Question

Aplicação de integral:
Calcule a área sob o gráfico de f:
$f(x)= x^{2} -5x+9, para 1 \leq x \leq 4$

Answer 1

Pela definição de integral, temos:

$A = \int\limits^4_1 {x^2 - 5x + 9} \, dx$

Portanto temos que:

$\int {x^2 - 5x + 9} \, dx = \dfrac{x^3}{3} - \dfrac{5x^2}{2} + 9x + C$

Logo:

$A = \int\limits^4_1 {x^2 - 5x + 9} \, dx \\ \\ \\ A = \left (\dfrac{x^3}{3} - \dfrac{5x^2}{2} + 9x \right )\limits^4_1 \\ \\ \\ A = \left (\dfrac{4^3}{3} - \dfrac{5*4^2}{2} + 9*4 \right ) - \left (\dfrac{1^3}{3} - \dfrac{5*1^2}{2} + 9*1 \right ) \\ \\ \\ A = \left (\dfrac{64}{3} - 40 + 36 \right ) - \left (\dfrac{1}{3} - \dfrac{5}{2} + 9 \right ) \\ \\ \\ A = \dfrac{64}{3} - 40 + 36 - \dfrac{1}{3} + \dfrac{5}{2} - 9 \\ \\ \\ A = \dfrac{63}{3} - 13 + \dfrac{5}{2}$

$A = 21 - 13 + \dfrac{5}{2} \\ \\ \\ A = 8 + \dfrac{5}{2} \\ \\ \\ A= \dfrac{16 + 5}{2} \\ \\ \\ A = \dfrac{21}{2} = 10,5$