Question

Обчислити площу обмежену лініями, у=-2/х, у=х-3

Answer 1

Ответ:

$\int\limits_{1}^{2}(3 - \frac{2}{x} - x)dx$

$f(x) = 3 - \frac{2}{x} - x$

$0 = 3 - \frac{2}{x} - x$

$0 = 3 - \frac{2}{x} - x.x≠0$

$3 - \frac{2}{x} - x = 0$

$\frac{3x - 2 - {x}^{2} }{x} = 0$

$3x - 2 - {x}^{2} = 0$

$- {x}^{2} + 3x - 2 = 0$

${x}^{2} - 3x + 2 = 0$

${x}^{2} - x - 2x + 2 = 0$

$x(x - 1) - 2(x - 1) = 0$

$(x - 1)(x - 2) = 0$

$x - 1 = 0 \\ x - 2 = 0$

$x = 1 \\ x≠0 \\ x = 2$

$x = 1 \\ x = 2$

${x}_{1}=1, {x}_{2}=2$

$A=|\int\limits_{1}^{2}3-\frac{2}{x}-xdx|$

$A = | \int3 - \frac{2}{x} - xdx |$

$A = | \int3dx - \int \frac{2}{x}dx - \int{xdx}|$

$A = |3x - 2ln( |x| ) - \frac{ {x}^{2} }{2} |$

$A = |(3x - 2ln( |x|) - \frac{ {x}^{2} }{2} ) {|}^{2}_{1}|$

$A = |3 \times 2 - 2ln( |2| ) - \frac{ {2}^{2} }{2} - (3 \times 1 - 2ln( |1| ) - \frac{ {1}^{2} }{2} )|$

$A = |6 - 2 ln(2) - 2 - (3 - 2 ln(1) - \frac{1}{2} ) |$

$A = |4 - 2 ln(2) - (3 - 2 \times 0 - \frac{1}{2} ) |$

$A = |4 - 2 ln(2) - (3 - 0 - \frac{1}{2} ) |$

$A = |4 - 2 ln(2) - (3 - \frac{1}{2}) |$

$A = |4 - 2 ln(2) - \frac{6 - 1}{2} |$

$A = |4 - 2 ln(2) - \frac{5}{2} |$

$A = | \frac{8 - 5}{2} - 2 ln(2) |$

$A = | \frac{3}{2} - 2ln(2) |$

$A = \frac{3}{2} - 2ln(2)$

$A≈0.1137$