Solve Algebra Problem: Integral of (10-x^.5)^2

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Homework Statement



http://www.wolframalpha.com/input/?i=integral+of+(10-x^.5)^2

Homework Equations





The Attempt at a Solution



so at the bottom of the page, it ends up converting

(1/6)(√x-10)3(3√x+10)

to

(x/6)(3x-80√x+600)

The problem is, when I try the simplification by myself, i always end up with 3x^2-80x^1.5+600x-10000. I realize I can simplify by x giving me

x(3x-80x^.5+600-10000/x)

Why does the answer given not have this extra 10000/x at the end?

Thanks
 
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That's not quite correct:$$\frac{(\sqrt{x}-10)^3(3\sqrt{x}+10)}{6}+c$$ ... gets simplified to: $$\frac{x(3x-80\sqrt{x}+600)}{6}+c$$ ... where ##c## is an arbitrary constant.

When you try the simplification by yourself, you get: $$3x^2-80x^{1.5}+600x-10000$$. Notice that the last 10000 is a constant? It just gets included in the c. (note - I think you left a 1/6 off the front.)

It's easy to miss - wolfram writes "+ arbitrary constant" in pale grey.
 
Simon Bridge said:
That's not quite correct:$$\frac{(\sqrt{x}-10)^3(3\sqrt{x}+10)}{6}+c$$ ... gets simplified to: $$\frac{x(3x-80\sqrt{x}+600)}{6}+c$$ ... where ##c## is an arbitrary constant.

When you try the simplification by yourself, you get: $$3x^2-80x^{1.5}+600x-10000$$. Notice that the last 10000 is a constant? It just gets included in the c. (note - I think you left a 1/6 off the front.)

It's easy to miss - wolfram writes "+ arbitrary constant" in pale grey.

ok i see, i kind of assumed that you had to integrate the 10000 first which wouldn't leave it as the C, but i understand now. Thanks!
 
Please post calculus problems in the Calculus & Beyond section, not in the Precalc Math section. I am moving this thread.
 
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