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

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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` ...

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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.`

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