I missed the negative sign before the two for the last line. It is sheer coincidence that the integrals I set up happen to yield the same answer as the one on your answer sheet.
I got 41/6. But the answer states 31/6!
This is shouldn't to be, by virtue, a straight to the point simple problem because of the abstraction. I don't fare very well in applicational calculus.
However, it's ridiculous that I'm always working out 41/6 but which the sheet states 31/6.
I typed the question word for word.
You always integrate from left to right. That is how the definite integral is defined. If you integrate from right to left, you get negative the correct area.
So, if you integrate from -1 to -2, you are integrating from a bigger number to a smaller number. That is going right to left. You mean it should be the integral of x^2-3x from -2 to -1? Yes. That is correct.
The region bounded by y=3x and y=x^2 from x=0 to x=3 is NOT part of the region bounded by
y=x^2, y=3x, x=-2, and x=-1
Note that x=0 and x=3 are both bigger than x=-1, which is the right bound of the two lines (x=-2 and x=-1). So, if you include that area, you are included area that is NOT bounded by the given functions and lines.
We might be confused by a case of semantics or stipulative definition.
Shouldn't the bigger value be at the top of the sigma?
Edit: In essence, if I move along the x-axis from right to left, I would obtain a negative value-small value minus larger value. So therefore, adding a minus sign outside of the sigma resolve this problem since area can never be negative much like a scalar value.
In moving from left to right, we would proceed the algebraic operation as per usual since we will always obtain a positive value
Consider the trpezoid bouded but small base 1 +3 = 4 these are the coordinates from the parabola and the line. It is +3 because we are using the distance between points.
The long base is 4 + 6 = 10
area = [(10 + 4)/2]*1 = 7 = 42/6
This area is slightly larger than 41/6 because the curvature of the pararable