Política de uso de dados

Ao navegar por este site, você concorda com a política de uso de dados.

Política de uso de dados

Ao navegar por este site, você concorda com a política de uso de dados.

| | | |

Cálculo I

4 Integração 4.2 Propriedades de integração 4.4 Integração por substituição

Ajude a manter o site livre, gratuito e sem propagandas. Colabore!

4.3 Regras Básicas de Integração

A integral indefinida de uma função $f$ em relação a $x$ é

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int f(x)\,dx=% F(x)+C,

(4.116)

onde $F$ é uma primitiva de $f$ , i.e. $F^{\prime}=f$ , e $C$ é uma constante indeterminada. Na sequência, vamos estudar as regras básicas para o cálculo de integrais.

4.3.1 Integral de Função Potência

Com base na derivada de função potência, podemos afirmar que

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int x^{r}\,dx% =\frac{x^{r+1}}{r+1}+C,\leavevmode\nobreak\ r\neq-1.

(4.117)

De fato, para $r\neq-1$ , temos

	$\displaystyle F(x)=\frac{x^{r+1}}{r+1}+C,$		(4.118)
	$\displaystyle F^{\prime}(x)=(r+1)\frac{x^{r}}{r+1}=x^{r}.$		(4.119)

Exemplo 4.3.1.

Estudamos os seguintes casos:

a)

$\int x\,dx=\frac{x^{2}}{2}+C.$ (4.120)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(x, x)
```
x**2/2
```
b)

$\displaystyle\int\frac{1}{x^{2}}\,dx=\int x^{\color[rgb]{0,0,1}\definecolor[% named]{pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}-2}\,dx$ (4.121)

$\displaystyle\text{}\qquad=\frac{x^{{\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}-2}+1}}{{\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}-2}+1}+C=-\frac{1}{x}+C.$ (4.122)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(1/x**2, x)
```
-1/x
```

Exemplo 4.3.2.

Vamos calcular

\int_{-1}^{1}x^{2}\,dx.

(4.123)

Da regra da potência (4.117), temos

\int x^{2}\,dx=\frac{x^{3}}{3}+C.

(4.124)

Logo, do teorema fundamental do cálculo, temos

	$\displaystyle\int_{-1}^{1}x^{2}\,dx=\left.\frac{x^{3}}{3}\right\|_{-1}^{1}$		(4.125)
	$\displaystyle\text{}\qquad=\frac{1^{3}}{3}-\frac{(-1)^{3}}{3}$		(4.126)
	$\displaystyle\text{}\qquad=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}.$		(4.127)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(x**2, (x, -1, 1))

2/3

4.3.2 Regra da Multiplicação por Constante

Seja $k$ uma constante. Então, temos a seguinte regra da multiplicação por constante

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int k\cdot f(% x)\,dx=k\cdot\int f(x)\,dx

(4.128)

De fato, se $F$ é uma primitiva de $f$ , então pela regra da multiplicação por constante para derivadas, temos

	$\displaystyle(k\cdot F)^{\prime}=k\cdot F^{\prime}$		(4.129)
	$\displaystyle\text{}\qquad=k\cdot f,$		(4.130)

i.e. $k\cdot F$ é primitiva de $k\cdot f$ .

Exemplo 4.3.3.

Estudamos os seguintes casos:

a)

$\displaystyle\int 2x\,dx=2\int x\,dx$ (4.131)

$\displaystyle\text{}\qquad=2\left(\frac{x^{2}}{2}+C\right)$ (4.132)

$\displaystyle\text{}\qquad=x^{2}+2C$ (4.133)

$\displaystyle\text{}\qquad=x^{2}+C.$ (4.134)

Aqui, fizemos um abuso de linguagem ao assumir $2C=C$ . Isso pode ser feito, pois $C$ denota uma constante indeterminada e, multiplicá-la por dois continua sendo indeterminada e constante. Vamos fazer este tipo de simplificação de notação várias vezes ao longo do texto.

