Antiderivative and Integral Calculus: Solving for the Correct Answer

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Homework Statement


\int (X-1) \sqrt{X} dx


The Attempt at a Solution


It is multiple choice, I believe the answer is either (2/5)x^(5/2)-(2-3)x^(3/2)+c or
(1/2)x^2+2x^(3/2)-x+c

I have tried to find the derivatives of both these answers yet neither of them gave me the correct antidrivatives. I must be doing something wrong. To do it the right way I have done the u's but I just can't figure it out.
 
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\int(x-1)\sqrt xdx

What was your first step?

Hint: Distribute the \sqrt x
 
I distributed the \sqrt{x} and replaced with u's I then got

\int u-u^(1/2) This let's see where the -2/3x^(3/2) comes from but I still don't understand where the 2/5x^(5/2) comes from.
 
You don't need to replace with u's. What are the rules for multiplying variables with the same base with exponents?
 
Thank you, I forgot how to distribute for a second.
 
This seems a fairly straightforward case of multiplying out the brackets and solving using the sum rule of integrals:

\int \left(f \pm g\right) \,dx = \int f \,dx \pm \int g \,dx\rightarrow

\int (x-1)\sqrt{x}\,dx\rightarrow \int x^{\frac{3}{2}}-x^{\frac{1}{2}}\,dx=\frac{2}{5}x^{\frac{5}{2}}-\frac{2}{3}x^{\frac{3}{2}}+c

No need to use the u unless the question asks you to? Or am I missing something here?
 
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I'm quite surprised they have multiple-choice anti derivative questions! I mean, if one doesn't really know how to integrate it they can just differentiate every option and see which one matches.
 
Gib Z said:
I'm quite surprised they have multiple-choice anti derivative questions! I mean, if one doesn't really know how to integrate it they can just differentiate every option and see which one matches.

It sounds like calculus for dummies. :smile: I've never heard of multiple choice exams either? Not in A' Level or anywhere else?
 
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