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# integration of term, to power of fractions? Watch

1. I had thought that was 2x^3/2; and from there I had 2(4)^3/2 then divided that 3; then subtract the value of 2(1)^3/2 then divided by 3.

I ended up with 3; exactly, when in reality, the answer specified is 4.6667 (4 and two thirds.

Where have I gone wrong
2. (Original post by apronedsamurai)

I had thought that was 2x^3/2; and from there I had 2(4)^3/2 then divided that 3; then subtract the value of 2(1)^3/2 then divided by 3.

I ended up with 3; exactly, when in reality, the answer specified is 4.6667 (4 and two thirds.

Where have I gone wrong
The integral of is not . What do you get when you differentiate with respect to x?
3. 2 x^{3/2}.Was a value I had gotten from an online integral calculator :|

http://www.integral-calculator.com/#...und=4&lbound=1

I had thought that 1/2+1 would give me 3/2 so that the answer would x^3/2
4. (Original post by apronedsamurai)
2 x^{3/2}.Was a value I had gotten from an online integral calculator :|

http://www.integral-calculator.com/#...und=4&lbound=1

I had thought that 1/2+1 would give me 3/2 so that the answer would x^3/2
Your power of x is right. It's the constant I object to, and the integral calculator objects to
5. Im not being pedantic, but it was the integral calculator that gave me the constant of 2!

Ok, I have been doing the calculations manually thus far; i.e. without the use of the integration function on my Casio scientific calculator, and so now I know how to properly use them, I am golden

However, I would still like to be able to compute these calculations manually, in case they should crop up in a non-calculator setting.

With that in mind, where did I go wrong?
6. (Original post by apronedsamurai)
Im not being pedantic, but it was the integral calculator that gave me the constant of 2!

Ok, I have been doing the calculations manually thus far; i.e. without the use of the integration function on my Casio scientific calculator, and so now I know how to properly use them, I am golden

However, I would still like to be able to compute these calculations manually, in case they should crop up in a non-calculator setting.

With that in mind, where did I go wrong?

Although your integral product is wrong, your working out is somehow correct, by dividing by 3. You had which is not 3. Arithmetic error.

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7. Yes. I got x^2/3 over 2/3.

I then performed the calculation on that assumption; and given that the upper limit was four, I assumed then that meant 4^2/3 divided by 2/3
8. (Original post by apronedsamurai)
Yes. I got x^2/3 over 2/3.

I then performed the calculation on that assumption; and given that the upper limit was four, I assumed then that meant 4^2/3 divided by 2/3

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9. I did.

What am I missing?

let us say x^3. Using your notation, that would be x(3+1) over (3+1) so x^4 /4, yes?
10. (Original post by apronedsamurai)
I did.

What am I missing?

let us say x^3. Using your notation, that would be x(3+1) over (3+1) so x^4 /4, yes?

In your OP you said you got for the integral, which isn't what you said in your last post. It seems you just divided by 3 later on, which gives you the correct value, but you seem to have made an arithmetic mistake.

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11. Can you just please be more direct and tell me where I am going wrong?
12. (Original post by apronedsamurai)
Im not being pedantic, but it was the integral calculator that gave me the constant of 2!
No. It gave you the constant .
13. (Original post by apronedsamurai)
Can you just please be more direct and tell me where I am going wrong?
I was on my phone, which I'm an ape at using. In your OP, to reiterate, you said, was , which was incorrect. However, when evaluating the definite integral, you divided by 3, which is correct. However, again, you made an arithmetic mistake.

I did write this out in my earlier post however.
14. Yeah, this is the bit that is throwing me off.

\dfrac{2(4)^{\frac{3}{2}}}{3}

How do you get that equal to 16 /3 ?

8 x 3/2= 12 when I do it :/

Ah, right, so the order is 2 times the () value; then the index is inverted. But why has the index been inverted here? Why is it 8 x 2/3 instead of 8 x 3/2?
15. (Original post by apronedsamurai)
Yeah, this is the bit that is throwing me off.

How do you get that equal to 16 /3 ?
When you have something like, , the way to calculate it is via firstly, taking the mth root of a, then putting that result to the power of n. However, this can be done in any order, up to you, this is the way I'd do it.

In your example, a would be 4, n would be 3, and m would be 2. Hence, you square root 4, then cube it.

16. Thank you for being so patient with me. I must admit integration is an area I flounder with, horribly.

Sorry for being so dense.
17. (Original post by apronedsamurai)
Thank you for being so patient with me. I must admit integration is an area I flounder with, horribly.

Sorry for being so dense.
Don't be silly, it's my pleasure. Practice makes perfect.

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