1. ## Exponential Integral Questions

1. The curve y = √(ex +1) is rotated about the x-axis from x = 0 to x = 1. Find the exact volume of the solid formed.

2. Find the exact area enclosed between the curve y = e2x and the lines y = 1 and x = 2.

Thanks

2. ## Re: Exponential Integral Questions

1)
y^2 = e^x + 1
Volume = pi* integral(0 to 1) (e^x + 1) dx
= pi* [e^x + x] (0 to 1)
= pi* (e + 1) - e^0
= pi*e units cubed

2) At y = 1 we have 1 = e^2x
ln1 = 2x
x = 0

So we integrate from 0 to 2
A = integral(0 to 2) (e^2x) dx
= [e^2x / 2] (0 to 2)
= e^4 / 2 - 1/2
= 1/2 (e^4 - 1) units squared

Edit: yeah minus the rectangle as 1729 said
Edit: I got my second answer off yahoo answers (not copy pasted but I copied their answer) rip got betrayed

3. ## Re: Exponential Integral Questions

Originally Posted by pikachu975
1)
y^2 = e^x + 1
Volume = pi* integral(0 to 1) (e^x + 1) dx
= pi* [e^x + x] (0 to 1)
= pi* (e + 1) - e^0
= pi*e units cubed

2) At y = 1 we have 1 = e^2x
ln1 = 2x
x = 0

So we integrate from 0 to 2
A = integral(0 to 2) (e^2x) dx
= [e^2x / 2] (0 to 2)
= e^4 / 2 - 1/2
= 1/2 (e^4 - 1) units squared

Yeah I got the same answer for question 1 but I thought the textbook had a different answer - appears it was right all along! (my bad)

I got the same for question 2 as well but the textbook answers say 1/2(e^4 - 5) units squared. (I'm sure it says that)

Thoughts?

4. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
Yeah I got the same answer for question 1 but I thought the textbook had a different answer - appears it was right all along! (my bad)

I got the same for question 2 as well but the textbook answers say 1/2(e^4 - 5) units squared. (I'm sure it says that)

Thoughts?
\noindent The answer you have is A = \frac{1}{2}(e^4-1) square units. This is the area bound by the curve, x = 0 and x = 2 \textit{and} the x-axis. However, the question does not mention the x-axis. \\\\ When we graph all given boundaries, including both lines (and it is important to graph these, otherwise it becomes difficult to see), we see that the required area A is given by \\\\ A = \frac{1}{2}(e^4-1) - 2 \times 1. \\\\ We needed to subtract the area of the rectangle bounded by y = 1, x=2 and both axes as this is not included in the region described in the question.\\\\ Thus, \\ \begin{align*} A &= \frac{1}{2}e^4 - \frac{1}{2} - 2 \\ &= \frac{1}{2}e^4 - \frac{5}{2} \\ &= \frac{1}{2}(e^4 - 5) \text{ square units, as required}\end{align*}

5. ## Re: Exponential Integral Questions

Originally Posted by 1729
The textbook is correct.

\noindent The answer you have is A = \frac{1}{2}(e^4-1) square units. This is the area bound by the curve, x = 0 and x = 2 \textit{and} the x-axis. However, the question does not mention the x-axis. When we graph all given boundaries, we see that the required area is A = \frac{1}{2}(e^4-1) - 2 \times 1. We need to subtract the rectangle below the line y = 1. Thus, \\ \begin{align*} A &= \frac{1}{2}e^4 - \frac{1}{2} - 2 \\ &= \frac{1}{2}e^4 - \frac{5}{2} \\ &= \frac{1}{2}(e^4 - 5) \text{ square units, as required}\end{align*}
Thanks!

6. ## Re: Exponential Integral Questions

Two more questions:

1. Find stat points and inflexions for the curve y=(In(x) - 1)3 and determine their nature.

2.(3x+3)/(x2-9) = 1/(x+3) + 2/(x-3)

Hence find ∫(3x+3)/(x2-9) dx

EDIT: For question number 2, I got:

In(x) + (1/2)In(x-3) + C

7. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
Two more questions:

1. Find stat points and inflexions for the curve y=(In(x) - 1)3 and determine their nature.

2.(3x+3)/(x2-9) = 1/(x+3) + 2/(x-3)

Hence find ∫(3x+3)/(x2-9) dx

EDIT: For question number 2, I got:

In(x) + (1/2)In(x-3) + C

It is 2.

