The integral of the function ∫ 3 + x√x from -1 to 4 results in a complex number due to the square root term, which is only defined for x ≥ 0. The evaluation yields 139/5 - 2i/5, confirming that the integral's result includes an imaginary component. This highlights the importance of understanding the domain of functions when performing definite integrals, particularly with square roots. The discussion also references Wolfram Alpha for verification of the solution.
PREREQUISITESStudents studying calculus, particularly those dealing with integrals involving square roots, and educators looking to clarify the implications of domain restrictions in mathematical functions.
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?
Because of the term with the square root, the integrand is defined only for x ≥ 0. What is the complete statement of the problem?Painguy said:Homework Statement
∫ 3+ x√x from -1 to 4
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?