Integrate e^x^2 + 2e^x^2x^2: Solution Explained

In summary: You can also do it in plain text, just by using parentheses correctly. When you write e^x^2 that could mean ##e^{x^2}## or ##(e^x)^2 = e^{2x}##. In plain text you would write the first as e^(x^2) and the second as (e^x)^2.
  • #1
chwala
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Homework Statement


##∫e^x^2 + 2e^x^2x^2 dx##[/B]

Homework Equations

The Attempt at a Solution


i let## u= x^2, ⇒ du = 2x dx, ⇒∫e^x^2 dx = e^x^2/2x ## is this correct? by using integration by parts.... i am getting ##xe^x^2-e^x^2/2x##
 
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  • #2
chwala said:

Homework Statement


##∫e^x^2 + 2e^x^2x^2 dx##[/B]

Homework Equations

The Attempt at a Solution


i let## u= x^2, ⇒ du = 2x dx, ⇒∫e^x^2 dx = e^x^2/2x ## is this correct? by using integration by parts.... i am getting ##xe^x^2-e^x^2/2x##

You can verify whether you are correct by using wolframalpha.com. Also, your code has an error.
 
  • #3
You may use integration by parts on the first term.
$$\int e^{x^2} \, dx=xe^{x^2}-\int x(2xe^{x^2}), dx$$
The equation above is sufficient for your problem.
Hope it helps.
 
  • #4
Abhishek Sethi said:
You may use integration by parts on the first term of the equation.
$$\int e^{x^2} \, dx=xe^{x^2}-\int x(2xe^{x^2}), dx$$
The equation above is sufficient for your problem.
Hope it helps.
i can see you are using integration by parts, what is your ##u, du, v, dv##?
 
  • #5
chwala said:
i can see you are using integration by parts, what is your ##u, du, v, dv##?
I am not sure what you exactly mean by ##u,v,du,dv##. To put it more clearly,
$$\int e^{x^2} \, dx=(e^{x^2})\int 1\ dx-\int x\frac{d(e^{x^2})}{dx} dx$$
 
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  • #6
Abhishek Sethi said:
I am not sure what you exactly mean by ##u,v,du,dv##. To put it more clearly,
$$\int e^{x^2} \, dx=(e^{x^2})\int 1\ dx-\int x\frac{d(e^{x^2})}{dx} dx$$
this is something "new " to me in my understanding integration by parts is ##∫udv= uv - ∫vdu## unless you have used a different approach, your method can't be described as integration by parts as stated in your opening statement.
 
  • #7
i can see you picked ## u= e^x^2, dv=1 ⇒v=x, u'=2xe^x^2 ##
 
  • #8
Thanks Abishek...greetings from Africa
 
  • #9
i don't know when you continue with the integration of the second term the function is getting complicated i.e if you ∫2x^2e^x^2 dx or are we picking again
dv = 1, u= 2x^2e^x^2?
 
  • #10
on ∫x.2xe^x^2 dx i let dv= x^2⇒ v=(x^3)/3, u= e^x^2 ⇒ du = 2x.e^x^2 thus
∫x.2xe^x^2 dx = (2e^x^2.x^3)/3-2 ∫(x^3)/3.2x.e^x^2 dx without simplification...this to me gets more complicated and no final answer is possible...
 
  • #11
lol lol lol i was blind but now i can see...

∫(e^x^2+2e^x^2x^2)dx= xe^x^2- ∫2e^x^2x^2dx + ∫2e^x^2x^2dx + k
= xe^x^2 + k

chikhabi from Africa lol
 
  • #12
chwala said:

Homework Statement


## ∫ e^x^2 + 2e^x^2x^2 dx##
##\int e^{x^2} + 2e^{x^2}x^2 dx## (fixed by SammyS.)

Homework Equations



The Attempt at a Solution


I let## u= x^2, ⇒ du = 2x dx, ⇒∫e^x^2 dx = e^x^2/2x ## is this correct? by using integration by parts.... i am getting ##xe^x^2-e^x^2/2x##
It's good to see you trying to use LaTeX, but you are having some problems.

Is this what you mean? ##\displaystyle\ \int \left( e^{x^2} + 2e^{x^2}x^2 \right) dx\ ## It's difficult to tell with all of those run-on exponents.
 
  • #13
yes that is what i mean...how do i type that in latex?
 
  • #14
chwala said:
yes that is what i mean...how do i type that in latex?
Easiest way: Hit "Reply" link under my post. See the coding.

Otherwise, for some LaTeX here in PF that you don't such access to: Right click on the LaTeX expression. Click on "Show Math As". Click on "TeX Commands". A pop-up window should appear with the LaTeX code.

The \displaystyle LaTeX command you see in my code gives larger format when doing "in-line" LaTeX using the ##\ \text{## ... ##}\ ## delimiters.

Also, look up LaTeX tutorial here in Physics Forums.
 
  • #15
##\left| \frac a b \right|## am trying practise on latex...
##\left[ 1 - \left( \frac a x \right)^2 \right]^{-1/3}##
##\int x^2e^x \, dx##
##{ \displaystyle\ \int \left( e^{x^2} + 2e^{x^2}x^2 \right) dx\ }##​
 
Last edited:
  • #16
chwala said:
yes that is what i mean...how do i type that in latex?

You can also do it in plain text, just by using parentheses correctly. When you write e^x^2 that could mean ##e^{x^2}## or ##(e^x)^2 = e^{2x}##. In plain text you would write the first as e^(x^2) and the second as (e^x)^2.
 
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1. What is the meaning of "Integrate e^x^2 + 2e^x^2x^2"?

The term "integrate" refers to finding the antiderivative of a function. In this case, we are finding the antiderivative of e^x^2 + 2e^x^2x^2.

2. Why is it important to integrate e^x^2 + 2e^x^2x^2?

Integrating a function allows us to find the area under the curve of the function, which has many practical applications in fields such as physics, engineering, and economics.

3. How do you solve the integral of e^x^2 + 2e^x^2x^2?

To solve this integral, we use techniques such as u-substitution, integration by parts, or trigonometric substitution. The specific method used may vary depending on the complexity of the function.

4. Can you explain the steps involved in finding the antiderivative of e^x^2 + 2e^x^2x^2?

To find the antiderivative of e^x^2 + 2e^x^2x^2, we first use the power rule to find the antiderivative of e^x^2. Then, we use the chain rule to find the antiderivative of 2e^x^2x^2. Finally, we add the two antiderivatives together to get the overall solution.

5. Are there any real-life applications of integrating e^x^2 + 2e^x^2x^2?

Yes, there are many real-life applications of integrating this function. For example, it can be used to calculate the displacement of an object under the influence of a changing force, or to determine the profit of a business based on changing market conditions.

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