# Antiderivative and Intergrals

1. Apr 3, 2008

### viper2308

1. The problem statement, all variables and given/known data
$$\int$$ (X-1) $$\sqrt{X}$$ dx

3. 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.

2. Apr 3, 2008

### rocomath

$$\int(x-1)\sqrt xdx$$

What was your first step?

Hint: Distribute the $$\sqrt x$$

3. Apr 3, 2008

### viper2308

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

$$\int$$ u-u^(1/2) This lets see where the -2/3x^(3/2) comes from but I still don't understand where the 2/5x^(5/2) comes from.

4. Apr 3, 2008

### Snazzy

You don't need to replace with u's. What are the rules for multiplying variables with the same base with exponents?

5. Apr 3, 2008

### viper2308

Thank you, I forgot how to distribute for a second.

6. Apr 4, 2008

### Schrodinger's Dog

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?

Last edited: Apr 4, 2008
7. Apr 4, 2008

### Gib Z

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.

8. Apr 4, 2008

### Schrodinger's Dog

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