New Calculus student needs help with integrals

[jackchen]

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

 

[gb7nash]

What have you tried so far?

 

[jackchen]

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! =)

 

[gb7nash]

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.

 

[jackchen]

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! =)

 

[gb7nash]

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:

$$\frac{x^{1.6}+26}{x} = x^{0.6}+\frac{26}{x}$$

Do you see how to integrate this now?

 

[jackchen]

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

 

[gb7nash]

Do you know the power rule for integration? If not...

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

 

[jackchen]

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

 

[gb7nash]

That looks much better.

 

[jackchen]

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 =)

 

[Dick]

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??

 

[jackchen]

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!

 

[jackchen]

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!

 

[Dick]

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.

 

[Dick]

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?

 

[jackchen]

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.

 

[jackchen]

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! =)

 

[Dick]

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.

 

[Dick]

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?