Why is the definite integral between negative limits of a function with positive values negative even though the area bound by the x-axis is positive? for example the integral of y=x^2 between x=-2 and x=-1

Referring back to the definition of an integral, it is the sum of small elements on the x-axis (dx) multiplied by the value of the function at that point (y) commonly expressed as the sum of ydx. Since x is negative in this region, so is dx, resulting in all of the elements of the sum being negative. A useful way of remembering this is to think about the problem graphically, and what quadrant our function crosses:x,y > 0 Both ydx (and in turn the integral in this quadrant) is positivex > 0 > y Both ydx (and in turn the integral in this quadrant) is negativex < 0 < y Both ydx (and in turn the integral in this quadrant) is negativex,y < 0 Both ydx (and in turn the integral in this quadrant) is positive
(which is case 3 for the function y=x2 for negative limits)

Answered by Maths tutor

2928 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Differentiate y=x^(-1/2)-x


Find a solution for the differential equation dy/dx=exp(-y)*sin2x which passes through the origin.


Points A and B have coordinates (–2, 1) and (3, 4) respectively. Find the equation of the perpendicular bisector of AB and show that it may be written as 5x +3 y = 10.


Find the general solution of the differential equation: d^2x/dt^2 + 5dx/dt + 6x = 2cos(t) - sin(t)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences