integration question (1 Viewer)

[velox]

I forgot what my teacher told me about this question. Cos when i do it i get a maths error on my calc.

 

[Estel]

hmm it should work:
2.3/5.[8^(5/3)]
= 192/5

(-8)^(5/3) works on the calculator... what are you putting in that gets you math error?

 

[velox]

Doesnt work on mine =/ (-8)x^y(5/3) is what im putting in. Maths error appears.

 

[Estel]

well (-8)^(5/3) = (-2)^5 = -32
Consider an investment in a new calculator

 

[velox]

I dont think it works on yours either.

 

[Meldrum]

Doesnt work on mine =/ (-8)x^y(5/3) is what im putting in. Maths error appears.

Dude, you've got to work out what 5/3 is before you put it in. It's stupid, I know, but it's a fact.

 

[Captain pi]

I forgot what my teacher told me about this question. Cos when i do it i get a maths error on my calc.

It appears your calculator refuses to extract fractional roots from negative numbers. Recall that a fractional power (x)^(m/n) is x^m·(nth root of x). Suppose x is negative and n is even, this will result in a complex answer, which your calculator cannot evaluate.

You may need to invest in a new calculator; however, I know not which. Another boredmember may be able to assist you.

(This calculation could be performed manually, with ease. )

 

[Slidey]

I dont think it works on yours either.

If you really must use a calculator, simply pretend the number is positive, perform the operation on the calculator, and THEN make it negative if necessary.

E.g.: (-64)^(5/3)

Pretend it is positive: 64^(5/3)=4^5=1024.

Since both 5 and 3 are odd, it will be negative: (-64)^(5/3) = -1024

 

[Slidey]

On that note, somebody tell me the answer to this:

((x^2)^1/2)⁴

Now this:

((x^2)⁴)^1/2

Now this:

(x^2)^4/2

My answer:

1) x^4

2) +x^4

3) ?

 

[Slidey]

Oh BTW, the trick your teacher told you is this:

If it is an EVALUATE question, then:

Integral of f(x) from -a to a is:
2*Integral (f(x)) from 0 to a, if f(x) is an even function
0 if f(x) is an odd function.

If it is an area question, it's two lots of the integral from 0 to a fro both even and odd.

And f(x)=x^2/3 is even in this case. Test: f(-x)=(-x)^2/3=-cubrt(x)^2=cubrt(x^2)=f(x) .'. even.

If it was odd you'd get f(-x)=-f(x). You can also get cases where the function neither odd nor even, such as f(x)=x^2+x

 

[Estel]

Slide they're all x^4

 

[Slidey]

Justification?

 

[shafqat]

indice laws?

 

[Slidey]

Let me explain it a different way. How would you do this question:

Solve simplify y^2=u, where u=x^2.

What states that you must do all arithmetic on exponents first?

Would you answer "all real x" to "what is the domain of x/x"? I think not.

 

[Estel]

((x^2)^1/2)^4
= |x|^4 = x^4

((x^2)^4)^(1/2)
= (x^8)^(1/2) = x^4 ... I remember having discussions on the definition of x^(1/2)... leads to no end but misery. Have your +- if it makes you happy.

(x^2)^(4/2)
=(x^2)^2
=x^4

sqrt(u) = sqrt(x^2) = x... what's the problem?

edit: you've changed your post.
y^2 = u = x^2 y = +-x
or y^2 = u y = +-rtu = +-x...
But that's different from saying y = x^(1/2)...

 

[Slidey]

Yes, I know it's a pedantic point, but I'm still interested in the answer nonetheless.

I am not arguing that it is plus/minus, but wondering if there's a reason it isn't plus/minus.

Hmm. Oh well.

 

[Estel]

"x^(1/2) has the property that when you square it, you get x"
"x^(1/2) denotes the positive square root of x"

I'm yet to see an answer on this...
ahh well

quote from wolfram: Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of 9 are -3 and 3, since (-3)^2 = (3)^2 = 9 . Any nonnegative real number x has a unique nonnegative square root r; this is called the principal square root and is written r=x^(1/2) or r = rtx

which contradicts the first statement in my post.
Let the fun continue =p

 

[velox]

Dude, you've got to work out what 5/3 is before you put it in. It's stupid, I know, but it's a fact.

i know but still, i thought the calculator would work it out.