# Fundamental Theorem of Calculus: `int_1^4 (2/sqrt x) dx` S4,1(2/(x^1/2))dx = 2/(x^3/2) = 2/(4^3/2) - 2/1 = -7/4

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Luca B. | Certified Educator

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You should evaluate the definite integral using the fundamental theorem of calculus such that:

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

You need to convert the square root into a power such that:

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

You should use the negative power property such that:

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

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

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

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

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

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

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

Hence, evaluating the given definite integral, using the fundamental theorem of calculus, yields `int_1^4 2/(sqrt x) dx = 4.`

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