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I need to calculate the length of a curve $y=2\sqrt{x}$ from $x=0$ to $x=1$.

So I started by taking $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$, and then doing substitution: $\left[u = 1+\frac{1}{x}, \text{d}u = \frac{-1}{x^2}\text{d}x \Rightarrow -\text{d}u = \frac{1}{x^2}\text{d}x \right]^1_0 = -\int\limits^1_0 \sqrt{u} \,\text{d}u$ but this obviously will not lead to the correct answer, since $\frac{1}{x^2}$ isn't in the original formula.

Wolfram Alpha is doing a lot of steps for this integration, but I don't think that many steps are needed.

How would I start with this integration?

  • 0
    Then it's not also $-\int\limits^1_0 \sqrt{u} \,\text{d}u$, Since $dx=-x^2 du$.2012-05-28

9 Answers 9

11

Here's something you might try. Note that the length of that arc will be the same as the length of the same arc, reflected over the line $y=x$. That is, the arc $y=x^2/4$, from $x=0$ to $x=2$.

  • 0
    @TheGuyOfDoom: Glad you approve!2012-05-31
10

Substitute $u=\sqrt{1+\frac{1}{x}}=\sqrt{\frac{x+1}{x}}$. Then $x=\frac{1}{u^2-1}$, so $dx=-\frac{2u}{(u^2-1)^2}du$, which makes: $\int\sqrt{1+\frac{1}{x}}\,{dx}=-\int \frac{2u^2}{(u^2-1)^2}\,{du}$ Can you continue from here?

5

A good thing to note here is that $y = 2 \sqrt x$ is the same as $x = \frac{y^2}{4}$. So by 'swapping the order of integration' (sort of), you calculate a much easier integral. But do remember that the domain for $x$ here is $[0,1]$, but for $y$ it's $[0,2]$.

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  • 0
    @Cameron: Thanks for that -2012-05-28
3

$u=1+\frac{1}{x}\Longrightarrow x\to 0^+\longrightarrow u\to +\infty\,\,,\,\,x=1\longrightarrow u = 2$ so with the new integration limits we get $\int_\infty^2-\sqrt{u}\,\left(-\frac{du}{(u-1)^2}\right)=\int_2^\infty\frac{\sqrt{u}}{(u-1)^2}\,du$

2

Put $x= \tan^{2}\theta$, then you have the integral as \begin{align*} \int_{0}^{1} \sqrt{1+\frac{1}{x}} \ dx &= \int_{0}^{\pi/4} \sqrt{\frac{1+\tan^{2}\theta}{\tan^{2}\theta}} \cdot 2\tan\theta \cdot\sec^{2}\theta \ d\theta \\\ &= 2 \cdot\int_{0}^{\pi/4} \sec^{3}\theta \ d\theta \end{align*}

Now integrate this function by parts. Take $u = \sec\theta$ then $du = \sec\theta \cdot \tan\theta$ and $dv = \sec^{2}\theta$. Then you have $v = \tan\theta$, so \begin{align*} \int_{0}^{\pi/4} \sec^{3}\theta \ d\theta &= (\sec\theta\cdot\tan\theta)\:\biggl|_{0}^{\pi/4} - \int_{0}^{\pi/4} \sec\theta \cdot \tan^{2}\theta \ d\theta \\\ &= \frac{1}{\sqrt{2}} -\int_{0}^{\pi/4} \sec^{3}\theta \ d\theta + \int_{0}^{\pi/4} \sec\theta \ d\theta \\\ &= \frac{1}{2} \cdot \biggl\{ \frac{1}{\sqrt{2}} + \int_{0}^{\pi/4} \sec\theta \ d\theta \:\biggr\} \\\ &= \frac{1}{2\sqrt{2}} + \frac{1}{2} \cdot \bigl(\:\log(\sec\theta +\tan\theta)\bigr)_{0}^{\pi/4} \\\ &= \frac{1}{2\sqrt{2}} + \frac{1}{2} \cdot \log(\sqrt{2}+1) \end{align*}

1

You can try $x = \cot^2(\theta), dx = -2\cot(\theta) \csc^2(\theta)$. This substitution comes from knowing that $\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \dfrac{1}{\cot^2(\theta)}$. The final integral is quite simple depending upon your comfort with trigonometry.

1

Note that the integrand $\sqrt{1 + {1 \over x}}$ decreases from infinity to $\sqrt{2}$ as $x$ goes from $0$ to $1$. The area under the graph is therefore equal to the area of the box $[0,1] \times [0,\sqrt{2}]$ plus the area under the graph of the inverse function $g(y)$ to $\sqrt{1 + {1 \over x}}$ from $y = \sqrt{2}$ to $y = \infty$. Note that $g(y) = {1 \over y^2 - 1}$. So the answer is $\sqrt{2} + \int_{\sqrt{2}}^{\infty} {1 \over y^2 - 1}\,dy$ This integral is easily computed, using partial fractions for example. The result is $\sqrt{2} + {1 \over 2}\ln\bigg({y - 1 \over y + 1}\bigg)\bigg|_{\sqrt{2}}^{\infty}$ $=\sqrt{2} - {1 \over 2}\ln\bigg({\sqrt{2} - 1 \over \sqrt{2} + 1}\bigg)$

0

You can use integrate by parts: \begin{align} I&=\int_0^1\sqrt{1+\frac{1}{x}}dx=x\sqrt{1+\frac{1}{x}}|_0^1-\int_0^1x\frac{-\frac{1}{x^2}dx}{2\sqrt {1+\frac{1}{x}}} =\sqrt{x(1+x)}|_0^1+\int_0^1\frac{d\sqrt{x}}{\sqrt {1+x}}\\ &=\sqrt{2}+\ln(\sqrt{x}+\sqrt{x+1})|_0^1 =\sqrt{2}+\ln(1+\sqrt{2}) \end{align}

-3

Notice that the integrand is a differential binomial, and then you may apply "Integration of differential binomial" (P. L. Chebyshev) and you're immediately done. See here.

  • 0
    (The link you had was not working, so I updated it.)2018-08-04