Question

метод интегрирования по частям

Answer 1

$\int (4 - x) {2}^{x} dx = \int 4 \times {2}^{x} dx - \int x \times {2}^{x} dx = \frac{4 \times {2}^{x} }{ ln(2) } - \int x \times {2}^{x} dx = \frac{4 \times {2}^{x} }{ ln(2) } - I$

$I = \int x \times {2}^{x} dx \\ u = x, \: dv = {2}^{x} dx \\ du = dx, \: v = \int {2}^{x} dx = \frac{ {2}^{x} }{ ln(2) }$

$I = uv - \int vdu = \frac{{2}^{x} x}{ ln(2) } - \int \frac{ {2}^{x} }{ ln(2) } dx = \frac{ {2}^{x} x}{ ln(2) } - \frac{ {2}^{x} }{ ln {}^{2} (2) }$

$\int(4 - x) {2}^{x} dx = \frac{4 \times {2}^{x } }{ ln(2) } - \frac{ {2}^{x}x }{ ln(2) } + \frac{ {2}^{x} }{ ln {}^{2} (2) } = \frac{ {2}^{x} }{ ln(2) } (4 + \frac{1}{ ln(2) } - x) = \frac{ {2}^{x} }{ ln(2) } ( \frac{4 ln(2) + ln(e) }{ ln(2)} - x ) = \frac{ {2}^{x} }{ ln(2) } ( \frac{ ln(16e) }{ ln(2) } - x) = \frac{ {2}^{x} }{ ln(2) } ( log_{2}(16e) - x)$