Помогите решить примеры по интегралам срочно

Question 1

Question 2

Решите задачу:

$\int\limits_1^2 {\frac{2}{x^2}} \, dx = 2 \int\limits_1^2 {x^{-2}} \, dx = -2x^{-1}|_1^2 = -\frac{2}{x}|_1^2 = -\frac{2}{2}+\frac{2}{1x} = -1+2 = 1;$

$\int\limits_1^2 {\sqrt{x}} \, dx = \int\limits_1^2 {x^\frac{1}{2}} \, dx = \frac{2}{3}x^\frac{3}{2}|_1^2 = \frac{2}{3}\sqrt{x^3}|_1^2 = \frac{2}{3}\sqrt{2^3}-\frac{2}{3}\sqrt{1^3} = \frac{4}{3}\sqrt{2}-\frac{2}{3};$

$\int\limits_{-1}^0 {x-\frac{1}{2}} \, dx = \int\limits_{-1}^0 {x} \, dx - \frac{1}{2}\int\limits_{-1}^0 {} \, dx = \frac{x^2}{2}|_{-1}^0-\frac{x}{2}|_{-1}^0 = \frac{0^2}{2}-\frac{(-1)^2}{2}-\frac{0}{2}+\frac{-1}{2}\\=-\frac{x}{2}-\frac{x}{2}=-1;$

$\int\limits_1^2 {x^{-3}} \, dx = \frac{x^{-2}}{-2}|_1^2 = -\frac{1}{2x^2}|_1^2 = -\frac{1}{2\cdot2^2}+\frac{1}{2\cdot1^2} = -\frac{1}{8}+\frac{1}{2} = \frac{3}{8};$

$\int\limits_{-2}^{-1} {5x^3} \, dx = \frac{5x^4}{4}|_{-2}^{-1} = \frac{5\cdot(-2)^4}{4}-\frac{5\cdot(-1)^4}{4} = \frac{5\cdot(-2)^4}{4}-\frac{5\cdot(-1)^4}{4} = 20- \frac{5}{4}=18,75$