\noindent \begin{align*} \int\frac{3x+3}{x^2-9}dx &= \int\left[\frac{1}{x+3} + \frac{2}{x-3}\right]dx \\ &= \int\frac{1}{x+3}dx + 2\int\frac{1}{x-3}dx \\ &= \ln{(x+3)} + 2\ln{(x-3)} + c \end{align*}

$\noindent Note the difference between the following \\ \int\frac{a}{x}dx = a\int\frac{1}{x}dx = a\ln{x} + c \\ whereas \int\frac{1}{ax}dx = \frac{1}{a}\int\frac{1}{x}dx = \frac{1}{a}\ln{x} + c, where a is some constant$

8. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
Two more questions:

1. Find stat points and inflexions for the curve y=(In(x) - 1)3 and determine their nature.

2.(3x+3)/(x2-9) = 1/(x+3) + 2/(x-3)

Hence find ∫(3x+3)/(x2-9) dx
$\noindent 1. Let f(x) = \left(\ln x -1\right)^{3}, so f'(x) = 3\left(\ln x -1\right)^{2}\cdot \frac{1}{x} = 3\cdot \frac{\left(\ln x -1\right)^{2}}{x}, using the chain rule. Differentiating again via the quotient rule, we have$

\begin{align*}f''(x) &= 3\cdot \frac{2\cdot \left(\ln x -1\right)\cdot \frac{1}{x} \cdot x -\left(\ln x -1\right)^{2}\cdot 1}{x^{2}} \\ &= 3\cdot \frac{2\left(\ln x -1\right) -\left(\ln x -1\right)^{2}}{x^{2}}\end{align*}

Can you get it from here? (If you're not sure about how to solve the equations f'(x) = 0 and f"(x) = 0, you can use an intermediary substitution u = ln(x) – 1.)

9. ## Re: Exponential Integral Questions

Originally Posted by 1729
It is 2
Yes, I would assume the answer is right but I'm unsure as to what I did wrong...

10. ## Re: Exponential Integral Questions

Originally Posted by InteGrand
$\noindent 1. Let f(x) = \left(\ln x -1\right)^{3}, so f'(x) = 3\left(\ln x -1\right)^{2}\cdot \frac{1}{x} = 3\cdot \frac{\left(\ln x -1\right)^{2}}{x}, using the chain rule. Differentiating again via the quotient rule, we have$

\begin{align*}f''(x) &= 3\cdot \frac{2\cdot \left(\ln x -1\right)\cdot \frac{1}{x} \cdot x -\left(\ln x -1\right)^{2}\cdot 1}{x^{2}} \\ &= 3\cdot \frac{2\left(\ln x -1\right) -\left(\ln x -1\right)^{2}}{x^{2}}\end{align*}

Can you get it from here?
Based on that, I got (e,0) as a min TP but apparently it's a point of inflexion?

11. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
Yes, I would assume the answer is right but I'm unsure as to what I did wrong...
It's a 2, because a primitive of 2/(x – 3) is 2*ln|x – 3|.

12. ## Re: Exponential Integral Questions

Originally Posted by InteGrand
It's a 2, because a primitive of 2/(x – 3) is 2*ln|x – 3|.
Oh right, you need a 2 to make 1 equal 2 *facepalm*

13. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
Based on that, I got (e,0) as a min TP but apparently it's a point of inflexion?
How did you find it to be a local minimum?

14. ## Re: Exponential Integral Questions

Originally Posted by InteGrand
How did you find it to be a local minimum?
With the first derivative test

15. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
Based on that, I got (e,0) as a min TP but apparently it's a point of inflexion?
$\noindent f''(e) = 3 \times \frac{2(\ln{e}-1)-(\ln{e}-1)^2}{e^2} = 3 \times 0 = 0, as \ln{e} = 1 \\\\ Thus, f'(e) = f''(e) = 0$

Originally Posted by boredofstudiesuser1
With the first derivative test
$\noindent As 2 < e < 3 we can use x = 2 and x = 3 in the first derivative test to confirm the inflexion point. \\\\f'(2) = 3 \times \frac{(\ln{2}-1)^2}{2} \approx 0.14123... \\ f'(3) = 3 \times \frac{(\ln{3}-1)^2}{3} \approx 0.00972...\\\\As f'(e-\epsilon) > 0, f'(e) = 0 and f'(e+\epsilon) > 0, and f''(e) = 0 then there is an inflexion point at (e,0)$

16. ## Re: Exponential Integral Questions

Originally Posted by 1729
$\noindent f''(e) = 3 \times \frac{2(\ln{e}-1)-(\ln{e}-1)^2}{e^2} = 3 \times 0 = 0, as \ln{e} = 1 \\\\ Thus, f'(e) = f''(e) = 0$
Yeah it appears to work that way.

17. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
With the first derivative test
If you use the first derivative test, you should be able to see that f'(x) is positive just to the left of e and also just to the right of e, so it's neither a local minimum nor a maximum. It is a stationary point of inflection.

(Note that the numerator of f'(x) is always positive for x in the domain and x not equal to e, so the sign of f'(x) is just the sign of the denominator (which is x), so in any neighbourhood of e, we have f'(x) > 0, for x not equal to e.)

18. ## Re: Exponential Integral Questions

This is a concept based question.

If I need to integrate a normal exponential with the base being a number other than e, can I follow this formula?:

∫n^(ax) = (n^ax)/a(loge(n))

If not, could anyone provide a sorta formula you could use for integrating logarithms and exponentials with a base other than e?

19. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
This is a concept based question.

If I need to integrate a normal exponential with the base being a number other than e, can I follow this formula?:

∫n^(ax) = (n^ax)/a(loge(n))
$\noindent (Assuming you are integrating with respect to x) Yes, that's valid, for n > 0 constant and a a non-zero constant.$

20. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
This is a concept based question.

If I need to integrate a normal exponential with the base being a number other than e, can I follow this formula?:

∫n^(ax) = (n^ax)/a(loge(n))
$\noindent Well \frac{d}{dx}a^x = a^x \ln{a}. \\\\ a^x = \int a^x \ln{a}dx. \\ As \ln{a} is a constant, then a^x = \ln{a} \int a^{x}dx \Rightarrow \int a^{x}dx = \frac{a^x}{\ln{a}}+c \\\\ So this works for simple functions and for a > 0. If we have a composite function, where u = f(x) then \\\\ \frac{d}{dx}a^u = u'a^u\ln{a}. \\\\ a^u = \int u'a^u \ln{a}dx. In your case, u = nx, n \neq 0. Thus u' = n and therefore both u' = n and \ln{a} are constants. Thus, \\\\ a^u = n\ln{a} \int a^udx \Rightarrow \int a^u dx = \frac{a^u}{n \ln{a}}. Substituting u = nx we get what you proposed, \\\\ \int a^{nx}dx = \frac{a^{nx}}{n \ln{a}}$

21. ## Re: Exponential Integral Questions

Originally Posted by 1729
$\noindent Well \frac{d}{dx}a^x = a^x \ln{a}. Multiply both sides by dx \\\\ a^x = \int a^x \ln{a}dx. \\ As \ln{a} is a constant, then a^x = \ln{a} \int a^{x}dx \Rightarrow \int a^{x}dx = \frac{a^x}{\ln{a}}+c$
How would I do this for a logarithm with a base other than e?

22. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
How would I do this for a logarithm with a base other than e?
What do you mean? (Do what?)

23. ## Re: Exponential Integral Questions

Originally Posted by InteGrand
What do you mean? (Do what?)
Is there a formula I can follow for integrating logarithms with a base other than e?

24. ## Re: Exponential Integral Questions

Originally Posted by boredofstudiesuser1
Is there a formula I can follow for integrating logarithms with a base other than e?
What you were asking for before was integrating exponentials rather than logarithms. The way to do this has been shown above by 1729 and you also gave a formula for it.

Integrating logarithms from scratch is not within the realms of 2U (only 4U).

Also you only would need to know how to do these for base e logarithms, because you can convert an arbitrary-based logarithm into log base e by a simple rescaling (use change-of-base formula).

25. ## Re: Exponential Integral Questions

Originally Posted by InteGrand
What you were asking for before was integrating exponentials rather than logarithms. The way to do this has been shown above by 1729 and you also gave a formula for it.

Integrating logarithms from scratch is not within the realms of 2U (only 4U).

Also you only would need to know how to do these for base e logarithms, because you can convert an arbitrary-based logarithm into log base e by a simple rescaling (use change-of-base formula).
Right ok, thank you very much for all your help.

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