Double Integrals F807D1

1. **Problem Statement:** (i) Evaluate the double integral $$\iint_D e^{-x-2y} \, dA$$ where $$D$$ is the region in the first quadrant bounded by $$y \leq x$$. (ii) Evaluate the double integral $$\int_0^4 \int_{x/2}^1 \frac{1}{1+y^3} \, dy \, dx$$. --- 2. **Integral (i) Setup:** The region $$D$$ is in the first quadrant where $$y \leq x$$. Since both $$x$$ and $$y$$ are non-negative, the limits are: - $$x$$ from $$0$$ to $$\infty$$ (since no upper bound given) - $$y$$ from $$0$$ to $$x$$. So the integral becomes: $$\int_0^{\infty} \int_0^x e^{-x-2y} \, dy \, dx$$ --- 3. **Integral (i) Inner Integral:** Evaluate the inner integral with respect to $$y$$: $$\int_0^x e^{-x-2y} \, dy = e^{-x} \int_0^x e^{-2y} \, dy$$ Calculate the inner integral: $$\int_0^x e^{-2y} \, dy = \left[-\frac{1}{2} e^{-2y} \right]_0^x = -\frac{1}{2} e^{-2x} + \frac{1}{2}$$ So the inner integral is: $$e^{-x} \left( \frac{1}{2} - \frac{1}{2} e^{-2x} \right) = \frac{e^{-x}}{2} - \frac{e^{-3x}}{2}$$ --- 4. **Integral (i) Outer Integral:** Now integrate with respect to $$x$$: $$\int_0^{\infty} \left( \frac{e^{-x}}{2} - \frac{e^{-3x}}{2} \right) dx = \frac{1}{2} \int_0^{\infty} e^{-x} dx - \frac{1}{2} \int_0^{\infty} e^{-3x} dx$$ Calculate each integral: $$\int_0^{\infty} e^{-x} dx = \left[-e^{-x}\right]_0^{\infty} = 1$$ $$\int_0^{\infty} e^{-3x} dx = \left[-\frac{1}{3} e^{-3x}\right]_0^{\infty} = \frac{1}{3}$$ So the outer integral is: $$\frac{1}{2} (1) - \frac{1}{2} \left( \frac{1}{3} \right) = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}$$ --- 5. **Integral (i) Final Answer:** $$\boxed{\frac{1}{3}}$$ --- 6. **Integral (ii) Setup:** Evaluate: $$\int_0^4 \int_{x/2}^1 \frac{1}{1+y^3} \, dy \, dx$$ --- 7. **Integral (ii) Inner Integral:** The inner integral is with respect to $$y$$: $$\int_{x/2}^1 \frac{1}{1+y^3} \, dy$$ This integral does not have a simple elementary antiderivative, so we leave it as is for now. --- 8. **Integral (ii) Change Order of Integration:** The region of integration is: - $$0 \leq x \leq 4$$ - $$x/2 \leq y \leq 1$$ Rewrite the limits for $$y$$ first: - $$0 \leq y \leq 1$$ - For each fixed $$y$$, $$x$$ varies from $$0$$ to $$2y$$ (since $$x/2 \leq y \Rightarrow x \leq 2y$$) So the integral becomes: $$\int_0^1 \int_0^{2y} \frac{1}{1+y^3} \, dx \, dy$$ --- 9. **Integral (ii) Inner Integral with respect to $$x$$:** Since the integrand does not depend on $$x$$: $$\int_0^{2y} \frac{1}{1+y^3} \, dx = \frac{1}{1+y^3} \times (2y - 0) = \frac{2y}{1+y^3}$$ --- 10. **Integral (ii) Outer Integral:** Now integrate with respect to $$y$$: $$\int_0^1 \frac{2y}{1+y^3} \, dy$$ --- 11. **Integral (ii) Simplify and Substitute:** Let $$u = 1 + y^3$$, then $$du = 3y^2 dy$$. Rewrite the integral: $$\int_0^1 \frac{2y}{1+y^3} dy = 2 \int_0^1 \frac{y}{1+y^3} dy$$ We want to express $$y dy$$ in terms of $$du$$, but $$du = 3y^2 dy$$, so this substitution is not straightforward. Instead, use partial fraction or another substitution. --- 12. **Integral (ii) Alternative substitution:** Rewrite the denominator: $$1 + y^3 = (1 + y)(1 - y + y^2)$$ But this complicates the integral. Try substitution: $$t = y^3 \Rightarrow dt = 3y^2 dy$$ Rewrite $$y dy$$ in terms of $$dt$$: $$y dy = \frac{y dy}{1}$$ but no direct substitution. --- 13. **Integral (ii) Use substitution $$t = y^3$$:** Rewrite the integral: $$I = 2 \int_0^1 \frac{y}{1+y^3} dy = 2 \int_0^1 \frac{y}{1+t} \frac{dt}{3y^2} = \frac{2}{3} \int_0^1 \frac{1}{(1+t) y} dt$$ But $$y = t^{1/3}$$, so: $$I = \frac{2}{3} \int_0^1 \frac{1}{(1+t) t^{1/3}} dt = \frac{2}{3} \int_0^1 \frac{t^{-1/3}}{1+t} dt$$ --- 14. **Integral (ii) Final form:** $$I = \frac{2}{3} \int_0^1 \frac{t^{-1/3}}{1+t} dt$$ This integral is a known special function (the Beta or hypergeometric function), so the exact elementary form is complicated. Numerical approximation or special functions are needed for exact value. --- 15. **Summary:** - (i) The value of the integral is $$\frac{1}{3}$$. - (ii) The integral reduces to $$\frac{2}{3} \int_0^1 \frac{t^{-1/3}}{1+t} dt$$ which can be evaluated numerically or expressed in terms of special functions. --- **Final answers:** (i) $$\boxed{\frac{1}{3}}$$ (ii) $$\boxed{\frac{2}{3} \int_0^1 \frac{t^{-1/3}}{1+t} dt}$$ (numerical or special function evaluation required)