Area of triangle using double integrals

Question

I have one (rather simple) problem, but I'm stuck and can't figure out what I'm constantly doing wrong. I need to calculate area of triangle with points at $(0,0)$, $(t,0)$, $(t,\frac{t}{2})$. In other words triangle under function $y=\frac{x}{2}$, for $x\in [0,t]$ I thought it is calculated with $$ \int_0^t \int_0^\frac{t}{2} dudv$$ But it turns out that this equals to $\frac{t^2}{2}$, when obviously this area is $\frac{t\times\frac{t}{2}}{2} = \frac{t^2}{4}$. What am I doing wrong here? I need to calculate it this way, not with single integral, or geometrically.

Answer 1

The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$\int_0^t\int_0^{\frac x2}\,\mathrm dy\,\mathrm dx.$$

Answer 2

I think you made a mistake, the bounds on the inner integral should not be constant: $$ \int_0^t \int_0^{v/2} dudv = \int_0^t \frac{v}{2} dv = \left. \frac{v^2}{4} \right|_0^t = \frac{t^2}{4}. $$