Интегралы вычислить определенные

Question 1

Question 2

кек

Question 3

1. $A = \int\limits^2_{-1}{x-3e^{x}+2} \, dx = \int\limits^2_{-1}{f(x)} \, dx = F(2) - F(-1)\\$
$f(x) = x-3e^{x}+2 \\ F(x) = \int{f(x)} \, dx = \int{x-3e^{x}+2} \, dx = \\$
$= \int{x} \, dx -3 \int{e^{x}} \, dx + \int{2} \, dx = \\$
$\frac{x^2}{2} -3e^x + 2x + C\\$
$A = F(2) - F(-1) = (\frac{2^2}{2} - 3e^2 + 2 \times 2) - (\frac{{-1}^{2}}{2} - 3e^{-1} - 2) = \\$
$= 2 - 3e^2 + 4 - \frac{1}{2} + 3e^{-1} + 2 = 7 \frac{1}{2} +3(e^{-1} - e^{2})$

2. $A = \int\limits^2_{-2}{\frac{(x^2 - 1)^2}{x^2}} \, dx = \int\limits^2_{-2}{f(x)} \, dx = F(2) - F(-2) \\$
$f(x) = \frac{(x^2 - 1)^2}{x^2}\\ F(x) = \int{f(x)} \, dx = \int{\frac{(x^2 - 1)^2}{x^2}} \, dx = \\$
$= \int{\frac{(x^2 - 1)^2}{x^2}} \, dx = \int{\frac{x^4 - 2x^2 +1}{x^2}} \, dx = \\$
$= \int{x^2} \, dx - 2\int{} \, dx + \int{\frac{1}{x^2}} \, dx \\ = \frac{x^3}{3} -2x - \frac{1}{x}+C$
$A = (\frac{2^3}{3} - 2*2 - \frac{1}{2}) - (\frac{(-2)^3}{3} -2*(-2) - \frac{1}{-2}) = \frac{16}{3} - 8 - 1 = \frac{16}{3} - 9$

3. $A = \int\limits^{\pi}_{0} {1 - Cos(6x)} \, dx = \int\limits^{\pi}_{0} {f(x)} \, dx = F(\pi) - F(0) \\$
$f(x) = 1 - Cos(6x)\\ F(x) = \int{1 - Cos(6x)} \, dx = \int{} \, dx - \int{Cos(6x)} \, dx = \\ = x - \frac{Sin(6x)}{6} + C\\$
$A = (\pi - \frac{1}{6} * Sin(6\pi)) - (0 - \frac{1}{6} * Sin(0)) = \pi$

4. Решим путем внесением под знак дифференциала (какой нижний предел??)
$A = \int\limits^{\pi}_{?} {\sqrt{1-cos(x)}sin(x)} \, dx = \int\limits^{\pi}_{?} {f(x)} \, dx = F(\pi) - F(0) \\$
$f(x) = \sqrt{1-cos(x)}sin(x) \\ F(x) = \int {\sqrt{1-cos(x)}sin(x)} \, dx \\$
$sin(x)dx = - d(cos(x))\\$
$F(x) = \int {\sqrt{1-cos(x)}sin(x)} \, dx \\ = - \int {\sqrt{1-cos(x)}} \, d(Cos(x)) = \\$
$\frac{2}{3}(1 - Cos(x))^{\frac{3}{2}} + C = F(x)$
$A = F(\pi) - F(?)$