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The function $y = f(x)$, restricted on the domain $ 0 < x < 1$ and satisfying

$$y^{5}+y^{4} + x = 0,$$

seems to be well-defined and smooth. So how does one integrate this thing? That is, what is $\int_{0}^{1} f(x) dx$?

Of course, one can use Newton's method to approximate $f(x)$ for any given value of $x$ and numerically integrate the result. But this feels uninspiring. So I was wondering if there is a general "trick" or insight to integrating algebraic functions.

michaelkovarik
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  • Not sure why you think the definite integral should be especially easy to compute when you can't even write down what the function itself is. – MPW Oct 21 '14 at 21:50

2 Answers2

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As MPW says, in general there is no reason to expect that the integral should be easy to compute. However, you can do the following. Differentiate the equation once to obtain $$0=5y'y^4+4y'y^3+1=y'y^3(y+4(y+1))+1. $$ Then use the equation to write $$ y+1 = -\frac{x}{y^4}$$ and plug this into the above to get $$-1= y'y^3(y+4(-\frac{x}{y^4}))=\frac{y'y^5-4xy'}{y},$$ or $$ y=4xy'-y'y^5.$$ Now you're in business: $$ \int_0^1 yd x=\int_0^1(4xy'-y'y^5)d x=4y(1)-4\int_0^1yd x -\frac{1}{6} y(1)^6.$$ Solving for the desired quantity you get $$ \int_0^1 yd x=-\frac{1}{30}y(1)^6+\frac{4}{5}y(1). $$

Teri
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This is @Teri: 's idea.

Given $$F(y) + x =0$$ differentiate and get $$F'(y)\, d y + d x = 0$$ and so $$y\, d x = - y\, F'(y)\, d y$$ and now integrate: $$\int_a^b y\, d x = \int_{y(a)}^{y(b)}-y F'(y) dy $$

orangeskid
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