Question

Let \( T \) denote the triangle in the \( xy \)-plane bounded by the \( x \)-axis and the lines \( y = x \) and \( x = 1 \). The value of the double integral (over \( T \)) \[ \iint_T (5 - y) \, dx \, dy \] is equal to ............. (rounded off to two decimal places).

Hint:

When evaluating double integrals over triangular regions, carefully determine the limits of integration based on the geometry of the region.

Answer 1

Step 1: Set up the region of integration.
The region \( T \) is the triangle with vertices at \( (0, 0) \), \( (1, 1) \), and \( (1, 0) \). The bounds for the integral are: - \( x \) ranges from 0 to 1, - For a given \( x \), \( y \) ranges from 0 to \( x \). Thus, the double integral becomes: \[ \int_0^1 \int_0^x (5 - y) \, dy \, dx. \] Step 2: Perform the inner integral.
First, integrate with respect to \( y \): \[ \int_0^x (5 - y) \, dy = \left[ 5y - \frac{y^2}{2} \right]_0^x = 5x - \frac{x^2}{2}. \] Step 3: Perform the outer integral.
Now, integrate with respect to \( x \): \[ \int_0^1 \left( 5x - \frac{x^2}{2} \right) \, dx = \left[ \frac{5x^2}{2} - \frac{x^3}{6} \right]_0^1 = \frac{5}{2} - \frac{1}{6} = \frac{15}{6} - \frac{1}{6} = \frac{14}{6} = \frac{7}{3}. \] Final Answer: \[ \boxed{2.33}. \]