Antiderivative and Integral Calculus: Solving for the Correct Answer

[viper2308]

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.

 

[rocomath]

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

What was your first step?

Hint: Distribute the $$\sqrt x$$

 

[viper2308]

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.

 

[Snazzy]

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

 

[viper2308]

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

 

[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?

 

[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.

 

[Schrodinger's Dog]

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. I've never heard of multiple choice exams either? Not in A' Level or anywhere else?