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"?

### 3 Answers | Add Yours

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.`

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.

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...

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes