# Area under a curve-positive and negative area cancellation

Gold Member

## Homework Statement

So this is a problem I've been facing while finding the area under some curves, for example finding the area bounded by sinx and cosx between pi/4 and 5pi/4. integrating sinx-cosx with these limits would result in the positive and negative areas cancelling out, how do I get the modulus of this area. In basic functions like sinx I would just double the area I found between 0 and pi to calculate upto 2pi but it seems more troubling to do here.
Do I need to integrate the modulus of the function, then I'd have to do extra work and find where sinx-cosx changes sign

## Homework Equations

All pertaining to calculus

## The Attempt at a Solution

Given above within the problem

Mark44
Mentor

## Homework Statement

So this is a problem I've been facing while finding the area under some curves, for example finding the area bounded by sinx and cosx between pi/4 and 5pi/4.
What does "bounded by sinx and cosx" mean?
Is this your integral? ##\int_{\frac \pi 4}^{\frac{5\pi} 4} \sin(x) - \cos(x) dx##

Krushnaraj Pandya said:
integrating sinx-cosx with these limits would result in the positive and negative areas cancelling out, how do I get the modulus of this area. In basic functions like sinx I would just double the area I found between 0 and pi to calculate upto 2pi but it seems more troubling to do here.
Do I need to integrate the modulus of the function, then I'd have to do extra work and find where sinx-cosx changes sign
You need to determine the intervals on which ##\sin(x) - \cos(x)## is positive, and those on which this expression is negative.

As a simpler example, ##\int_0^{2\pi} \sin(x) dx = 0##, but if you want to find the area bounded by the graph of this function, you need the absolute value of the portion that lies below the horizontal axis. It's the same idea with the function you want to integrate.

Krushnaraj Pandya said:

## Homework Equations

All pertaining to calculus

## The Attempt at a Solution

Given above within the problem

Gold Member
What does "bounded by sinx and cosx" mean?
Is this your integral? ∫5π4π4sin(x)−cos(x)dx∫π45π4sin⁡(x)−cos⁡(x)dx\int_{\frac \pi 4}^{\frac{5\pi} 4} \sin(x) - \cos(x) dx
yes it is, I got the correct answer as well.
You need to determine the intervals on which sin(x)−cos(x)sin⁡(x)−cos⁡(x)\sin(x) - \cos(x) is positive, and those on which this expression is negative.

As a simpler example, ∫2π0sin(x)dx=0∫02πsin⁡(x)dx=0\int_0^{2\pi} \sin(x) dx = 0, but if you want to find the area bounded by the graph of this function, you need the absolute value of the portion that lies below the horizontal axis. It's the same idea with the function you want to integrate.
Got it, thank you very much :D