The force is simply the weight of the water displaced. You should have a theorem that proves that. So integrate to find the volume and multiply by the density of water. But if you have a triangle, the volume is zero unless it has some thickness. Then it's easy.
Added: if it is completely submerged, the force is simply the volume of the shape. For a right triangle with legs of 3 and 7, the area is 21/2 and the volume would be $\frac{21t}{2}$, where $t$ is the thickness. Then the force would be $\frac{21t}{2}62.4$ pounds upward.
Added2: If there is no thickness, you are considering everything to have unit thickness. The 62.5 is the density of water in lb/ft^3 (though it is really 62.43 max unless you have salt water). The reason you don't need an integral is that you said it was submerged 1 foot below the surface. It's like if you asked in integral calculus for the area under $y=1-x$ from 0 to 2. If the upper limit were below 1, I might do an integral, but with limit above 1 I would just find the area of the triangle.
But maybe you meant the lower corner is 1 foot down, so the triangle is partially submerged. You still need the area of what is in the water and your expression integrating $dA$ is to do exactly that. You can either use similar triangles to see that the submerged shape has height 1 foot and width $\frac{7}{3}$ feet for an area of $\frac{7}{6}$ foot, or integate $\int_0^1\frac{7}{3}xdx$ get the same as the volume element is $dx$ high and $\frac{7}{3}x$ wide (determined by the same similar triangles I used in the first).