Question

1) Calcule a seguinte integral definida:

$\int\limits^4_1 {{(5-2t+3t^2)}} \, dt$

Answer 1

Temos o seguinte:

$\int\limits^4_1 {5 - 2t+3t^2} \, dt$

Assim:

$\int {5 - 2t+3t^2} \, dt = \int {5} \, dt - \int {2t} \, dt + \int {3t^2} \, dt \\ \\ = 5\int {1} \, dt - 2\int {t} \, dt + 3\int {t^2} \, dt \\ \\ = 5t - 2 \cdot \frac{t^2}{2} + 3 \cdot \frac{t^3}{3} \\ \\ = 5t - t^2 + t^3 + C$

Logo aplicando os limites:

$(5t - t^2 + t^3)\limits^4_1 \\ \\ = (5 \cdot 4 - 4^2 + 4^3) - (5 \cdot 1 - 1^2 + 1^3) \\ \\ = (20 - 16 + 64) - (5 - 1 + 1) \\ \\ = 68 - 5 \\ \\ = 63$

Answer 2

$\displaystyle\mathsf{\int\limits_{1}^{4}(5-2t+3{t} ^{2})dt = \left[5t-{t}^{2}+{t}^{3}\right] _{1}^{4}} \\\mathsf{5.4 - {4}^{2}+ {4}^{3} - (5 - 1 + 1) } \\\mathsf{=20 - 16 + 64 - 5 = 63}$