Question

integrala de la 0 la 1 din xfx=7/4 unde fx=e^x+3x^2​

Answer 1

$f(x) = {e}^{x} + 3 {x}^{2}$

$\int_{0}^{1} xf(x) \: dx= \frac{7}{4}$

$\int_{0}^{1}xf(x) \: dx = \int_{0}^{1}x( {e}^{x} + 3 {x}^{2} ) \: dx = \int_{0}^{1}(x {e}^{x} + 3 {x}^{3} ) \: dx$

$= \int_{0}^{1}x {e}^{x} \: dx + \int_{0}^{1}3 {x}^{3} \: dx = \int_{0}^{1}x {e}^{x} \: dx + 3\int_{0}^{1} {x}^{3} \: dx$

$= \int_{0}^{1}x {e}^{x} \: dx + 3 \times \frac{ {x}^{3 + 1} }{3 + 1} | _{0}^{1} = \int_{0}^{1}x {e}^{x} \: dx + 3 \times \frac{ {x}^{4} }{4} |_{0}^{1}$

$= \int_{0}^{1}x {e}^{x} \: dx + \frac{3 {x}^{4} }{4} |_{0}^{1} = \int_{0}^{1}x {e}^{x} \: dx + \frac{3 \times {1}^{4} }{4} - \frac{3 \times {0}^{4} }{4}$

$= \int_{0}^{1}x {e}^{x} \: dx + \frac{3}{4} = 1 + \frac{3}{4} = \frac{4}{4} + \frac{3}{4} = \frac{4 + 3}{4} = \frac{7}{4}$

$\star)\int_{0}^{1}x {e}^{x} \: dx = {e}^{x} (x - 1) |_{0}^{1} = {e}^{1} (1 - 1) - {e}^{0} (0 - 1) = 0 - 1 \times ( - 1) = 1$

$\int x {e}^{x} \: dx = g(x) \times h(x) - \int h(x) \times g'(x) \: dx$

$= x {e}^{x} - \int {e}^{x} \times 1 \: dx = x {e}^{x} - \int {e}^{x} \: dx = x {e}^{x} - {e}^{x} = {e}^{x} (x - 1)$

$g(x) = x = > g'(x) = x' = 1$

$h'(x) = {e}^{x} = > h(x) = \int h'(x) \: dx = \int {e}^{x} \: dx = {e}^{x}$