Area

[Loki123]

Is this correct? One thing that I am sure is not correct is the line on the graph I marked yellow. Why isn't that part there? If I say x^2=4, x can be 2 or - 2. So why If I say y=x^1/2 and if x can be 4, why can't y be 2 or - 2?

 

[JeffM]

The yellow line is irrelevant because it is the curve describing [imath]y = - \sqrt{x}.[/imath] Remember that the square root function is defined to be non-negative.

 

[Loki123]

The yellow line is irrelevant because it is the curve describing [imath]y = - \sqrt{x}.[/imath] Remember that the square root function is defined to be non-negative.

If I say x^2=4, x can be 2 or - 2. So why If I say y=x^1/2 and if x can be 4, why can't y be 2 or - 2?

 

[Deleted member 4993]

If I say x^2=4, x can be 2 or - 2. So why If I say y=x^1/2 and if x can be 4, why can't y be 2 or - 2?

y² = x → y = √x ................ only positive values by definition.

 

[JeffM]

The WHOLE POINT of a function is that it never gives an ambiguous answer. It is never one-to-many.

So it is absolutely true that

[math](-4)^2 = 16 = (4)^2 .[/math]
But when we define the square root as a FUNCTION, we have to choose which of those we mean. So, one definition of the square root function is

[math]f(x) = \sqrt{x^2} \iff f(x) = |x|[/math]

 

[lookagain]

So why If I say y=x^1/2 and . . .

Loki123, when you type \(\displaystyle \ x^{\tfrac12} \ \) in this horizontal style, you must put this fractional exponent inside
grouping symbols, such as x^(1/2), because what you typed is interpreted as \(\displaystyle \ \dfrac{x^1}{2} \ \) due to the
Order of Operations.

 

[Loki123]

The WHOLE POINT of a function is that it never gives an ambiguous answer. It is never one-to-many.

So it is absolutely true that

[math](-4)^2 = 16 = (4)^2 .[/math]
But when we define the square root as a FUNCTION, we have to choose which of those we mean. So, one definition of the square root function is

[math]f(x) = \sqrt{x^2} \iff f(x) = |x|[/math]

so if f(x)=sqrt(4) f(x)=2?

 

[JeffM]

so if f(x)=sqrt(4) f(x)=2?

Yes though I would prefer to say

[math]f(x) = \sqrt{x} \implies f(4) = 2.[/math]

 

[pka]

Area is found by an integral: [imath]\displaystyle\int_0^1 {\left[ {\sqrt x - {x^3}} \right]dx = \mathop {\left[ {\frac{2}{3}{x^{\frac{3}{2}}} - \frac{1}{4}{x^4}} \right]}\nolimits_{x = 0}^{x = 1} }[/imath]
That seems to be your first answer, why now doubt it is correct?
The only area created by the two curves, [imath]\sqrt{x}~\&~x^3[/imath] is that from [imath]x=0\text{ to }x=1[/imath]!
Th upper curve being [imath]\sqrt{x}[/imath] and the lower curve being [imath]x^3[/imath].
[imath][/imath][imath][/imath][imath][/imath]