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: Identify the vertices.
Let the vertices be
\[ A\left(-\frac{1}{2},\frac{1}{2}\right), \quad B(1,2), \quad C(1,-1) \]

Step 2: Use centroid property for linear function.
For a linear function \(f(x,y)=x\), the integral over a triangle is
\[ \iint_R x\,dA=(\text{Area of triangle})\times (\text{x-coordinate of centroid}) \]

Step 3: Find x-coordinate of centroid.
\[ \bar{x}=\frac{x_1+x_2+x_3}{3} \]
\[ \bar{x}=\frac{-\frac{1}{2}+1+1}{3} \]
\[ \bar{x}=\frac{\frac{3}{2}}{3}=\frac{1}{2} \]

Step 4: Find area of triangle.
The points \(B(1,2)\) and \(C(1,-1)\) form a vertical side. Its length is
\[ BC=2-(-1)=3 \]

Step 5: Find perpendicular distance from \(A\) to line \(x=1\).
\[ \text{distance}=1-\left(-\frac{1}{2}\right)=\frac{3}{2} \]

Step 6: Calculate area.
\[ \text{Area}=\frac{1}{2}\times 3\times \frac{3}{2} \]
\[ \text{Area}=\frac{9}{4} \]

Step 7: Calculate the double integral.
\[ \iint_R x\,dA = \frac{9}{4}\times \frac{1}{2} = \frac{9}{8} = 1.125 \]
Rounded off to one decimal place,
\[ \boxed{1.1} \]