Question

3) Calcule a seguinte integral definida:

$\int\limits^1_0 {(3+x \sqrt{x} )} \, dx$

Answer 1

Temos o seguinte:

$\int\limits^1_0 { 3 + x \sqrt{x} } \, dx$

Assim:

$\int { 3 + x \sqrt{x} } \, dx = \int { 3 + \sqrt{x^3} } \, dx = \int { 3 + x^{ \frac{3}{2} } } \, dx = \int { 3 } \, dx + \int {x^{ \frac{3}{2} } } \, dx \\ \\ \int { 3 } \, dx + \int {x^{ \frac{3}{2} } } \, dx = 3x + \frac{x^{ \frac{3}{2} + 1 }}{ \frac{3}{2} + 1 } = 3x + \frac{x^{ \frac{5}{2} }}{ \frac{5}{2} } = 3x + \frac{2}{5} \cdot \sqrt{x^5} + C$

Logo aplicando os limites:

$(3x + \frac{2}{5} \cdot \sqrt{x^5} )\limits^1_0 \\ \\ =(3 \cdot 1 + \frac{2}{5}\cdot \sqrt{1^5} ) - (3\cdot 0 + \frac{2}{5} \cdot \sqrt{0^5} ) \\ \\ = (3 + \frac{2}{5} ) - 0 \\ \\ = \frac{17}{5}$

Answer 2

$\mathsf{x\sqrt{x}={x}^{\frac{3}{2}}}$

$\displaystyle\mathsf{\int\limits_{0}^{1}(3 + x\sqrt{x})=\int\limits_{0}^{1}(3+{x}^{\frac{3}{2}})dx} \\ = \mathsf{\left[3x+\dfrac{2}{5}{x}^{\frac{5}{2}} \right]_{0}^{1}}$

$\mathsf{3.1+\dfrac{2}{5}{1}^{\frac{5}{2}}=3+\dfrac{2}{5}=\dfrac{17}{3}}$