Integrating x+2y over Triangle with Vertices (0,0), (2,0), (0,3)

Calculus 3 · Axiom Academy

EXAMPLE Integrating x+2y over a Triangle Set up the limits from the hypotenuse and evaluate — both integration orders give the same value. Evaluate , where R is the triangle with vertices (0,0) , (2,0) , and (0,3) . Region R . The dashed strip shows a Type I slice: for a fixed x , y runs from the x -axis up to the hypotenuse. Nice work — you evaluated over the triangle, and confirmed it two ways. Sketch first: plotting R shows the legs lie on the axes and the only nontrivial boundary is the hypotenuse. Get the hypotenuse: through (2,0) and (0,3) the intercept form gives (equivalently ). Inner bounds depend on the outer variable: Type I uses with ; Type II uses with . Order is a free check: and both give 8 — matching answers is strong evidence the setup is right. Whenever a region has one slanted edge, read its equation off two vertices and let the inner limits be functions of the outer variable.

This is the written version of the interactive lesson above. See the full Calculus 3 course.