Double Integral: Finding Triangle Area

[Icosahedron]

Homework Statement

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

The Attempt at a Solution

<br /> \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)

 

[HallsofIvy]

Homework Statement

<br /> \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

<br /> \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.)

 

[HallsofIvy]

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