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The function is: $$ f(x) =2\ln(4-x^2)$$ and the length of the arc is from $-1$ to $1$.

So i know the formula for calculating the arc length is$$\int_{a}^{b}\sqrt{1+(f'(x))^2} dx$$ the derivative of the function $f(x)$ is, if i am correct $$2\frac{d}{dg}\ln(g)\frac{d}{dx}(4-x^2)$$ with $g = (4-x^2)$. So$$ f'(x)= -\cfrac{4x}{4-x^2}$$and $$(f'(x))^2=\cfrac{16x^2}{(4-x^2)^2}$$ and the formula becomes: $$\int_{-1}^{1}\sqrt{1+\cfrac{16x^2}{(4-x^2)^2}}dx$$ the problem is when i use the $u$ substitution of $$u=1+\cfrac{16x^2}{(4-x^2)^2}$$ And i want to make new boundaries for the integration $i$ have $2$ the same numbers (both are 2,77). I don't know what i have done wrong but the answer must be $$4\ln(3)-2$$

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The substitution you're suggesting won't help simplify your problem.

$$1+\frac{16x^2}{(4-x^2)^2}$$

$$=\frac{(4-x^2)^2 + 16x^2}{(4-x^2)^2}$$ $$=\frac{(4+x^2)^2}{(4-x^2)^2}$$

This will help solve.

To the question you have posed, this is happening because the function is even. And if $f(x)$ is an even function, then:

$$\int_{-1}^{1} f(x) dx = 2 \int_{0}^{1} f(x) dx$$

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It turns out that$$(\forall x\in[-1,1]):\sqrt{1+f'(x)^2}=\frac{4+x^2}{4-x^2}=\frac2{x+2}-\frac2{x-2}-1$$and my guess is that you can take it from here.

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When you do a $u$ substitution, you have to make sure that $u$ is an injective function over the limits of integration. The way you've defined your $u$, however, $u(-x)=u(x)$, so it's not injective. When you substitute $du$ in for $dx$, you should find that $dx = \frac {du}{h'(x)}$, where $h(x) = 1+\cfrac{16x^2}{(4-x^2)^2}$. But at $x=0$, $h'(0)=0$, making this an improper integral. You can fix this by splitting the integral into an integral from $-1$ to $0$ (strictly speaking, to be rigorous you have to take the integral from $-1$ to $-\epsilon$ and take the limit as $\epsilon$ goes to zero, but you can get away with fudging on that point), and another integral from $0$ to $1$. $u$ will then be injective within each integral. And since $u$ is symmetric, you can simply take one integral and double the answer you get.

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