1. Calculate $$\int_0^1 (3 - 2x) \, dx$$ $$= \left[3x - x^2\right]_0^1 = (3(1) - 1^2) - (0) = 3 - 1 = 2$$ 2. Calculate $$\int_{-3}^{-2} y (y + 1)^2 \, dy$$ Expand: $$y(y^2 + 2y + 1) = y^3 + 2y^2 + y$$ Integrate: $$\int y^3 + 2y^2 + y \, dy = \frac{y^4}{4} + \frac{2y^3}{3} + \frac{y^2}{2}$$ Evaluate from -3 to -2: $$\left(\frac{(-2)^4}{4} + \frac{2(-2)^3}{3} + \frac{(-2)^2}{2}\right) - \left(\frac{(-3)^4}{4} + \frac{2(-3)^3}{3} + \frac{(-3)^2}{2}\right)$$ $$= \left(\frac{16}{4} - \frac{16}{3} + 2\right) - \left(\frac{81}{4} - 54/3 + \frac{9}{2}\right) = (4 - 5.3333 + 2) - (20.25 - 18 + 4.5) = 0.6667 - 6.75 = -6.0833$$ 3. Calculate $$\int_0^a (\sqrt{a} - \sqrt{t})^2 \, dt$$ Expand: $$a - 2\sqrt{a}\sqrt{t} + t$$ Rewrite: $$a - 2\sqrt{a} t^{1/2} + t$$ Integrate termwise: $$\int_0^a a \, dt = a t \big|_0^a = a^2$$ $$\int_0^a -2\sqrt{a} t^{1/2} dt = -2\sqrt{a} \cdot \frac{2}{3} t^{3/2} \big|_0^a = -\frac{4}{3} a^{1/2} a^{3/2} = -\frac{4}{3} a^2$$ $$\int_0^a t \, dt = \frac{t^2}{2} \big|_0^a = \frac{a^2}{2}$$ Sum: $$a^2 - \frac{4}{3} a^2 + \frac{a^2}{2} = a^2 \left(1 - \frac{4}{3} + \frac{1}{2}\right) = a^2 \left(\frac{6}{6} - \frac{8}{6} + \frac{3}{6}\right) = a^2 \cdot \frac{1}{6} = \frac{a^2}{6}$$ 4. Calculate $$\int_2^5 (5x - 2)(7x + 5) \, dx$$ Expand: $$35x^2 + 25x - 14x - 10 = 35x^2 + 11x - 10$$ Integrate: $$\int 35x^2 + 11x - 10 \, dx = \frac{35x^3}{3} + \frac{11x^2}{2} - 10x$$ Evaluate from 2 to 5: $$\left(\frac{35(5)^3}{3} + \frac{11(5)^2}{2} - 10(5)\right) - \left(\frac{35(2)^3}{3} + \frac{11(2)^2}{2} - 10(2)\right)$$ $$= \left(\frac{35 \, 125}{3} + \frac{11 \, 25}{2} - 50\right) - \left(\frac{35 \, 8}{3} + \frac{11 \, 4}{2} - 20\right)$$ $$= \left(1458.33 + 137.5 - 50\right) - \left(93.33 + 22 - 20\right) = 1545.83 - 95.33 = 1450.5$$ 5. Calculate $$\int_1^3 \frac{m^2 - 2}{m^4} \, dm = \int_1^3 (m^{-2} - 2m^{-4}) \, dm$$ Integrate: $$\int m^{-2} dm = -m^{-1}$$ $$\int -2m^{-4} dm = 2/3 m^{-3}$$ Sum: $$-m^{-1} + \frac{2}{3} m^{-3}$$ Evaluate from 1 to 3: $$\left(-\frac{1}{3} + \frac{2}{3 \cdot 27}\right) - \left(-1 + \frac{2}{3}\right) = \left(-\frac{1}{3} + \frac{2}{81}\right) - \left(-\frac{1}{3}\right) = \left(-0.3333 + 0.0247\right) + 0.3333 = 0.0247$$ 6. Calculate $$\int_{-\frac{1}{2}}^{\frac{1}{2}} (3x + 1)^7 \, dx$$ Substitute $$u = 3x + 1$$, $$du = 3 dx$$, $$dx = \frac{du}{3}$$ Change limits: When $$x = -\frac{1}{2}$$, $$u = 3(-\frac{1}{2}) + 1 = -\frac{3}{2} + 1 = -\frac{1}{2}$$ When $$x = \frac{1}{2}$$, $$u = 3(\frac{1}{2}) + 1 = \frac{3}{2} + 1 = \frac{5}{2}$$ Integral becomes: $$\int_{-\frac{1}{2}}^{\frac{5}{2}} u^7 \frac{du}{3} = \frac{1}{3} \int_{-\frac{1}{2}}^{\frac{5}{2}} u^7 du = \frac{1}{3} \left[ \frac{u^8}{8} \right]_{-\frac{1}{2}}^{\frac{5}{2}} = \frac{1}{24} \left( \left(\frac{5}{2}\right)^8 - \left(-\frac{1}{2}\right)^8 \right)$$ Calculate powers: $$\left(\frac{5}{2}\right)^8 = \frac{5^8}{2^8} = \frac{390625}{256}$$ $$\left(-\frac{1}{2}\right)^8 = \frac{1}{256}$$ Difference: $$\frac{390625}{256} - \frac{1}{256} = \frac{390624}{256}$$ Final: $$\frac{1}{24} \times \frac{390624}{256} = \frac{390624}{6144} = 63.5$$ 7. Calculate $$\int_0^2 y^2 \sqrt{1 + y^3} \, dy$$ Substitute $$u = 1 + y^3$$, $$du = 3y^2 dy$$, so $$y^2 dy = \frac{du}{3}$$ Change limits: When $$y=0$$, $$u=1$$ When $$y=2$$, $$u=1 + 8 = 9$$ Integral becomes: $$\int_1^9 \sqrt{u} \frac{du}{3} = \frac{1}{3} \int_1^9 u^{1/2} du = \frac{1}{3} \left[ \frac{2}{3} u^{3/2} \right]_1^9 = \frac{2}{9} (9^{3/2} - 1^{3/2})$$ Calculate powers: $$9^{3/2} = (\sqrt{9})^3 = 3^3 = 27$$ Final: $$\frac{2}{9} (27 - 1) = \frac{2}{9} \times 26 = \frac{52}{9}$$