Sure! Let's tackle each part step by step. --- ## 1. **Find the value of \( k \)** The joint pdf must integrate to 1 over the entire support: \[ \int_^1 \int_^1 k(x^2 + y^2) \, dx \, dy = 1 \] Let’s compute the integral: \[ \int_^1 \int_^1 (x^2 + y^2) \, dx \, dy = \int_^1 \left[ \int_^1 x^2 \, dx + \int_^1 y^2 \, dx \right] dy \] But \(\int_^1 y^2 \, dx = y^2 \cdot \int_^1 dx = y^2 \cdot (1-) = y^2\). So: \[ \int_^1 \int_^1 (x^2 + y^2) dx\,dy = \int_^1 \left( \int_^1 x^2 dx + y^2 \right) dy \] Calculate \(\int_^1 x^2 dx = \frac{1}{3}\): \[ = \int_^1 \left(\frac{1}{3} + y^2\right) dy = \frac{1}{3} \int_^1 dy + \int_^1 y^2 dy = \frac{1}{3}(1) + \left[\frac{y^3}{3}\right]_^1 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] So, \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. **Find \(P(X + Y \leq 1)\)** We want: \[ P(X+Y \leq 1) = \int_^1 \int_^{1-y} k(x^2 + y^2) dx\,dy \] Plug in \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_^1 \int_^{1-y} (x^2 + y^2) dx\,dy \] Evaluate the inner integral: \[ \int_^{1-y} (x^2 + y^2) dx = \int_^{1-y} x^2 dx + \int_^{1-y} y^2 dx = \left[\frac{x^3}{3}\right]_^{1-y} + y^2[x]_^{1-y} = \frac{(1-y)^3}{3} + y^2(1-y) \] Now, integrate with respect to \( y \): \[ P = \frac{3}{2} \int_^1 \left[\frac{(1-y)^3}{3} + y^2(1-y)\right] dy \] Expand: \[ = \frac{3}{2} \int_^1 \left( \frac{(1-y)^3}{3} + y^2 - y^3 \right) dy \] \[ = \frac{3}{2} \left[ \frac{1}{3}\int_^1 (1-y)^3 dy + \int_^1 y^2 dy - \int_^1 y^3 dy \right] \] Compute each term: - \(\int_^1 (1-y)^3 dy = \left[\frac{(1-y)^4}{-4}\right]_^1 = \left(\frac{}{-4}\right) - \left(\frac{1}{-4}\right) = \frac{1}{4}\) - \(\int_^1 y^2 dy = \frac{1}{3}\) - \(\int_^1 y^3 dy = \frac{1}{4}\) So: \[ P = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left( \frac{1}{12} + \frac{1}{3} - \frac{1}{4} \right) \] Get common denominator (12): \[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} = \frac{2}{12} = \frac{1}{6} \] \[ P = \frac{3}{2} \cdot \frac{1}{6} = \frac{1}{4} \] --- ## 3. **Find \( E[X] \) and \( E[Y] \)** By symmetry, both will be the same. \[ E[X] = \int_^1\int_^1 x f(x,y) dx dy = \int_^1\int_^1 x \cdot k(x^2 + y^2) dx dy \] Plug \(k = \frac{3}{2}\): \[ = \frac{3}{2} \int_^1\int_^1 x(x^2 + y^2) dx dy = \frac{3}{2} \int_^1\int_^1 (x^3 + x y^2) dx dy \] Let’s split and solve: - \(\int_^1 x^3 dx = \frac{1}{4}\) - \(\int_^1 x dx = \frac{1}{2}\) - \(\int_^1 y^2 dy = \frac{1}{3}\) So, \[ \int_^1 (x^3 + x y^2) dx = \frac{1}{4} + y^2 \cdot \frac{1}{2} \] Now integrate over \(y\): \[ E[X] = \frac{3}{2} \int_^1 \left( \frac{1}{4} + \frac{y^2}{2} \right) dy = \frac{3}{2} \left[ \frac{1}{4} \cdot 1 + \frac{1}{2} \cdot \frac{1}{3} \right] = \frac{3}{2} \left( \frac{1}{4} + \frac{1}{6} \right) \] \[ = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] Similarly, \(E[Y] = \frac{5}{8}\) due to symmetry. --- ## **Summary of Answers** 1. **\( k = \frac{3}{2} \)** 2. **\( P(X+Y \leq 1) = \frac{1}{4} \)** 3. **\( E[X] = E[Y] = \frac{5}{8} \)** If you need any step explained further, let me know!