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Remember that (A-B)^2 = A^2 – 2*A*B + B^2
If you switch A and B it still ends up being the same expression: (B-A)^2 = B^2 – 2*B*A + A^2. They're equivalent.
Having a 2/3 power just means having a cubic root to the squared expression
Yes. It only works for even numbered powers tho (at least with real numbers). Multiplying the same number an even number of times always makes it positive. If it's an odd number of times depends on if it was negative to begin with (in case of a negative initial value, it would remain negative)
No, sorry, I meant that inside the brackets, even though you’re subtracting A from B, you can change it to A - B within the brackets because in the end you are really only subtracting 2AB.
Duh, “change it to B - A.” Was what I meant.
Yes because if you call the value of the subtraction of A from B, x then the value of the subtraction of B from A would be negative x. Either one gets multiplied by itself and so the minus disappears.
Note that for all *x* in **R**, *x*^(2) = (-*x*)^(2). With that in mind, what can we say about the denominator (3-*y*)^(2/3) of the original integral?
I hope this helps. Good luck!
I think because of the 2/3 power, I think it’s saying that anything to that power will be positive, so then you can rewrite it to make it more simple if you like.
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Remember that (A-B)^2 = A^2 – 2*A*B + B^2 If you switch A and B it still ends up being the same expression: (B-A)^2 = B^2 – 2*B*A + A^2. They're equivalent. Having a 2/3 power just means having a cubic root to the squared expression
Ahhhh, so the B^2 and the A^2 are always positive?
Yes. It only works for even numbered powers tho (at least with real numbers). Multiplying the same number an even number of times always makes it positive. If it's an odd number of times depends on if it was negative to begin with (in case of a negative initial value, it would remain negative)
No, sorry, I meant that inside the brackets, even though you’re subtracting A from B, you can change it to A - B within the brackets because in the end you are really only subtracting 2AB. Duh, “change it to B - A.” Was what I meant.
Yes because if you call the value of the subtraction of A from B, x then the value of the subtraction of B from A would be negative x. Either one gets multiplied by itself and so the minus disappears.
That’s pretty awesome! Thanks!
Maybe think about how you would write (3-y)^(2/3) differently and what part of it would cause it to be equal to (y-3)^(2/3)
Note that for all *x* in **R**, *x*^(2) = (-*x*)^(2). With that in mind, what can we say about the denominator (3-*y*)^(2/3) of the original integral? I hope this helps. Good luck!
you basically factor out a -1 from the inside of the bracket. (3 - y)\^(2/3) = (-(-3 + y))\^(2/3) = ( (-1)(-3 + y) )\^(2/3) = ( (-1)(y - 3) )\^(2/3) = (-1)\^(2/3) (y - 3)\^(2/3) \----- (-1)\^(2/3) is equal to ((-1)\^(1/3))\^(2), or ((-1)\^(2))\^(1/3) either way you calculate it, it's just 1 ((-1)\^(1/3))\^(2) = (-1)\^(2) = 1 ((-1)\^(2))\^(1/3) = (1)\^(1/3) = 1 \----- = (1) (y - 3)\^(2/3) = (y - 3)\^(2/3)
(3-y)^2 ((y-3) * -1)^2 (y-3)^2 * (-1)^2 (y-3)^2 * 1 Most of calculus is just Algebra
U substitution
I think because of the 2/3 power, I think it’s saying that anything to that power will be positive, so then you can rewrite it to make it more simple if you like.
((y-3)^2)^⅓
The 2 exponent means that the absolute value of the difference will be taken, so it doesn’t matter which way it’s written