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(2*x, x)
```
x**2
```
b)

$\displaystyle\int\frac{1}{3}\sqrt{x}\,dx=\frac{1}{3}\int x^{\frac{1}{2}}\,dx$ (4.135)

$\displaystyle\text{}\qquad=\frac{1}{3}\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C$ (4.136)

$\displaystyle\text{}\qquad=\frac{1}{3}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+C$ (4.137)

$\displaystyle\text{}\qquad=\frac{2}{9}\sqrt{x^{3}}+C.$ (4.138)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(1/3*sym.sqrt(x), x)
```
0.222222222222222*x**(3/2)
```
c)

$\displaystyle\int_{0}^{1}-x^{2}\,dx=-\int_{0}^{1}x^{2}\,dx$ (4.139)

$\displaystyle\text{}\qquad=-\left[\frac{x^{3}}{3}\right]_{0}^{1}$ (4.140)

$\displaystyle\text{}\qquad=-\left(\frac{1}{3}-\frac{0}{3}\right)$ (4.141)

$\displaystyle\text{}\qquad=-\frac{1}{3}.$ (4.142)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(-x**2, (x, 0, 1))
```
-1/3
```

4.3.3 Regra da soma ou subtração

Se $f$ e $g$ são funções integráveis, então vale a seguinte regra da soma/subtração

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int\left[f(x)% \pm g(x)\right]\,dx=\int f(x)\,dx\pm\int g(x)\,dx.

(4.143)

De fato, sejam $F$ uma primitiva de $f$ e $G$ uma primitiva de $g$ . Temos

	$\displaystyle(F\pm G)^{\prime}=F^{\prime}\pm G^{\prime}$		(4.144)
	$\displaystyle\text{}\qquad=f\pm g,$		(4.145)

i.e. $F\pm G$ é primitiva de $f\pm g$ .

Exemplo 4.3.4.

Estudamos os seguintes casos:

a)

$\displaystyle\int{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{% 0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}x}+{% \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}1}\,dx={\color% [rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int x\,dx}+{% \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}\int dx}$ (4.146)

$\displaystyle\text{}\qquad={\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}\frac{x^{2}}{2}+C_{1}}+{\color[rgb]{1,0,0}% \definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\pgfsys@color@rgb@stroke{1}{0}{% 0}\pgfsys@color@rgb@fill{1}{0}{0}x+C_{2}}$ (4.147)

$\displaystyle\text{}\qquad=\frac{x^{2}}{2}+x+C.$ (4.148)

Aqui, $C_{1}$ , $C_{2}$ e $C=C_{1}+C_{2}$ denotam constantes indeterminadas.

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(x+1, x)
```
x**2/2 + x
```
b)

$\displaystyle\int{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{% 0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\sqrt{x}% }-{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}x}\,dx={\color% [rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int x^{\frac{% 1}{2}}\,dx}-{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}\int x\,dx}$ (4.149)

$\displaystyle\text{}\qquad={\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}\frac{2}{3}x^{\frac{3}{2}}}-{\color[rgb]{1,0,0}% \definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\pgfsys@color@rgb@stroke{1}{0}{% 0}\pgfsys@color@rgb@fill{1}{0}{0}\frac{x^{2}}{2}}+C.$ (4.150)

⬇

1import sympy as sym

2from sympy.abc import x

3f = sym.sqrt(x) - x

4sym.integrate(f, x)
```
2*x**(3/2)/3 - x**2/2
```

	$\displaystyle\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int(2x^{2}+3x% -1)\,dx=\int\left[2x^{2}+(3x-1)\right]\,dx$		(4.151)
	$\displaystyle\text{}\quad=\int 2x^{2}\,dx+\int 3x-1\,dx$		(4.152)
	$\displaystyle\text{}\quad=\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor% }{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}% \int 2x^{2}\,dx-\int 3x\,dx-\int\,dx$		(4.153)
	$\displaystyle\text{}\quad=2\int x^{2}\,dx+3\int x\,dx-\int\,dx$		(4.154)
	$\displaystyle\text{}\quad=\frac{2}{3}x^{3}+\frac{3}{2}x^{2}-x+C.$		(4.155)

⬇

1import sympy as sym

2from sympy.abc import x

3f = 2*x**2 + 3*x - 1

4sym.integrate(f, x)

2*x**3/3 + 3*x**2/2 - x

Exemplo 4.3.5.

Vamos calcular

\int_{0}^{1}x^{2}+1\,dx.

(4.156)

Temos

	$\displaystyle\int x^{2}+1\,dx=\int x^{2}\,dx+\int\,dx$		(4.157)
	$\displaystyle\text{}\qquad=\frac{x^{3}}{3}+x+C.$		(4.158)

Agora, do teorema fundamental do cálculo, temos

	$\displaystyle\int_{0}^{1}x^{2}+1\,dx=\left.\frac{x^{3}}{3}+x\right\|_{0}^{1}$		(4.159)
	$\displaystyle\text{}\qquad=\left(\frac{1}{3}+1\right)-\left(\frac{0^{3}}{3}+0\right)$		(4.160)
	$\displaystyle\text{}\qquad=\frac{4}{3}.$		(4.161)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(x**2 + 1, (x, 0, 1))

4/3

4.3.4 Integral de $x^{-1}$

Começamos lembrando que

\frac{d}{dx}\ln x=\frac{1}{x},\leavevmode\nobreak\ x>0.

