Question

Calcule as integrais:

$a)\large \displaystyle \text { \mathsf{ \left \int\limits^4_2 {(x^3-2)} \, dx \right } }$

$b)\large \displaystyle \text { \mathsf{ \left \int\limits^4_1 {3\sqrt{2x} } \, dx \right } }$

$c)\large \displaystyle \text { \mathsf{ \left \int\limits^1_0 {\frac{1}{\sqrt{x} } } \, dx \right } }$

$d)\large \displaystyle \text { \mathsf{ \left \int\limits^\pi _0 {(x+senx)} \, dx \right } }$

$e)\large \displaystyle \text { \mathsf{ \left \int\limits^2_0 {(x^2+5)^3} \, dx \right } }$

Answer 1

Resposta:

a) Encontremos uma primitiva da função $f(x) = x^3-2:$

$F(x) = \int {f(x)} \, dx\\\\ = \int {(x^3-2)} \, dx\\\\ = \frac{1}{4}x^4 -2x + C.$

Assim:

$\int\limits^4_2 {f(x)} \, dx = F(x)|_{_{2}}^{^{4}}\\\\ = (\frac{1}{4}\,.\,4^4 -2(4) + C) - (\frac{1}{4}\,.\,2^4 -2(2)+C)\\\\= (56 + C) - (0 + C)\\\\= \boxed{56}$

b) Encontremos uma primitiva da função $f(x) = 3\sqrt{2x}:$

$F(x) = \int {f(x)} \, dx\\\\= \int {(3\sqrt{2x}) } \, dx\\\\= \int {(3\sqrt{2}\,.\,\sqrt{x} ) } \, dx\\\\= 3\sqrt{2}\,.\,\int {\sqrt{x} } \, dx\\\\= 3\sqrt{2}\,.\,\int {x^{\frac{1}{2}} } \, dx\\\\ = 3\sqrt{2}\,.\,\frac{2}{3} x^{\frac{3}{2}}+ C\\\\ = 2\sqrt{2}x^{\frac{3}{2}}+ C.$

Assim:

$\int\limits^4_1 {f(x)} \, dx = F(x)|_{_{1}}^{^{4}}\\\\= (2\sqrt{2}\,.\,4^{\frac{3}{2}}+ C) - (2\sqrt{2}\,.\,1^{\frac{3}{2}}+ C)\\\\= 16\sqrt{2} - 2\sqrt{2}\\\\= \boxed{14\sqrt{2}}$

c) Encontremos uma primitiva da função $f(x) = \frac{1}{\sqrt{x} }:$

$F(x) = \int {f(x)} \, dx\\\\= \int {\frac{1}{\sqrt{x} } } \, dx\\\\= \int {x^{-\frac{1}{2}} } \, dx\\\\= 2x^{\frac{1}{2}}+ C\\\\ = 2\sqrt{x} + C.$

Perceba, neste caso, que $f(x)$ não é definida para $x = 0.$ Há, portanto, uma descontinuidade. Assim:

$\int\limits^1_0 {f(x)} } \, dx = \lim_{b \to 0^+} \int\limits^1_b {f(x) } } \, dx\\\\ = \lim_{b \to 0^+} F(x)|_{_{b}}^{^{1}}\\\\= \lim_{b \to 0^+}(2\sqrt{1} + C) - (2\sqrt{b} + C)\\\\= 2 - 0\\\\= \boxed{2}$

d) Encontremos uma primitiva da função $f(x) = x + senx:$

$F(x) = \int {f(x)} \, dx\\\\= \int {(x + senx)} \, dx\\\\= \int {x} \, dx + \int {senx} \, dx\\\\ = (\frac{1}{2}x^2 + C_1) + (-cosx + C_2)\\\\= \frac{1}{2}x^2 - cosx + C.$

Assim:

$\int\limits^\pi _0 {f(x)} \, dx = F(x)|_{_{0}}^{^{\pi }}\\\\= (\frac{1}{2}\pi ^2 - cos(\pi ) + C) - (\frac{1}{2}\,.\,0^2 - cos(0) + C)\\\\ = ( \frac{1}{2}\pi ^2 +1 + C) - (-1 + C)\\\\= \boxed{\frac{1}{2}\pi ^2 + 2}$

e) Encontremos uma primitiva da função $f(x) = (x^2 + 5)^3:$

$F(x) = \int {f(x)} \, dx\\\\= \int {(x^2+5)^2} \, dx\\\\= \int {(x^6 + 3x^4\,.\,5 + 3x^2\,.\,25 + 125)} \, dx \\\\= \int {(x^6 + 15x^4 + 75x^2 + 125)} \, dx\\\\ = \frac{1}{7}x^7 + 3x^5 + 25x^3 + 125x + C.$

Assim:

$\int\limits^2 _0 {f(x)} \, dx = F(x)|_{_{0}}^{^{2 }}\\\\= (\frac{1}{7}\,.\,2^7 + 3\,.\,2^5 + 25\,.\,2^3 + 125\,.\,2 + C) - ( \frac{1}{7}\,.\,0^7 + 3\,.\,0^5 + 25\,.\,0^3 + 125\,.\,0 + C)\\\\= \frac{128}{7} + 96 + 200 + 250\\\\= \boxed{\frac{3950}{7}}$