# Double integral

1. May 26, 2008

### Icosahedron

1. The problem statement, all variables and given/known data
$$\int_a^b\,dx\int_a^x\,dy$$ should give area of a triangle, I can’t see how.

3. The attempt at a solution

$$\int_a^b\,dx\,{(x-a)}$$ but then I won’t get 1/2 (ab)… (don't get the first brackets in latex between the two x)

Last edited: May 26, 2008
2. May 26, 2008

### HallsofIvy

Staff Emeritus
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"?

3. May 26, 2008

### EngageEngage

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.)

4. May 26, 2008

### HallsofIvy

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

5. May 26, 2008

### Icosahedron

Thank you! Have been starring at this for like two hours, now after your help it is crystal clear.

6. May 27, 2008

### Icosahedron

Wait! Nothing is crystal clear.

So I draw two vertical lines on the x axis at a and b. Then on the y axis an horizontal line at a and 45 degree line for x=y. That gives a beautiful triangle that has area of 1/2(b-a)^2.

But when I do the integral, I get 1/2(b-a)^2 - (b-a)a.

?????

7. May 27, 2008

### Nick89

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

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

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

which you should recognize as:
$$\frac{1}{2} (b - a)^2$$

8. May 27, 2008

### Icosahedron

Idiotic is I am, for some reason I plugged b-a in 1/2 x^2, not b and a each and substract them. Some while ago that I calculated definite integrals.

thanks Nick, HallsofIvy, EngageEngage