# Find the area enclosed by given curves: y=`sqrt(x)` , y=`(1/2)x` , x=9. I can find the intersections which are x=0 and x=4, then what is the point of giving us "x=9"?

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Top Answer

degeneratecircle | High School Teacher | (Level 2) Associate Educator

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The curves intersect at `x=4,` but that doesn't stop you from finding the area enclosed by the curves up to any (positive) `x`value.

Just split up the area into two integrals.

However, after reading the question again, I do agree that the wording, "enclosed" in particular, is a little confusing. Maybe it would have been better to say "find the area between the two curves...from `x=0` to ```x=9` ", or something similar.

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to evaluate the limits of integration solving the equation `sqrt x = x/2` , such that:

`sqrt x = x/2`

You need to raise to square both sides, such that:

`x = (x^2)/4 => (x^2)/4 - x = 0`

Factoring out x yields:

`x(x/4 - 1) = 0 => {(x = 0),(x/4 - 1 = 0):} => {(x = 0),(x/4 = 1):} => {(x = 0),(x = 4):} `

Since the limits of integration are `x = 0` and `x = 4` , you need to consider a value included in interval `[0,4],` hence `x = 1` .

You should notice that the function `sqrt x > x/2` for `x in [4,9].`

You need to evaluate both functions at `x = 1` , such that:

`y = sqrt x => y = sqrt 1 => y = 1`

`y = x/2 => y = 1/2`

Notice that `sqrt x > x/2` for `x in [0,4]` , hence, evaluating the area enclosed by the given curves, yields:

`int_0^4 (sqrt x - x/2) dx`

Using the property of linearity of integral yields:

`int_0^4 (sqrt x - x/2) dx = int_0^4 (sqrt x) dx - int_0^4 (x/2) dx`

`int_0^4 (sqrt x - x/2) dx = x^(1/2+1)/(1/2+1)|_0^4 - x^2/4|_0^4`

Using the fundamental theorem of calculus yields:

`int_0^4 (sqrt x - x/2) dx = (2/3)(4sqrt4 - 0) - (4^2/4 - 0)`

`int_0^4 (sqrt x - x/2) dx = 16/3 - 16/4`

`int_0^4 (sqrt x - x/2) dx = 16(1/3 - 1/4)`

`int_0^4 (sqrt x - x/2) dx = 16/12`

`int_0^4 (sqrt x - x/2) dx = 4/3`

Hence, evaluating the area enclosed by the given curve, using the limits of integration `x = 0` and `x = 4` , yields `int_0^4 (sqrt x - x/2) dx = 4/3.`

Educator Approved

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The area between the curves `y = sqrt x` and `y = x/2` has to be determined.

The graph of the two curves is shown below.

The value x = 9 is the upper limit to the definite integral which gives the area between the two curves.

The area between the curves is:

`int_0^4 sqrt x - x/2 dx + int_4^9 x/2 - sqrt x`

= `|x^(3/2)/(3/2) - x^2/4|_0^4 + |x^2/4 - x^(3/2)/(3/2)|_4^9`

= `(2/3)*(4^(3/2) - 0) - (1/4)*(4^2 - 0) + (1/4)*(9^2 - 4^2) - (2/3)*(9^(3/2) - 4^(3/2))`

= `16/3 - 4 + 65/4 - 38/3`

= `59/12`

As you can see this is the answer provided in your book.

user6978788 | Student, Undergraduate | (Level 1) Honors

Posted on

You need to evaluate the limits of integration solving the equation , such that:

You need to raise to square both sides, such that:

Factoring out x yields:

Since the limits of integration are and , you need to consider a value included in interval hence .

You should notice that the function for

You need to evaluate both functions at , such that:

Notice that for , hence, evaluating the area enclosed by the given curves, yields:

Using the property of linearity of integral yields:

Using the fundamental theorem of calculus yields:

Hence, evaluating the area enclosed by the given curve, using the limits of integration and , yields

I got 4/3 too, but the answer in the textbook gives 59/12. I guess that has something to do with the "x=9" given in the question but how? x=9 is not even in the area they enclosed...

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