How do you find the definite integral of #(sqrt(2x+1))dx# from #[1, 4]#? Calculus Introduction to Integration Formal Definition of the Definite Integral 1 Answer Konstantinos Michailidis Apr 24, 2016 It is #int sqrt(2x+1)dx=1/3*int (2x+1)'*sqrt(2x+1)dx=1/3*[2x+1]^(3/2)+c# Hence #int_1^4 sqrt(2x+1)dx=[1/3*[2x+1]^(3/2)]_1^4=9-sqrt(3) ~~ 7.2679# Answer link Related questions What is the Formal Definition of the Definite Integral of the function #y=f(x)# over the... How do you use the definition of the definite integral? What is the integral of dy/dx? What is an improper integral? How do you calculate the double integral of #(xcos(x+y))dr# where r is the region: 0 less than... How do you apply the evaluation theorem to evaluate the integral #3t dt# over the interval [0,3]? What is the difference between an antiderivative and an integral? How do you integrate #3x^2-5x+9# from 0 to 7? Question #f27d5 How do you evaluate the definite integral #int sqrtt ln(t)dt# from 2 to 1? See all questions in Formal Definition of the Definite Integral Impact of this question 7921 views around the world You can reuse this answer Creative Commons License