Integration is a process of finding the area under the graph

[thereddevils]

Homework Statement

Find the area bounded by y=1-x^2 and y=x-1 .

Homework Equations

The Attempt at a Solution

i know it can be done like this

\int^1_{-2} [(1-x^2)-(x-1)]dx

we sort of shift the x-axis below . From what i understand , integration is a process of finding the area under the graph but that way of doing is not area under the graph ?

 

[Mentallic]

Actually when you do the integral of f(x)-g(x) you're finding the integral of f(x) and then the integral of g(x) (both being areas) and you take away the second integral. This leaves you with the area between the curves.

http://thesaurus.maths.org/mmkb/media/png/AreaBetweenCurves.png"

Notice how the area between the 2 curves is the area of the top curve minus the area of the lower curve. This can be extended for curves above and below the x-axis but you'll need to draw a few examples of each case to prove it to yourself.

Remember that taking the integral from x=a to x=b of a curve that is below the x-axis (for that domain, a<=x<=b) gives you a negative result, even though the area is positive.

 

[thereddevils]

Actually when you do the integral of f(x)-g(x) you're finding the integral of f(x) and then the integral of g(x) (both being areas) and you take away the second integral. This leaves you with the area between the curves.

http://thesaurus.maths.org/mmkb/media/png/AreaBetweenCurves.png"

Notice how the area between the 2 curves is the area of the top curve minus the area of the lower curve. This can be extended for curves above and below the x-axis but you'll need to draw a few examples of each case to prove it to yourself.

Remember that taking the integral from x=a to x=b of a curve that is below the x-axis (for that domain, a<=x<=b) gives you a negative result, even though the area is positive.

thanks Mentallic , that's what i am confused with - when the curve is below and above the x-axis like the example here .

I can see that the area can be found by taking the integral of 1-x^2 minus the integral of x-1 . In the example you gave me , both f(x) and g(x) are above the x-axis , what if its both above and below ?

 

[Mark44]

That's why it's important to sketch a graph of the region whose area you want. Many times students overlook the importance of getting a good feel for the geometry of the situation and tend to immediately set up an integral and start in on it.

For example, if you want to find the area between the graph of y = sinx between 0 and 2pi, a naive student will set up the following integral:
\int_0^{2\pi} sin(x) dx

The problem with the integral above is that it evaluates to zero, which is not the area between the curve and the x-axis.

The correct formulation is
\int_0^{\pi} sin(x) dx + \int_{\pi}^{2\pi} -sin(x) dx

In the first interval, [0, pi], the graph of the sine function is above the x-axis, so the typical area element is sin(x)\Delta x. In the second interval, [pi, 2pi], the graph of the sine function is below the x-axis, so the typical area element is (0 -sin(x))\Delta x = -sin(x)\Delta x.

 

[Mentallic]

There is a property about integrals that should be obvious if you think about it.

\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx

This means if you split up the integral into integrals that are above and below the x-axis only, and remember that since the integral below the x-axis is negative, minus that gives you a positive value.

 

[thereddevils]

thanks Mark and Mentallic , i know its important to see the set up first before evaluating the integral .

But why is this working for this question correct ?

\int^1_{-2} [(1-x^2)-(x-1)]dx

the area under the graph for 1-x^2 and x-1 go beyond the x-axis in this case when it's integrated with respect to x . My teacher said we can sort of shift the x-axis downwards and integrate as usual . My question is why we can 'shift' the x-axis downwards ? Doesn't it change the properties of the graph and subsequently affect our result ?

But i have also tried and it turns out to be the correct answer .

 

[Mentallic]

Well it's hard to explain without actually being there to draw pictures for you to understand, which is what my teacher did when I was learning this, and what your teacher should have done.

let 1-x^2=f and x-1=g

\int_{-2}^1(f-g)dx=\int_{-2}^{-1}(f-g)dx+\int_{-1}^{1}(f-g)dx

Now let's look at the second integral first.

\int_{-1}^{1}(f-g)dx=\int_{-1}^{1}fdx-\int_{-1}^{1}gdx

\int_{-1}^{1}fdx is just the area above the x-axis that the parabola makes.

\int_{-1}^{1}gdx is the negative of the area that the line makes between x=-1, x=1 and the x-axis. It's negative because it lies below the x-axis.

Now since we are doing the integral of f-g, that is the area f makes with the x-axis minus the negative of the area g makes with the x-axis. That's just the two areas added together.

A similar idea applies for the last bit, \int_{-2}^{-1}(f-g)dx and as a consequence, if you take the area between two curves, it doesn't matter whether they're below, above or a mixture with the x-axis.

 

[thereddevils]

Well it's hard to explain without actually being there to draw pictures for you to understand, which is what my teacher did when I was learning this, and what your teacher should have done.

let 1-x^2=f and x-1=g

\int_{-2}^1(f-g)dx=\int_{-2}^{-1}(f-g)dx+\int_{-1}^{1}(f-g)dx

Now let's look at the second integral first.

\int_{-1}^{1}(f-g)dx=\int_{-1}^{1}fdx-\int_{-1}^{1}gdx

\int_{-1}^{1}fdx is just the area above the x-axis that the parabola makes.

\int_{-1}^{1}gdx is the negative of the area that the line makes between x=-1, x=1 and the x-axis. It's negative because it lies below the x-axis.

Now since we are doing the integral of f-g, that is the area f makes with the x-axis minus the negative of the area g makes with the x-axis. That's just the two areas added together.

A similar idea applies for the last bit, \int_{-2}^{-1}(f-g)dx and as a consequence, if you take the area between two curves, it doesn't matter whether they're below, above or a mixture with the x-axis.

thanks ! Very well explained , i am actually looking for this .