Antiderivative and Integral Calculus: Solving for the Correct Answer

In summary, the conversation is about finding the correct antiderivative for the integral \int (x-1)\sqrt{x}\,dx, with the options of (2/5)x^(5/2)-(2-3)x^(3/2)+c or (1/2)x^2+2x^(3/2)-x+c. The person is struggling with finding the correct answer and is discussing their thought process, including using the u-substitution method. Another person suggests using the sum rule of integrals and the conversation ends with a discussion about the use of multiple choice exams in calculus.
  • #1
viper2308
19
0

Homework Statement


[tex]\int[/tex] (X-1) [tex]\sqrt{X}[/tex] 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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  • #2
[tex]\int(x-1)\sqrt xdx[/tex]

What was your first step?

Hint: Distribute the [tex]\sqrt x[/tex]
 
  • #3
I distributed the [tex]\sqrt{x}[/tex] and replaced with u's I then got

[tex]\int[/tex] 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.
 
  • #4
You don't need to replace with u's. What are the rules for multiplying variables with the same base with exponents?
 
  • #5
Thank you, I forgot how to distribute for a second.
 
  • #6
This seems a fairly straightforward case of multiplying out the brackets and solving using the sum rule of integrals:

[tex]\int \left(f \pm g\right) \,dx = \int f \,dx \pm \int g \,dx\rightarrow[/tex]

[tex]\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[/tex]

No need to use the u unless the question asks you to? Or am I missing something here?
 
Last edited:
  • #7
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
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?
 

What is an antiderivative?

An antiderivative is the inverse operation of differentiation in calculus. It is a function that, when differentiated, gives the original function as its result. It is also known as an indefinite integral.

What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration, indicating the start and end points of the integration. It gives a numerical value as the result. An indefinite integral, on the other hand, has no specified limits and is represented by a constant of integration. It gives a general function as the result.

How do you find the antiderivative of a function?

To find the antiderivative of a function, you need to use the reverse rules of differentiation. These rules include the power rule, product rule, quotient rule, and chain rule. You can also use tables of integrals or software programs to find the antiderivative.

What is the relationship between antiderivatives and integrals?

The antiderivative and integral are inverse operations of each other. When you take the antiderivative of a function, you get a general function, and when you take the integral of a function, you get a numerical value. The definite integral can be calculated using the antiderivative.

What are the applications of antiderivatives and integrals?

Antiderivatives and integrals are used in various fields such as physics, engineering, economics, and statistics. They are used to calculate areas, volumes, and other quantities in real-life applications. They are also used to solve differential equations, which are essential in understanding natural phenomena.

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