`int_1^2((4 + u^2)/(u^3))du` Evaluate the integral.

Textbook Question

Chapter 5, 5.3 - Problem 40 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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

`int_a^b f(u) du = F(b) - F(a)`

`int_1^2 (4+u^2)/(u^3) du = int_1^2 4/(u^3) du + int_1^2 (u^2)/(u^3) du`

`int_1^2 (4+u^2)/(u^3) du = 4int_1^2 (u^(-3)) du + int_1^2 1/u du`

Using the formula` int u^n = (u^(n+1))/(n+1)+ c` yields:

`4int_1^2 (u^(-3)) du = 4(u^(-2))/(-2) = -2/(u^2)|_1^2 = -2(1/2^2 - 1/1^2)`

`4int_1^2 (u^(-3)) du =-2(1/4 - 1) = -2*(-3/4) = 3/2`

`int_1^2 1/u du = ln u|_1^2 = ln 2 - ln 1 = ln 2`

Hence, evaluating the definite integral, yields `int_1^2 (4+u^2)/(u^3) du = 3/2 + ln 2.`

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