New Calculus student needs help with integrals

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Homework Help Overview

The discussion revolves around integral calculus, specifically focusing on evaluating definite integrals and understanding the underlying concepts such as u-substitution and integration techniques. The original poster presents two integral problems with specified limits and seeks assistance in solving them.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods for evaluating the integrals, including substitution techniques and the application of integration rules. Questions arise regarding the use of u-substitution and the correct approach to integrating specific functions. Some participants express confusion about the steps involved in the integration process.

Discussion Status

The conversation is ongoing, with participants providing insights and corrections to each other's approaches. Some have offered guidance on integration techniques, while others are still grappling with the concepts. There is a mix of attempts at solving the integrals, and multiple interpretations of the problems are being explored.

Contextual Notes

Participants mention constraints such as the requirement to provide answers to two decimal places and the need for clarity on critical points in related calculus problems. There is also a focus on ensuring that the mathematical terminology and processes are accurately understood.

jackchen
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I'm a new Calculus student, and it's killing me! Please help! :(
Thank you

Homework Statement



Find to 2-decimal place of the following integrals.


Homework Equations



1. (Upper limit= 33, lower limit= 18)
1 / ([7x+8]^0.25) dx

2. (Upper limit = 33, lower limit = 5)
(x^1.6 + 26) / (x) dx


I am so at a loss, please help me. Thank you
 
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What have you tried so far?
 
For Question 1:
: [ln|(7x+8)^0.25|] (upper=33, lower=18)
Replace x with upper and lower, and subtract, I've got
3.931872942 - 3.402328159 = 0.529544783

For Question 2:
: (33^1.6 + 26) / 33 - (5^1.6 + 26) / 5
= 8.936954709 - 7.826527804 = 1.110426905

Am I right? Thanks! =)
 
jackchen said:
For Question 1:
: [ln|(7x+8)^0.25|] (upper=33, lower=18)
Replace x with upper and lower, and subtract, I've got
3.931872942 - 3.402328159 = 0.529544783

For Question 2:
: (33^1.6 + 26) / 33 - (5^1.6 + 26) / 5
= 8.936954709 - 7.826527804 = 1.110426905

Am I right? Thanks! =)

Unfortunately, no.

1) Do you know how to use u-substitution?

2) Divide both terms out by x.
 
Hello!

For 1) I don't know u-substitution.

For 2), what do you mean by dividing the terms by x?
I replaced the equation once with 33 (upper limit for x), and another time with 5 (lower limit for x), then substract them together.
What is the correct way?

Thanks! =)
 
jackchen said:
Hello!

For 1) I don't know u-substitution.

For 2), what do you mean by dividing the terms by x?
I replaced the equation once with 33 (upper limit for x), and another time with 5 (lower limit for x), then substract them together.
What is the correct way?

Thanks! =)

You can't do 1) if you don't know u-substitution. Here's a link if you need a refresher:

http://en.wikipedia.org/wiki/Integration_by_substitution

I don't know what you're doing with 2). You plug in the numbers after you integrate. By dividing, I mean:

[tex]\frac{x^{1.6}+26}{x} = x^{0.6}+\frac{26}{x}[/tex]

Do you see how to integrate this now?
 
Hello, please pardon me as I'm very weak in Math and Calculus.

For Question 2:
If it becomes x^0.6 + 26/x, according to my textbook "26/x" will become "26ln(x)"?

x^0.6 is supposed to be f'(x) right?
Am I supposed to convert it back to f(x)? How do I do that with the power being 0.6?

Thank you!
Jack
 
Thank you gb7nash!

I've worked out the following for question 2:
"(x^1.6 + 26) / x" will become "(x^1.6 / 1.6) + 26ln(x)"

Replacing x with the upper and lower limits:
((33^1.6 / 1.6) + 26ln(33)) - ((5^1.6 / 1.6) + 26ln(5))
I get: 258.9838875 - 50.05328511 = 208.9306018

Is this correct?

Thank you!
Jack
 
  • #10
That looks much better.
 
  • #11
Hello, for question 1, this is what I've worked out.

Original Question: 1 / (7x + 8)^0.25 dx (upper=33, lower=18)

This becomes:
==> (7x + 8) ^ -0.25 dx
==> ((7x + 8) ^ 0.75) / ( 7 ^ 0.75)
==> 4/21 [(7x + 8) ^ 0.75]

Replacing x with upper limit of 33, and lower limit 18
==> 4/21 [(7(33) + 8) ^ 0.75 - (7(18) + 8) ^ 0.75]
==> 4/21 [60.78528058 - 39.38479586]
==> 4/21 [21.40048472]
Final answer: 4.076282804

Am I correct for question 1?

Thank you =)
 
  • #12
jackchen said:
Hello, for question 1, this is what I've worked out.

Original Question: 1 / (7x + 8)^0.25 dx (upper=33, lower=18)

This becomes:
==> (7x + 8) ^ -0.25 dx
==> ((7x + 8) ^ 0.75) / ( 7 ^ 0.75)
==> 4/21 [(7x + 8) ^ 0.75]

Replacing x with upper limit of 33, and lower limit 18
==> 4/21 [(7(33) + 8) ^ 0.75 - (7(18) + 8) ^ 0.75]
==> 4/21 [60.78528058 - 39.38479586]
==> 4/21 [21.40048472]
Final answer: 4.076282804

Am I correct for question 1?

Thank you =)

Yes, you are. But your derivation isn't making me comfortable. The integral isn't ((7x + 8) ^ 0.75) / ( 7 ^ 0.75) and then you magically went to the correct 4/21 [(7x + 8) ^ 0.75], what happened??
 
