Question

The double integral of \( f(x,y)=x \) over the triangular region with vertices at \( \left(-\frac{1}{2},\frac{1}{2}\right) \), \( (1,2) \), and \( (1,-1) \) is rounded off to one decimal place.

Hint:

For a linear function over a triangular region, the double integral equals area of the triangle multiplied by the function value at the centroid.

Answer 1

Correct Answer: 1.1

Step 1: A shortcut for linear functions.
For $f(x,y)=x$, the integral over any triangle equals the area times the $x$ value of the centroid.

Step 2: Centroid $x$ value.
\[ \bar x=\frac{-\frac12+1+1}{3}=\frac{1}{2} \]

Step 3: Find a base.
Two vertices $(1,2)$ and $(1,-1)$ share $x=1$, so that vertical side has length $2-(-1)=3$.

Step 4: Find the height.
The third vertex $(-\tfrac12,\tfrac12)$ is at horizontal distance $1-(-\tfrac12)=\tfrac32$ from that side.

Step 5: Area.
\[ \text{Area}=\tfrac12\cdot 3\cdot \tfrac32=\tfrac94 \]

Step 6: Multiply.
\[ \iint_R x\,dA=\tfrac94\cdot\tfrac12=\tfrac98=1.125 \]
Rounded to one decimal, this is $1.1$.
\[ \boxed{1.1} \]