Question

2) Calcule a seguinte integral definida:

$\int\limits^2_1 { \frac{v^3+3v^5}{v^4} } \, dv$

Answer 1

Temos o seguinte:

$\int\limits^2_1 { \frac{v^3 + 3v^5}{v^4} } \, dv$

Assim:

$\int { \frac{v^3 + 3v^5}{v^4} } \, dv = \int { \frac{v^3}{v^4} + \frac{3v^5}{v^4} } \, dv = \int { \frac{1}{v} + 3v } \, dv = \int { \frac{1}{v} } \, dv + \int { 3v } \, dv \\ \\ \int { \frac{1}{v} } \, dv + \int { 3v } \, dv = ln(v) + 3 \cdot \frac{v^2}{2} + C$

Aplicando os limites:

$(ln(v) + 3 \cdot \frac{v^2}{2} )\limits^2_1 \\ \\ = (ln(2) + 3 \cdot \frac{2^2}{2}) - (ln(1) + 3 \cdot \frac{1^2}{2} ) \\ \\ = (ln(2) + 6) - (0 + \frac{3}{2} ) \\ \\ = ln(2) + 6 - \frac{3}{2} \\ \\ = ln(2) + \frac{9}{2}$

Answer 2

$\displaystyle\mathsf{\int\limits_{1}^{2}\dfrac{{v}^{3}+3{v}^{5}}{{v}^{4}}dv}\\\displaystyle\mathsf{\int\limits_{1}^{2}(\dfrac{1}{v}+3v)dv =\left[Ln|v|+\dfrac{3}{2}{v}^{2}\right]_{1}^{2} }$

$\mathsf{Ln(2)+\dfrac{3}{2}.{2}^{2}-(Ln(1)+\dfrac{3}{2}.{1}^{2})}\\=\mathsf{Ln(2)+\dfrac{12}{2}-\dfrac{3}{2}=Ln(2)+\dfrac{9}{2}}$