  • #13
I have another question 3.

Question: f(x) = x^4 + ax^2
When "a" = 191, there is more than 1 critical path.
Considering the x-coordinates of the critical path, find the greatest value of x to 2 decimal places.

My working:
==> f(x) = x^4 + (191)x^2
==> f'(x) = 4x^3 - (382)x = 0
==> 2x(2x^2 - 191) = 0
==> Therefore: 2x^2 = 191, solving it, final answer: x = 9.772410143

Is this right too?
Thank you so much!
 
  • #14
Dick said:
Yes, you are. But your derivation isn't making me comfortable. The integral isn't ((7x + 8) ^ 0.75) / ( 7 ^ 0.75) and then you magically went to the correct 4/21 [(7x + 8) ^ 0.75], what happened??

Hello, I'm sorry, it should be
(7x + 8) ^ 0.75 / (7(0.75))

I wrote wrongly, I'm sorry!
 
  • #15
jackchen said:
Hello, I'm sorry, it should be
(7x + 8) ^ 0.75 / (7(0.75))

I wrote wrongly, I'm sorry!

That's much better. Thanks. But you still may need to show more steps on how you did that. Like using u-substitution.
 
  • #16
jackchen said:
I have another question 3.

Question: f(x) = x^4 + ax^2
When "a" = 191, there is more than 1 critical path.
Considering the x-coordinates of the critical path, find the greatest value of x to 2 decimal places.

My working:
==> f(x) = x^4 + (191)x^2
==> f'(x) = 4x^3 - (382)x = 0
==> 2x(2x^2 - 191) = 0
==> Therefore: 2x^2 = 191, solving it, final answer: x = 9.772410143

Is this right too?
Thank you so much!

Partially. x=9.77.. is a critical point (not critical path) of f(x). But it's not the only solution of f'(x)=0. There are three of them. And I think you probably want to find the greatest value of f(x). Finding the 'greatest value of x' doesn't make much sense. And the greatest value of f(x) doesn't really work either, it has a minimum, it doesn't have a maximum. Are you sure that's the correct question?
 
  • #17
Dick said:
Partially. x=9.77.. is a critical point (not critical path) of f(x). But it's not the only solution of f'(x)=0. There are three of them. And I think you probably want to find the greatest value of f(x). Finding the 'greatest value of x' doesn't make much sense. And the greatest value of f(x) doesn't really work either, it has a minimum, it doesn't have a maximum. Are you sure that's the correct question?

Hello Dr Dick, the question is: "When a = 191, there is more than one critical point. Considering the x-coordinates of the critical points, what is the value of the greatest x-coordinate, giving your answer to 2 decimal places?"

Will my answer be correct now? Sorry I'm very weak in Calculus.
 
  • #18
Another question: Find the enclosed area between the curves, with the equations:
y = x^2 - 14x
y = 5.8x - x^2

My working
Using the rule: x = - (b/2a)
f(x) = x^2 - 14x
x = - (-14/2(1)) = 7
f(x) = 7^2 - 14(7) = -49
Minimum turning point = (7, -49)

g(x) = 5.8x - x^2
x = -(5.8/2(-1)) = 2.9
g(x) = 5.8(2.9) - (2.9^2) = 8.41
Maximum turning point = (2.9, 8.41)

f(x) = g(x)
==> x^2 - 14x = 5.8x - x^2
==> 2x^2 -19.8x = 0
==> 2x(x = 9.9) = 0
Therefore, x = 0 or x = 9.9

Solving with upperlimit = 9.9, lowerlimit = 0, top - bottom:
(5.8x - x^2) - (x^2 - 14x) dx
==> (19.8x - 2x^2) dx
==> [(19.8/2)x^2 - (2/3)x^3]
Replacing x = 9.9 and 0
==> (970.299 - 646.866) - (0)
Final Answer, area = 323.433

Is this correct?

Thank you! =)
 
  • #19
jackchen said:
Hello Dr Dick, the question is: "When a = 191, there is more than one critical point. Considering the x-coordinates of the critical points, what is the value of the greatest x-coordinate, giving your answer to 2 decimal places?"

Will my answer be correct now? Sorry I'm very weak in Calculus.

Yes, I think your answer is correct.
 
  • #20
jackchen said:
Another question: Find the enclosed area between the curves, with the equations:
y = x^2 - 14x
y = 5.8x - x^2

My working
Using the rule: x = - (b/2a)
f(x) = x^2 - 14x
x = - (-14/2(1)) = 7
f(x) = 7^2 - 14(7) = -49
Minimum turning point = (7, -49)

g(x) = 5.8x - x^2
x = -(5.8/2(-1)) = 2.9
g(x) = 5.8(2.9) - (2.9^2) = 8.41
Maximum turning point = (2.9, 8.41)

f(x) = g(x)
==> x^2 - 14x = 5.8x - x^2
==> 2x^2 -19.8x = 0
==> 2x(x = 9.9) = 0
Therefore, x = 0 or x = 9.9

Solving with upperlimit = 9.9, lowerlimit = 0, top - bottom:
(5.8x - x^2) - (x^2 - 14x) dx
==> (19.8x - 2x^2) dx
==> [(19.8/2)x^2 - (2/3)x^3]
Replacing x = 9.9 and 0
==> (970.299 - 646.866) - (0)
Final Answer, area = 323.433

Is this correct?

Thank you! =)

That's fine as well. But you didn't need your 'turning points' (I would call them critical points) to calculate the area, did you?
 

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