`int_1^9((x - 1)/sqrt(x))dx` Evaluate the integral.

Textbook Question

Chapter 5, 5.3 - Problem 29 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Hello!

Find the indefinite integral first:

`int((x-1)/sqrt(x))dx=int(x^(1/2)-x^(-1/2))dx=(2/3)*x^(3/2)-2*x^(1/2)+C.`

So the definite integral is equal to

`((2/3)*x^(3/2)-2*x^(1/2))|_(x=1)^(x=9)=((2/3)*27-6)-((2/3)-2)=12-2/3+2=13 and 1/3 approx 13.33.`

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