrite it for integration by parts




Integration by parts formula.



Substitute (sec$^2$($\theta$)-1) for tan$^2$($\theta$)


Distribute sec($\theta$)


Seperate into two integrals


Collect like integrals.


Pull out coefficient and Integrate sec($\theta$)

$$2\int(\frac{sec^3(\theta)}{2})d\theta=\frac{sec(\theta)tan(\theta)}{2}+\frac12 ln\lvert sec(\theta)+tan(\theta)\rvert$$

Divide by 2 on both sides of the equality sign

$$\int_2^6(\frac{sec^3(\theta)}{2})d\theta=\left[\frac{sec(\theta)tan(\theta)}{4}+\frac14 (ln\lvert sec(\theta)+tan(\theta)\rvert)\right]_2^6$$

Substitute from trig to algebraic

$$ \int_2^6(\frac {sec^3(\theta)}{2}) d\theta = \left[ \frac{(\sqrt{1+u^2})(u)}{4}+\frac14 (ln\lvert (\sqrt{1+u^2})+(u)\rvert) \right]_2^6 $$

$$ \left[ \frac{(\sqrt{1+(6)^2})(6)}{4}+\frac14 (ln\lvert (\sqrt{1+(6)^2})+(6)\rvert) \right]-\left[ \frac{(\sqrt{1+(2)^2})(2)}{4}+\frac14 (ln\lvert (\sqrt{1+(2)^2})+(2)\rvert) \right] $$

$$ \bbox[5px,border:2px solid red] {\left[ \frac{(3\sqrt{37})}{2}+\frac14 (ln\lvert (\sqrt{37})+6\rvert) \right]-\left[ \frac{\sqrt{5}}{2}+\frac14 (ln\lvert (\sqrt{5})+2\rvert) \right]} $$

$$ \left[ (9.12)+(0.62) \right]-\left[ (1.12)+(0.36) \right] $$

$$\bbox[5px,border:2px solid red]{\approx 8.0061098067}$$

Use the Distance Formula to approximate and double check answer. $$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ $(1,1)and(3,9)$


$$\approx 8.24621125124$$

  • 1
    $\begingroup$ Use a hyperbolic trigonometric substitution instead ($\sinh$), to simplify the calculation. Independently of that, you cannot check your answer using the distance formula: you habe calculated the length of a curved line, whereas the distance formula gives athe length of the straight segment connecting the two points... $\endgroup$ – b00n heT Mar 27 '17 at 21:01
  • $\begingroup$ MATLAB says eval(simplify(int(sqrt(1+4*x^2),1,3))) equals (approximately) 8.268145901063960. $\endgroup$ – Fabio Somenzi Mar 27 '17 at 21:05
  • $\begingroup$ @b00nheT Strictly speaking, you are right, but one can use the distance to realize that there's a problem: the length of the parabolic arc should not be less than the straight-line distance. I suspect that may be what the OP intended. $\endgroup$ – Fabio Somenzi Mar 27 '17 at 21:11
  • 2
    $\begingroup$ Why was this downvoted? $\endgroup$ – projectilemotion Mar 27 '17 at 21:12
  • $\begingroup$ @Fabio Somenzi. Indeed, you are correct as well. I admit not to have gone through all the calculation, but most definitely yours is an indicator of there being a mistake. $\endgroup$ – b00n heT Mar 27 '17 at 21:13

The exact expression you have obtained in the penultimate red box is correct, but the final numerical value in the last red box is not correct, and should be $8.268145901064$

When you do the substitution $2x=\sinh u$ this leads to $$L=\frac 12\int \cosh^2 udu=\frac 14(u+\sinh u\cosh u)$$

So when you apply the limits you are evaluating $$\frac 14(\operatorname{arsinh}6+6\sqrt{37}-\operatorname{arsinh}2-2\sqrt{5})$$


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