How do you find the definite integral for: #(1/x^(2))dx# for the intervals #[5, 6]#? Calculus Introduction to Integration Formal Definition of the Definite Integral 1 Answer Eddie Aug 4, 2016 # = 1/30# Explanation: #int_5^6 1/x^2 dx# #= int_5^6 d/dx (-1/x) dx# #= (-1/x)_5^6 = (1/x)_6^5# #= 1/5 -1/6 = 1/30# 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 2639 views around the world You can reuse this answer Creative Commons License