What is the area under the curve #y = 2x^{-3}# from 6 to 10? Calculus Introduction to Integration Integration: the Area Problem 1 Answer Alan P. Aug 28, 2015 Area under #y=2x^(-3)# from 6 to 10 is #4/225# Explanation: The area under the curve #y=2x^(-3)# from 6 to 10 is #int_6^10 2x^-3 dx# #= -1*x^-2|_6^10# #= -1*1/x^2|_6^10# #= (-1/10^2) - (-1/6^2)# #=1/36 -1/100# #=4/225# (after a bit of arithmetic) Answer link Related questions How do you find the area of a region using integration? How do you use integration to find area under curve? Why does integration find the area under a curve? How do I evaluate #int_0^5|x-5|dx# by interpreting it in terms of areas? How do you find the area of the parallelogram with vertices (4,5), (9, 9), (13, 10), and (18, 14)? How do you evaluate the integral of absolute value of (x - 5) from 0 to 10 by finding area? How do you find the area of the parallelogram with vertices k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3)? How do you find the area of the parallelogram with vertices: p(0,0,0), q(-5,0,4), r(-5,1,2), s(-10,1,6)? How do you evaluate #int5# between the interval [0,4]? What is a surface integral? See all questions in Integration: the Area Problem Impact of this question 2064 views around the world You can reuse this answer Creative Commons License