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Find the area enclosed by the three curves: y=`1/x` , y=x, y=`(1/4)x` , x>0.  

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Find the area enclosed by the three curves: y=`1/x` , y=x, y=`(1/4)x` , x>0.

 

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sciencesolve's profile pic

Posted (Answer #1)

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You need to evaluate the limits of integration `y = 1/x, y = x, y = (1/4)x` such that:

`1/x = (1/4)x => 1/x - x/4 = 0 => 4 - x^2 = 0 => x^2 = 4 => x = 2 (x = -2` is not valid since `x>0` )

`1/x = x => x^2 = 1 => x = 1 (x = -1` invalid)

`x = x/4 => 1 = 1/4` invalid

You need to find the area enclosed by the given curves, hence, you need to evaluate the following definite integral, such that:

`int_1^2 (x - 1/x - x/4)dx = int_1^2 x dx - int_1^2 1/x dx - int_1^2 x/4 dx`

`int_1^2 (x - 1/x - x/4)dx = x^2/2|_1^2 - ln x|_1^2 - x^2/8|_1^2`

Using the fundamental theorem of calculus yields:

`int_1^2 (x - 1/x - x/4)dx = 4/2 - 1/2 - ln 2 + ln 0 - 4/8 + 1/8`

Since `ln 1 = 0` yields:

`int_1^2 (x - 1/x - x/4)dx = 2 - 1 + 1/8 - ln 2`

`int_1^2 (x - 1/x - x/4)dx = 9/8 - ln 2`

Hence, evaluating the area enclosed by the given curves, using definite integral, yields `int_1^2 (x - 1/x - x/4)dx = 9/8 - ln 2.`

user6978788's profile pic

Posted (Answer #2)

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You need to evaluate the limits of integration ` ` such that:

` ` is not valid since ` ` )

` ` invalid)

` ` invalid

You need to find the area enclosed by the given curves, hence, you need to evaluate the following definite integral, such that:

1234567-1-2-3-4-5-6-712345-1-2-3-4-5

` `

` `

Using the fundamental theorem of calculus yields:

` `

Since ` ` yields:

` `

` `

Hence, evaluating the area enclosed by the given curves, using definite integral, yields ` `


The correct answer is ln2 and I get it now.

We should do it by 2 areas adding up instead of putting all 3 functions together to get 1 area.

`int_0^1(x-x/4)dx` +`int_1^2(1/x-x/4)dx `

thus we can get ln2

tonys538's profile pic

Posted (Answer #3)

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The area enclosed by the the curves y = 1/x, y = x and y = x/4 has to be determined while x >0.

The graph of the three curves is shown below:

It is seen in the graph provided above that the required area can be divided into two smaller pieces.

Where x lies in [0,1] the area between the curves y = x and y = x/4 has to be determined. This has to be added to the area between the curves y = 1/x and y = x/4 where x lies in [1, 2]

The integral giving the area is:

`int_0^1(x - x/4) dx + int_1^2(1/x - x/4) dx`

= `(3/4)*int_0^1 x + int_1^2 (1/x - x/4) dx`

= `|(3/4)*x^2/2|_0^1+ |ln x - x^2/8|_1^2`

= `(3/4)*1/2 + ln 2 - ln 1 - 4/8 + 1/8`

= `0 + ln 2 - ln 1`

= ln 2 as ln 1 = 0

The required area is ln 2

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