Double integral

Have you drawn a picture? Draw x and y coordinate axes. x must lie between x= a and x= b on the x-axis so draw two vertical lines there. y must lie between y= a and y= x. Draw those lines. What figure does that give? That's the region you are integrating over and "dydx" is the "differential of area".

What did you do so that you "won't get 1/2(ab)", wave a magic wand? Show what you did. What is the integral of (x- a) dx from a to b?

Now, what makes you think the area of the triangle should be (1/2)(ab)? Look at the picture you drew above. What are the "height" and "base"?

 

if you draw out a picture you will see that you will get a triangle but its area will NOT be 1/2(ab). if you compute the integral you will find a familiar formula that is slightly different from what you have (you will have to play around with it to get it into form.)

 

If you do the integral [itex]\int_a^b (x-a)dx[/itex] by "substituting" u= x- a, you will get exactly the same as that "familiar formula"!

 

If you do the integral correctly you should also end up with 1/2(b-a)^2. You probably made a mistake somewhere!

[tex]\int_a^b dx \, \int_a^x dy = \int_a^b dx (x-a) = \left[ \frac{1}{2}x^2 - ax \right]_a^b = ...[/tex]

If you don't make any mistakes you will end up with:
[tex]\frac{1}{2}a^2 - ab + \frac{1}{2}b^2[/tex]

which you should recognize as:
[tex]\frac{1}{2} (b - a)^2[/tex]