I'm in minor confusion why my answer is wrong as opposed towards the book's answer.
I'm asked to find the area enclosed by the equations,
If I compute the area under each equation, and also consider a sign test on both integrals seperately, then:-
Now if I consider another sign test,
The total area available under either equation, with limit , is . Then the area , enclosed by and will be less than . So in order for to be subtracted from it must have its sign reversed because it's negative.
However, the book has answer of
The enclosed area is greater than
I don't understand why my answer is wrong?
I've already done that.
If I can see signs could be an issue I have a habit to treat each integral seperately.
The majority of the area under y1 is negative.
I can't see why it's not immiterial if you choose to substract the areas individually or as a lumped integrand?
I've treated the problem as a lumped integrand and I've got the area enclosed as units.
But I don't understand how?
The area requested is part of the area under but this tells me its greater than because for the total area under for is .This is what I don't understand.
2. Normally you can avoid the use of absolute values if you don't integrate "over"(?) a zero. In your case both the straight line and the parabola have zeros in the given interval. That means you have to split the interval. The easiest way is of course to integrate the difference of functions as described by Plato.
3. In the attached sketch the green area is your first integral, the red area your second.
What's the computer package you used to build that graph?
That's another reason why I was going wrong. I didn't realise the area UNDER the x-axis turns itself out onto the outside of the curve.
Imagine drawing a line, from the parabola, vertically all the way down through x =2 and then stop until you reach the y value where x = -2, that's how I've always pictured the area under any -f(x) parabola.
This is why I was making the claim on how the segmented area could be possibly bigger than the area under the parabola.
But it all makes sense now.