1. ## Check my antiderivatives, please.

Would you mind confirming if these are correct? If not, please guide me in the right direction. I find that antiderivatives are tricky in the sense that it is difficult to go backwards when we've learned to find derivatives all along! Just learned them yesterday so I'm giving them a go. I apologize, but I still have yet to get the hang of the math tags.

Find the most general antiderivative:

1) f(x)=4sqrtx^3 + 3sqrt x^4

Answer: (4/17)x^(17/4) + (3/16)x^(16/3) + C

2) g(x)= (5-4x^3+2x^6) / (x^6)

Answer: -x^-5 - 2x^-2 + 2x + C

Find f:

1) f'(x)=2x - [(3) / (x^4)] where x > 0, f(1)=3

Answer: x^2 + x^-3 + 1

If any are incorrect, I'll post up a picture of my work so we can see where I went wrong. Thanks guys.

2. You should simplify the first one to:
4*x^1.5 + 3x^2

After that the integral is:
(8/5)x^(5/2) + x^3 + C

3. Originally Posted by Marconis
1) f'(x)=2x - [(3) / (x^4)] where x > 0, f(1)=3

Answer: x^2 + x^-3 + 1
Correct, you can always find dy/dx of your answers to check.

4. Originally Posted by MIT
You should simplify the first one to:
4*x^1.5 + 3x^2

After that the integral is:
(8/5)x^(5/2) + x^3 + C
Wait, how did you get that from what I had? What am I missing?

5. Originally Posted by Marconis
Wait, how did you get that from what I had? What am I missing?
Based upon how you wrote it:

4sqrt(x^3) + 3sqrt(x^4)

That would be 4(x^3)^(1/2) + 3(x^4)^(1/2)

Simplified: 4x^(3/2) + 3x^2

Did you mean that each was the 4th root and 3rd root respectively?
sqrt = square root

Even so the answer is incorrect:
integral[x^(3/4) + x^(4/3)] =(4/7) x^(7/4) + (3/7)x^(7/3)

6. Oh, oops. I added the exponents together instead of multiplying them initially.

7. Originally Posted by MIT
Based upon how you wrote it:

4sqrt(x^3) + 3sqrt(x^4)

That would be 4(x^3)^(1/2) + 3(x^4)^(1/2)

Simplified: 4x^(3/2) + 3x^2

Did you mean that each was the 4th root and 3rd root respectively?
sqrt = square root

Even so the answer is incorrect:
integral[x^(3/4) + x^(4/3)] =(4/7) x^(7/4) + (3/7)x^(7/3)
I did it again and got this answer : (4/7) x^(7/4) + (3/7)x^(7/3)

8. Originally Posted by Marconis
1) f(x)=4sqrtx^3 + 3sqrt x^4
Is this $\displaystyle \displaystyle f(x)= \sqrt[4]{x^3}+\sqrt[3]{x^4}$ or $\displaystyle \displaystyle f(x)= 4\sqrt{x^3}+3\sqrt{x^4}$ ?

9. Originally Posted by pickslides
Is this \displaystyle $\displaystyle f(x)= \sqrt[4]{x^3}+\sqrt[3]{x^4}$ or $\displaystyle \displaystyle f(x)= 4\sqrt{x^3}+3\sqrt{x^4}$ ?
It's the first one. I know I confused MIT. I didn't realize until later that the way I typed that would cause some confusion.

10. Hey guys, back again. Just want to have a final clarification check...we are moving on to method of substitution, so I just want to make sure I have this down before I continue.

Find f

f ''(x)= 2 + x^3 +x^6
f ' (x) = 2x + (1/4)x^4 + (1/7)x^7 + C

f(x) =x^2 + (1/20)x^5 + (1/56)x^8 + Cx + D

I have another question that I didn't get a chance to ask my professor. He assigned us his own problems in which we can practice substitution. My question is: What things should we look for in an antiderivative problem that would make us think, "Oh, we need to use substitution here!" I consider the way I've been doing them (such as the problem above) the "plain old way" to find the antiderivative. For that I recognize to do it because of simple exponents and what not. For substitution, I feel I'd be looking at the problem and not even get that I had to substitute.

11. Originally Posted by Marconis
I have another question that I didn't get a chance to ask my professor. He assigned us his own problems in which we can practice substitution. My question is: What things should we look for in an antiderivative problem that would make us think, "Oh, we need to use substitution here!"
I always look for a function and it's derivative in the same integral when applying subtitution.

I.e $\displaystyle \displaystyle \int (x^2+5x-11)^5(2x+5)~dx$ Using substitution saves you expanding out the trinomial here.

Or if you have seen calculus with trig functions $\displaystyle \displaystyle \int \sin^5x \cos x~dx$

Do you know how to solve these?

12. I just learned them today and am going to practice them soon. In the simplest sense, don't you just look for something to take the derivative of in respect to dx, that when you solved dx/du, you'd find a du to replace in the antiderivative? If you get what I am saying, and what I am saying is correct, I will take a shot at your example!

Was the problem I posted correct?

13. Originally Posted by pickslides
I always look for a function and it's derivative in the same integral when applying subtitution.

I.e $\displaystyle \displaystyle \int (x^2+5x-11)^5(2x+5)~dx$ Using substitution saves you expanding out the trinomial here.

Or if you have seen calculus with trig functions $\displaystyle \displaystyle \int \sin^5x \cos x~dx$

Do you know how to solve these?
For the first example you provided: I can already tell that taking the derivative of (x^2 + 5x -11) would give you (2x + 5). So, in my head I'd pick that to substitute. I'd then go to take the derivative of it: u=(x^2 + 5x -11). Solving du/dx gives you: du/dx=2x+5. Solving for du, you get du=2x+5dx. Ah ha! See that above. So, (2x+5) simply becomes du. Finding the antiderivative of, (u)^5du. Brings (u^6/6) + C which gives, (x^2 + 5x -11)^6 / 6 + C. Am I correct? This is how I learned it (so I hope I did it correctly). I am curious why the du vanishes?

14. Originally Posted by Marconis
For the first example you provided: I can already tell that taking the derivative of (x^2 + 5x -11) would give you (2x + 5). So, in my head I'd pick that to substitute. I'd then go to take the derivative of it: u=(x^2 + 5x -11). Solving du/dx gives you: du/dx=2x+5. Solving for du, you get du=2x+5dx. Ah ha! See that above. So, (2x+5) simply becomes du. Finding the antiderivative of, (u)^5du. Brings (u^6/6) + C which gives, (x^2 + 5x -11)^6 / 6 + C. Am I correct? This is how I learned it (so I hope I did it correctly). I am curious why the du vanishes?
$\displaystyle \displaystyle (2x+5)\,\mathbf{dx}$ becomes $\displaystyle \displaystyle du$...

The $\displaystyle \displaystyle du$ loosely means "take the antiderivative with respect to $\displaystyle \displaystyle u$", just like $\displaystyle \displaystyle dx$ loosely means "take the antiderivative with respect to $\displaystyle \displaystyle x$".

So after you've taken the antiderivative, you've done what the $\displaystyle \displaystyle du$ has told you to do, so you don't need to write it any more.

15. Originally Posted by Prove It
$\displaystyle \displaystyle (2x+5)\,\mathbf{dx}$ becomes $\displaystyle \displaystyle du$...
Oh yeah, right. Little brain fart there.

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