Thread: Integrate the following, definite integral

1. Integrate the following, definite integral

Attached is the problem that I can't solve.
Can some one look at it and tell me what I am doing wrong?

Did I do the antiderivative correct also??

This is where I F(b) - F(a) to get my answer.

DISREGARD the bottom statement on the sheet, where I sent it to my teacher to help me, if he replies.

Thanks.
Joanne

2. Re: Integrate the following, definite integral

There's nothing to see ...

3. Re: Integrate the following, definite integral

Attached - ???
(I can't see anything... maybe it's just me!)

4. Re: Integrate the following, definite integral

There's a mistake in the second step, it has to be:
$\displaystyle \int_{-4}^{-1} \left(\frac{2}{x^{\frac{4}{5}}}-\frac{x^{\frac{2}{3}}}{5}\right)dx$
$\displaystyle =2\int_{-4}^{-1}x^{\frac{-4}{5}}dx -\frac{1}{5}\int_{-4}^{-1}x^{\frac{2}{3}}dx$
...

5. Re: Integrate the following, definite integral

Siron,
The first fraction of x is suppose to be x^ 2/5.
OK so is this correct?

10/7(1/x^7/5) - 3/25(x)^5/3 ??
But when I key in -4 and -1 for x, I will get an error.

I heard you do something with the negatives on the limits?? perhaps?
Is my antiderivative correct? and how do you get around the error of this?

Thanks

6. Re: Integrate the following, definite integral

$\displaystyle -4^{3/5}$ is definitely real. It's about -2.3

$\displaystyle \left(\dfrac{10}{3}(-1)^{2/5} - 3(-1)^{5/3}\right) - \left(\dfrac{10}{3}(-4)^{2/5} - 3(-4)^{5/3}\right)$

IIRC you can switch around the limits but only if you take the negative of the whole integral: that is $\displaystyle \int^a_b = -\int^b_a$

7. Re: Integrate the following, definite integral

So remove the negatives on both numbers and put the negative outside the integral sign?

So the -2.3 above, you basically do the calculation leaving out the (-) sign and then add it after???

8. Re: Integrate the following, definite integral

Sorry, I read -4/5 for the exponent.
My calculator says: $\displaystyle (-4)^{\frac{3}{5}}=-2,2979...$

9. Re: Integrate the following, definite integral

Siron, how do you do that calculation. I keyed in exactly what you have there and I get an error? Is there something I am missing to calculate this?
I am not seeing this for some reason. Is it because the root is odd, it's going to be a negative number?? as the root is 5??

10. Re: Integrate the following, definite integral

These are the keys I used
Code:
( - 4 ) ^ (3 / 5 )
Ultimately though $\displaystyle (-4)^{3/5} = \sqrt[5]{(-4)^3} = \sqrt[5]{-64}$. Since the exponent is odd this should give a real number.

11. Re: Integrate the following, definite integral

Which calculator do you have? Ti-83? Notice there are two '-' signs, one that indicates a negative number and one that indicates the operation the 'difference between to numbers', you have to use the one that indicates negative numbers.

12. Re: Integrate the following, definite integral

Just a few comments to add: an odd root of a negative number, or the fractional power of a negative number, where the denominator of the fractional power is odd is NOT an imaginary number! For n odd, $\displaystyle (-a)^{m/n}= -(a^{m/n})$. Some (but not all) calculators do roots by taking logarithms which will not give correct answers for negative arguments. Really good calculators should be able to handle the "odd root of a negative power" but not all do.

And, finally, $\displaystyle \int_{-4}^{-1} f(x)dx$ and $\displaystyle \int_{-1}^{-4} f(x)dx$ are NOT the "same thing". One is the negative of the other: $\displaystyle \int_{-4}^{-1} f(x)dx= -\int_{-1}^{-4} f(x)dx$.

13. Re: Integrate the following, definite integral

HI EVERYONE, thanks for all the help. There was something I was not getting and now you have helped me. I go an answer of 3.2351. I had to key in information into my calculator a different way for it to work. I don't have a graphing calculator yet. This week I will tho.

So to HallsofIvy, when you pull out the - and switch a and b. When does this get done? I know you add the negative to the equation, but has to happen to do this???
THanks again