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By integration by parts, we know $$\int x\sqrt{5-x}dx=x\frac{-2}{3}(5-x)^{3/2}-\frac{4}{15}(5-x)^{5/2}+C$$

The answer for this question in the back of the book is $$\frac{2}{5}(5-x)^{5/2}-\frac{10}{3}(5-x)^{3/2}+C$$ How can I manipulate the first expression to get the second one?

Blue
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William Garske
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  • What did you try? https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question/27933#27933 – Anne Bauval Apr 20 '23 at 16:36
  • I'm not sure where to start which is my problem. – William Garske Apr 20 '23 at 16:46
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    Calculate the difference of the two expressions. You may set $y=5-x$ to simplify the notations (so $x=5-y$), – Anne Bauval Apr 20 '23 at 16:48
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    It may help to graph these two functions to see if they really are off by a constant, and, if so, what that constant is. – TomKern Apr 20 '23 at 23:22
  • When you see that extra $x$, just do $x=5-(5-x)$, then rearrange this subtraction apart will do your work. – Nanayajitzuki Apr 24 '25 at 15:24
  • Weirdly, when I evaluate this antiderivative, I get $\frac{-2}{15}(3x+10)(5-x)^{3/2}$. (It's not weird.) You could also differentiate both your and the book answers to see if you recover the same integrand (then the "differs by a constant" would "go away"). – Eric Towers Apr 24 '25 at 15:24

1 Answers1

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These expressions are in fact the same:

$$ \begin{align} \text{(your answer)}{}-\text{(book answer)} &= \left(x\frac{-2}{3}(5-x)^{3/2}-\frac{4}{15}(5-x)^{5/2}\right) - \left(\dfrac{2}{5}(5-x)^{5/2}-\dfrac{10}{3}(5-x)^{3/2}\right) \\ &= (5-x)^{5/2} \left(-\frac{4}{15} - \frac{2}{5}\right) + (5-x)^{3/2}\left(-\frac{2}{3}x + \frac{10}{3}\right) \\ &= (5-x)^{3/2} \left((5-x)\left(-\frac{2}{3}\right) + \left(\frac{2}{3}\right)(5-x)\right) \\ &= 0. \end{align}$$

Thus they are both valid ways to write the answer.

Robin
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    (It would even be fine if the difference were some constant instead of zero. (There is nothing wrong with this Answer.)) – Eric Towers Apr 24 '25 at 15:21