Integrating Exponentials with Roots that have Roots? (And other small Q's)

In summary, integrating exponentials with roots involves using the power rule and substitution to simplify the expression. The power rule states that the integral of x^n is equal to (1/(n+1))x^(n+1), where n is any real number except for -1. An example of integrating an exponential with a root is e^(2√x), which can be solved by rewriting the root as a fractional power and using the power rule. The general process for integrating expressions with multiple roots is to use substitution and then integrate the resulting expression. However, not all integrals with roots can be solved using the same method and may require more advanced techniques.
  • #1
kape
25
0
Hello, I have a few questions! I need clarification on certain points that were not very clear in my calculus book.

----------
Question 1:

I know that [tex] \int e^{ax} dx = \frac{1}{a} e^{ax} [/tex]

But how do you integrate [tex] \int e^{ax^2} dx [/tex]?


-----------
Question 2:

I know that integrating by parts is [tex] \int (something) dx= uv - \int vdu [/tex]

But what if there is a range?

If it is [tex] \int_{a}^{b} (something) dx [/tex] does it equal [tex] \left[ uv \right]_{a}^{b} - \int_{a}^{b} vdu [/tex] or does it simply equal [tex] uv - \int_{a}^{b} vdu [/tex]?


-----------
Question 3:

How do you integrate [tex] \int log_ax dx [/tex] and [tex] \int e^{ln|secx|} dx [/tex].

In fact, is [tex] e^{ln|secx|} [/tex] reducable?


-----------
Question 4:

I was taught that arcsinx exist only in the range [tex] \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] [/tex] and [tex] \left[ \frac{\pi}{2}, \frac{3\pi}{2} \right] [/tex] (I think because it fails the horizontal test if it isn't in those ranges)

If so, is it possible to integrate [tex] \int_{0}^{\pi} xarcsinx dx [/tex]? (If it is possible, is it because it isn't simply arcsinx but xarcsinx?)


-----------
Question 5:

I am having a lot of problems integrating fuctions with exponents etc that have complex roots. My elementary calculus is shaky at best and I'm taking Advanced Engineering Mathematics (Kreyzig) - I have to. Can anyone recommend me any links or books that may help me?


------------------
Reply to HallsofIvy:

Thank you for your reply! I have a question about your reply on question 1: In my Adv Eng Maths (Kreyzig) book, one of the questions is how to integrate [tex] \int xe^{x^2/2} [/tex] and the answer is [tex] e^{x^2/2} + C [/tex] but I don't understand how to do it!
 
Last edited:
Physics news on Phys.org
  • #2
Answer to question 1: You don't. That integral,
[tex]\int e^{ax^2}dx[/tex]
is well known not to have an elementary integral. In fact, precisely that integral (with a= -1) is important in Statistics and it's integral is defined to be "Erf(x)", the error function.

If it were
[tex]/int xe^{ax^2}dx[/tex]
then you could make the substitution [itex]u= e^{ax^2}[/itex] and have the xdx already for du= 2xdx.

Question 2, Yes, just plug the limits of integration into the formula.

Question 3 seems to have disappeared.
 
Last edited by a moderator:
  • #3
Notice that [tex]\frac{d}{dx}\left(\frac{x^2}{2}\right)=x[/tex]
So it is actually [tex]\int{f'\left(x\right)e^{f\left(x\right)}=e^{f\left(x\right)}+c[/tex]
 
  • #4
Thank you for answering questions 1 & 2.. I think I understand.

Sorry question 3 was deleted, don't quite know how that happened.

Also, I have one more question: (which is kind of similar to question 3)


----------
Question 6

How do you integrate:

[tex] \int \frac{1}{x^a} dx [/tex]

[tex] \int \frac{1}{a^x} dx [/tex]

[tex] \int \frac{1}{a^{bx}} dx [/tex]

[tex] \int \frac{1}{a^{bx^{c}}} dx [/tex]

Should I have learned this somewhere? I don't see these in the integral tables or rules..
 
Last edited:
  • #5
[tex]\frac{1}{x^a}= x^{-a}[/tex]
Use the power rule.

[tex]\frac{1}{a^x}= a^{-x}[/tex]
Make the substitution u= -x and then use
[tex]\int a^x dx= \frac{a^x}{ln a}[/tex]

Same thing:
[tex]\frac{1}{a^{bx}}= a^{-bx}[/tex]
Make the substitution u= -bx.

[tex]\int \frac{1}{a^{bx^c}}dx[/tex]
depends strongly on what c is. There is no general anti-derivative.
 

1. How do you integrate exponentials with roots?

Integrating exponentials with roots involves using the power rule and substitution to simplify the expression. First, rewrite the root as a fractional power. Then, use the power rule to integrate the exponential term. Finally, substitute the original variable back in and solve for the integral.

2. What is the power rule for integrating exponentials?

The power rule states that the integral of x^n is equal to (1/(n+1))x^(n+1), where n is any real number except for -1. For example, the integral of 2x^3 would be (1/4)x^4 + C.

3. Can you give an example of integrating an exponential with a root?

Sure, let's say we have the integral of e^(2√x). First, we rewrite the root as a fractional power, giving us the integral of e^(2x^(1/2)). Then, we use the power rule to integrate the exponential term, giving us (1/2)e^(2x^(1/2)+1). Finally, we substitute the original variable back in and add a constant, giving us the final answer of (1/2)e^(2√x) + C.

4. What is the general process for integrating expressions with multiple roots?

The general process for integrating expressions with multiple roots is to use substitution to simplify the expression. First, identify the innermost root and substitute a new variable for it. Then, continue substituting until all the roots have been eliminated. Finally, integrate the resulting expression and substitute the original variables back in.

5. Can all integrals with roots be solved using the same method?

No, not all integrals with roots can be solved using the same method. Some may require more advanced techniques such as integration by parts or trigonometric substitution. It is important to familiarize yourself with various integration methods and choose the most appropriate one for each specific integral.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
333
  • Calculus and Beyond Homework Help
Replies
4
Views
728
  • Calculus and Beyond Homework Help
Replies
15
Views
775
  • Calculus and Beyond Homework Help
Replies
16
Views
552
  • Calculus and Beyond Homework Help
Replies
8
Views
754
  • Calculus and Beyond Homework Help
Replies
5
Views
327
  • Calculus and Beyond Homework Help
Replies
7
Views
927
  • Calculus and Beyond Homework Help
Replies
1
Views
524
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
Back
Top