Question

Срочно помогите с интервалами!!!!
Заранее спасибо

Answer 1

1. $\int\limits^2_{-1} \, dx = x|\frac{2}{-1} = 2+1=3$

2. $\int\limits^4_1 {(3-2x)} \, dx = 3x-2\frac{x^2}{2} |\frac{4}{1} = 12-16-3+1=-6$

3. $\int\limits^1_{-1} {x^2} \, dx = \frac{x^3}{3} |\frac{1}{-1} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

4. $\int\limits^{\frac{\pi }{2} }_{\frac{\pi }{2} } {cosx} \, dx = sinx\frac{\frac{\pi }{2} }{\frac{\pi }{6} } = sin\frac{\pi }{2} - sin\frac{\pi }{6} = 1-\frac{1}{2} = \frac{1}{2}$

5. $\int\limits^7_2 {\frac{1}{x^2} } \, dx = \frac{1}{x} \frac{7}{2} = \frac{1}{7} - \frac{1}{2} = -\frac{5}{14}$

6. $\int\limits^{\frac{\pi }{4} }_0 {\frac{1}{cos^2x} } \, dx = \frac{1}{cosx} |\frac{\frac{\pi }{4} }{0} = \frac{1}{cos\frac{\pi }{4} } - \frac{1}{cos0} = \frac{2}{\sqrt{2} } -1$

7. $\int\limits^3_1 {2x} \, dx = 2\frac{x^2}{2} |\frac{3}{1} = 9-1=8$

8. $\int\limits^4_1 {(x^2-6x+9)} \, dx = \frac{x^3}{3} -6\frac{x^2}{2} + 9x|\frac{4}{1} = \frac{64}{3} - 48+36-\frac{1}{3} +3-9=3$