# Double integral

Icosahedron

## Homework Statement

$$\int_a^b\,dx\int_a^x\,dy$$ should give area of a triangle, I can’t see how.

## 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)

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Homework Helper

## Homework Statement

$$\int_a^b\,dx\int_a^x\,dy$$ should give area of a triangle, I can’t see how.
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".

## 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)
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"?

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.)

Homework Helper
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"!

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

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.

?

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:
$$\frac{1}{2} (b - a)^2$$

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