Definite integral with complex result?

Painguy
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Homework Statement


∫ 3+ x√x from -1 to 4


Homework Equations





The Attempt at a Solution


∫ 3+ x√x from -1 to 4 = 3x+(2(x^5/2))/5 evaluated from -1 to 4

(12 + (2(4^5/2))/5) +(3 +(2(-1)^5/2)/5) ?

15+64/5 +(2(-1^(5/2))/5)

139/5 -(2(-1^(5/2)))/5

139/5 -2i/5

is that right?
 
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Painguy said:

Homework Statement


∫ 3+ x√x from -1 to 4


Homework Equations





The Attempt at a Solution


∫ 3+ x√x from -1 to 4 = 3x+(2(x^5/2))/5 evaluated from -1 to 4

(12 + (2(4^5/2))/5) +(3 +(2(-1)^5/2)/5) ?

15+64/5 +(2(-1^(5/2))/5)

139/5 -(2(-1^(5/2)))/5

139/5 -2i/5

is that right?

Yes:

http://www.wolframalpha.com/input/?i=integrate+3+++x*sqrt(x)+from+-1+to+4
 
Painguy said:

Homework Statement


∫ 3+ x√x from -1 to 4
Because of the term with the square root, the integrand is defined only for x ≥ 0. What is the complete statement of the problem?

Although there is a version of the square root function whose domain includes negative numbers, this is usually not presented in calculus courses.
Painguy said:

Homework Equations





The Attempt at a Solution


∫ 3+ x√x from -1 to 4 = 3x+(2(x^5/2))/5 evaluated from -1 to 4

(12 + (2(4^5/2))/5) +(3 +(2(-1)^5/2)/5) ?

15+64/5 +(2(-1^(5/2))/5)

139/5 -(2(-1^(5/2)))/5

139/5 -2i/5

is that right?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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