Question

1. Change the order of the integration in the following double integral:
∫_(√3)^1dx ∫_(-√(4-x^2 ))^0ydy

2. Evaluate the following integral:

∫(3x-5)/√(9+6x-3x^2 ) dx

Answer 1

$1) \ \int_{\sqrt3}^1 \ dx \int_{-\sqrt{4-x^2}}^0 \ dy = \\ -\int_1^{\sqrt3} \ dx \int_{-\sqrt{4-x^2}}^0 \ dy\\ 1 \le x \le \sqrt3\\ -\sqrt{4-x^2} \le y \le 0\\ y = -\sqrt{4-x^2}\\ x^2+y^2 = 4 \\ x = \sqrt{4-y^2}\\ the\ upper\ semicircle\ with\ the\ center\ at\\ the\ point\ O (0; 0)\ and\ the\ radius\ 2.\\ \text{When changing the order of integration,}\\ \text{ our region must be divided into two subregions:}\\ D = D_1+D_2\\ D_1: 1 \le x \le \sqrt3, \ -1 \le y \le 0 \\ D_2: 1 \le x \le \sqrt{4-y^2}, \ -\sqrt3 \le y \le -1 \\ \int_{\sqrt3}^1 \ dx \int_{-\sqrt{4-x^2}}^0 \ dy = \\ \int_{-\sqrt3}^{-1} \ dy \int_1^{\sqrt{4-y^2}} \ dx + \int_{-1}^{0} \ dy \int_1^{\sqrt{3}} \ dx\\ answer: \ \int_{-\sqrt3}^{-1} \ dy \int_1^{\sqrt{4-y^2}} \ dx + \int_{-1}^{0} \ dy \int_1^{\sqrt{3}} \ dx\\ \\ \ \\ 2) \int \dfrac{3x-5}{\sqrt{9+6x-3x^2}}dx = \\ = \int \dfrac{3x-3}{\sqrt{9+6x-3x^2}}dx - \int \dfrac{2}{\sqrt{9+6x-3x^2}}dx = \\ = -\int \dfrac{1}{2\sqrt{9+6x-3x^2}}d(9+6x-3x^2) -\\ + \dfrac{2}{\sqrt3}*\int \dfrac{1}{\sqrt{4-(1-x)^2}}d(1-x) = \\ \dfrac{2}{\sqrt3} sin^{-1}(\dfrac{1}{2} - \dfrac{x}{2}) - \sqrt{9+6x-3x^2} +C)\\ answer: \ \dfrac{2}{\sqrt3} sin^{-1}(\dfrac{1}{2} - \dfrac{x}{2}) - \sqrt{9+6x-3x^2} +C$