# It's wrong the solution to definite integral from Mathematica 11.3? [closed]

I'm solving the exercise 23 from 4.8 section from "The Calculus 7th Leithold" (I use the Spanish edition "El Cálculo 7"), I write the solution in (physical) notebook, was 8*sqrt(2)/3, but when run in Mathematica:

Integrate[Sqrt[x - 1], {x, 1, 3}]


the Mathematica's solution is 4*sqrt(2)/3.

Take some time to find the problem with my solution, but I don't understand, then find the solution from the book, is the same that mine, please can tell me what happens?

## closed as off-topic by Henrik Schumacher, MarcoB, Carl Woll, Daniel Lichtblau, m_goldbergFeb 19 at 1:28

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• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Henrik Schumacher, MarcoB, Carl Woll, Daniel Lichtblau, m_goldberg
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• $\frac{4}{3} \sqrt{2}$ is correct. One has $\int_1^3 \sqrt{x-1} \, \mathrm{d} x = \int_0^2 \sqrt{x} \, \mathrm{d} x = \left[ \frac{2}{3} x^{\frac{3}{2}}\right]^{x=2}_{x=0} = \frac{4}{3} \sqrt{2}$. – Henrik Schumacher Feb 18 at 20:12
• Hi Henrik Schumacher, please try with [1,3] instead [0,2], and you'll see that is the double your answer. And don't forget is x-1, no x. – Rigby K. Feb 18 at 20:36
• To elaborate on Henrik's comment, here are the steps Wolfram|Alpha gives. – Chip Hurst Feb 18 at 20:40
• Sorry, the answer is correct, thanks to all. – Rigby K. Feb 18 at 20:50

Well, with the substitution w=(x-1)^(1/2) so that dx=2w dw the integral becomes Integrate[2 w^2, {w, 0, Sqrt[2]}] which is 4 Sqrt[2]/3. I guess I agree with Mathematica on this one. The book and its problem solutions are written by a person (or people) who may make mistakes.