# Math Help - Antidifferentiation

1. ## Antidifferentiation

Hello, I am having trouble finding the indicated integral on these two problems.

{ (e^x/2 + x times square root of x) dx

and

{ x^1.1 (1/3x - 1) dx

For the first one, I had 1(1/2)e^x = 1/2e^x + (1/1+1) (1/(1/2)+1) which doesn't appear to be correct, I also can't figure out the second one. If someone could illustrate how these integrals are found, Thank you.

2. Hello,

Assuming that i read it correctly...

$\int \frac{e^x}{2}+x \underbrace{\sqrt{x}}_{x^{1/2}} dx$

$=\frac{1}{2} \int e^x dx + \int x^{3/2} dx$

$=\frac{1}{2} e^x+\frac{2}{5} x^{5/2}$

3. Could you explain this a little more through? I am having trouble mostly with the second term, I don't understand how this turns out to be 2/5 x^5/2 Thanks...

4. Hm,

The antiderivative for $x^n$ is $\frac{x^{n+1}}{n+1}$

Here, how much is n ?

I'm sorry, this is true i've skipped some steps

5. Well the way my professor said to do these, was to put the constants in front, (1/2)? and do 1 over p+1 times x^p+1 x^p dx = 1/p+1 X^p+1 So relating this back to this problem, we have x times the square root of x, so the x term would be (1/2)x^2 and the square root is (1/1.5)x^1.5 Is this correct? This still doesn't lead to 2/5x^5/2

6. The x term is x^(3/2) because it's x^1*x^(1/2)=x^(1+1/2)=x^(3/2)

Here, p=3/2 -> p+1=5/2... But there are details i haven't understood in what you said oO

7. This is what I am getting for the entire problem:

1/2e^x + (1/2x^2)(1/(3/2)x^3/2) When I differentiate each one to check my answer, they do in fact differentiate back to the original terms in the problem, so that is what is confusing me, Hope I have explained this well. I just looked closer at your work, and you added the exponents on the x terms, I was multiplying them, but they should be added right?

8. They should be added, yes

$a^b a^c=a^{b+c}$

$a^{bc}=(a^b)^c=(a^c)^b$

9. Thanks Moo, it was that small part that had me hung up. How would this problem be intergrated:

{ x^1.1 (1/3x -1) Would I do the same thing as we just did in the previous problem? I started doing it and got (1/2.1)x^2.1 I know the second term we will have the divide by zero error, so this goes to lnx if I am not mistaken? The last term would be (1/2)x^2 but this is all being multiplied...

10. When there is a product, you can't integrate part by part... You can when it's :
- a multiplication between a constant and a function
- sum of functions

Develop it..

$\frac{x^{1.1}}{3x} = \frac{x^{1.1-1}}{3}$

So the integral is $\int x^{0.1}-x^{1.1} dx$

And do it just as before

11. Ok, so if I understand you correctly, it looks like you multiplied x^1.1 by 1/3x, giving us x^1.1 / 3x. The exponent on -1 is 1 so you subtracted it from the exponent 1.1 ? What happened to the denominator of 3 in these terms? I don't understand why x^.1 is subtracting x^1.1 Could you explain this?

12. Originally Posted by kdogg121
Ok, so if I understand you correctly, it looks like you multiplied x^1.1 by 1/3x, giving us x^1.1 / 3x. The exponent on -1 is 1 so you subtracted it from the exponent 1.1 ? What happened to the denominator of 3 in these terms? I don't understand why x^.1 is subtracting x^1.1 Could you explain this?
$\frac{a^b}{a^c}=a^{b-c}$