Determine the area of the given region under the curve   y=1/x^4 http://www.webassign.net/larson/4_04-36.gif

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You need to determine the  area of the region under the curve `y = 1/x^4` , over the interval  `[1,2],`  such that:

`A = int_1^2 f(x) dx => A = int_1^2 (1/x^4) dx`

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

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

`int_1^2 (x^(-4)) dx = -1/(3x^3)|_1^2`

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You need to determine the  area of the region under the curve `y = 1/x^4` , over the interval  `[1,2],`  such that:

`A = int_1^2 f(x) dx => A = int_1^2 (1/x^4) dx`

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

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

`int_1^2 (x^(-4)) dx = -1/(3x^3)|_1^2`

You need to use the fundamental theorem of calculus to evaluate the definite integral such that:

`int_(x_1)^(x_2) f(x) dx = F(x_2) - Fx_1)`

Reasoning by analogy yields:

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

`int_1^2 (1/x^4) dx = -1/24 + 1/3`

`int_1^2 (1/x^4) dx = (-1+8)/24`

`int_1^2 (1/x^4) dx = 7/24`

Hence, evaluating the area of the region under the given curve, over the interval `[1,2], ` yields `A = int_1^2 (1/x^4) dx = 7/24.`

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