Definite Integrals Homework Solutions | Math Help

In summary, Homework Statement 1 has a problem notational issue with x's and u's being in the same integral and the use of "+C". Homework Statement 2 has a problem with calculating the indefinite integral and taking the difference.
  • #1
A_Munk3y
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Homework Statement


I have to find the definite integrals of some problems...
I did them in paint because i do not know how to do it on the forum so sorry if that is a problem

Here is the problem and my attempted answer.
I think i got the first one right, but the second one, someone told me that i should be getting -ln(2) not ln(2), but i don't see how i should be getting -ln2

here it is
2zolmba.jpg
 

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  • #2
Hello again! There's a couple things wrong notationally with the first problem. You shouldn't really have both u's and x's in the same integral, although your simplification is correct. Also, the whole third line...bad! This is a definite integral (that means is has limits of integration, the 0 and 4). When you integrate, you don't need the "+C". Think about it as you are going to get a definite answer-a number. The "+C" is used in indefinite integrals (ones without limits of integration) to signify that after integrating, you get a family of functions with the same derivative. For example [itex] x^3 [/itex] and [itex] x^3 +4 [/itex] both have the same derivative, so if you had [itex] \int 3x^2 dx [/itex] how do you know which one you would get back out? That is why you put the +C. Same for the second example.
 
  • #3
When you go from ln|ln(x)|+C to plugging in e-1 and e-2 you swap which one is subtracted from which
 
  • #4
Juanriq said:
Hello again! There's a couple things wrong notationally with the first problem. You shouldn't really have both u's and x's in the same integral, although your simplification is correct. Also, the whole third line...bad! This is a definite integral (that means is has limits of integration, the 0 and 4). When you integrate, you don't need the "+C". Think about it as you are going to get a definite answer-a number. The "+C" is used in indefinite integrals (ones without limits of integration) to signify that after integrating, you get a family of functions with the same derivative. For example [itex] x^3 [/itex] and [itex] x^3 +4 [/itex] both have the same derivative, so if you had [itex] \int 3x^2 dx [/itex] how do you know which one you would get back out? That is why you put the +C. Same for the second example.

I don't really agree that it's terrible to have x's and u's in the same integral. As long as you're aware that you need to get rid of one or the other before evaluating anything it's not going to cause any trouble.

Also, a completely valid strategy is to calculate the indefinite integral, then just take the difference. The +C's cancel (in fact, when you calculate the definite integral you're really doing this even if you don't notice)
 
  • #5
Right, I didn't notice that the +C was dropped when the evaluating the LOI. My bad
 
  • #6
Office_Shredder said:
When you go from ln|ln(x)|+C to plugging in e-1 and e-2 you swap which one is subtracted from which

Office_Shredder said:
I don't really agree that it's terrible to have x's and u's in the same integral. As long as you're aware that you need to get rid of one or the other before evaluating anything it's not going to cause any trouble.

Also, a completely valid strategy is to calculate the indefinite integral, then just take the difference. The +C's cancel (in fact, when you calculate the definite integral you're really doing this even if you don't notice)

Juanriq said:
Right, I didn't notice that the +C was dropped when the evaluating the LOI. My bad

oh, i see where i messed up with the switching on e-1 and the e-2.
Ok, I'm getting -ln2 now since it will be 0-ln2 = -ln2 :approve:

but I'm still confused... did i do anything else wrong? Or does it seem ok?

2i944yg.jpg
 
Last edited:
  • #7
Looks good!
 
  • #8
One quibble: In some of the work for your first integral and all of it for the second integral you are connecting the intermediate expressions with "implies" (==>). Use "equals" (=) instead. ==> should be used between statements such as equations or inequalities. For example,
x2 = 4 ==> x = 2 or x = -2 (typically equations or inequalities).
 
  • #9
oh, ok i get ya.. :)
Alright thanks guys
 

Related to Definite Integrals Homework Solutions | Math Help

1. What is a definite integral?

A definite integral is a mathematical concept that represents the accumulation of infinitesimally small quantities over an interval. It is denoted by the symbol ∫ and is used to calculate the total area under a curve or the total value of a function over a specific interval.

2. How do I solve a definite integral?

To solve a definite integral, you need to follow two steps: 1) evaluate the indefinite integral of the function, and 2) substitute the upper and lower limits of the interval into the indefinite integral and subtract the values. The result will be the solution to the definite integral.

3. What are the properties of definite integrals?

There are several properties of definite integrals, including linearity, symmetry, and the fundamental theorem of calculus. Linearity states that the integral of a sum of functions is equal to the sum of the integrals of each individual function. Symmetry states that the integral of an odd function over a symmetric interval is equal to 0. The fundamental theorem of calculus states that the derivative of the integral of a function is equal to the original function.

4. When should I use definite integrals?

Definite integrals are typically used when you need to calculate the area under a curve or the total value of a function over a specific interval. They are also useful for solving problems involving rates of change, such as calculating displacement or velocity.

5. How can I check my answer to a definite integral?

To check your answer to a definite integral, you can use the properties of definite integrals to verify that your solution is correct. You can also use a graphing calculator or online tool to graph the function and visually compare the area under the curve to your calculated solution.

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