(4.162)

Para $x<0$ , usamos a regra da cadeia

	$\displaystyle\frac{d}{dx}\ln(-x)=\frac{1}{-x}\cdot(-x)^{\prime}$		(4.163)
	$\displaystyle\text{}\qquad=-\frac{1}{x}\cdot(-1)$		(4.164)
	$\displaystyle\text{}\qquad=\frac{1}{x}.$		(4.165)

Ou seja, temos que

\frac{d}{dx}\ln|x|=\frac{1}{x},

(4.166)

donde concluímos que a integral de $x^{-1}$ é

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int\frac{1}{x% }\,dx=\ln|x|+C.

(4.167)

Exemplo 4.3.6.

	$\displaystyle\int_{1}^{e}x^{-1}\,dx=\int_{1}^{e}\frac{1}{x}\,dx$		(4.168)
	$\displaystyle\text{}\qquad=\left[\ln\|x\|\right]_{1}^{e}$		(4.169)
	$\displaystyle\text{}\qquad=\ln\|e\|-\ln\|1\|$		(4.170)
	$\displaystyle\text{}\qquad=1-0=1.$		(4.171)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(1/x, (x, 1, sym.E))

4.3.5 Integral da Função Exponencial Natural

Da derivada da função exponencial natural, temos

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int e^{x}\,dx% =e^{x}+C.

(4.172)

Exemplo 4.3.7.

Vamos estudar os seguintes casos:

a)

$\displaystyle\int e^{2+x}\,dx=\int e^{2}e^{x}\,dx$ (4.173)

$\displaystyle\text{}\qquad=e^{2}\int e^{x}\,dx$ (4.174)

$\displaystyle\text{}\qquad=e^{2}e^{x}+C$ (4.175)

$\displaystyle\text{}\qquad=e^{2+x}+C.$ (4.176)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(sym.exp(2+x), x)
```
exp(x + 2)
```
b)

$\displaystyle\int_{0}^{\ln 2}e^{x}\,dx=\left.e^{x}\right|_{0}^{\ln 2}$ (4.177)

$\displaystyle\text{}\qquad=e^{\ln 2}-e^{0}$ (4.178)

$\displaystyle\text{}\qquad=2-1$ (4.179)

$\displaystyle\text{}\qquad=1.$ (4.180)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(sym.exp(x), (x, 0, sym.log(2)))
```
1
```

4.3.6 Integrais de Funções Trigonométricas

No que lembramos que

\frac{d}{dx}\cos(x)=-\operatorname{sen}(x)

(4.181)

temos que a integral da função seno é

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int% \operatorname{sen}(x)\,dx=-\cos(x)+C.

(4.182)

Exemplo 4.3.8.

Estudamos os seguintes casos:

a)

$\displaystyle\int 2\operatorname{sen}(x)\,dx=2\int\operatorname{sen}(x)dx$ (4.183)

$\displaystyle\text{}\qquad=-2\cos(x)+C$ (4.184)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(2*sym.sin(x), x)
```
-2*cos(x)
```
b)

$\displaystyle\int_{-\pi}^{\pi}\operatorname{sen}(x)\,dx=\left.-\cos(x)\right|_% {-\pi}^{\pi}$ (4.185)

$\displaystyle\text{}\qquad=-\cos(\pi)-[-\cos(-\pi)]$ (4.186)

$\displaystyle\text{}\qquad=1-1=0$ (4.187)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(sym.sin(x), (x, -sym.pi, sym.pi))
```
0
```

Também, lembramos que

\frac{d}{dx}\operatorname{sen}(x)=\cos(x),

(4.188)

donde temos que a integral da função cosseno

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int\cos(x)\,% dx=\operatorname{sen}(x)+C.

(4.189)

Exemplo 4.3.9.

Estudamos os seguintes casos:

a)

$\displaystyle\int\frac{1}{2}\cos(x)\,dx=\frac{1}{2}\int\cos(x)\,dx$ (4.190)

$\displaystyle\text{}\qquad=\frac{1}{2}\operatorname{sen}(x)+C.$ (4.191)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(1/2*sym.cos(x), x)
```
0.5*sin(x)
```
b)

$\displaystyle\int_{-\pi}^{\pi}\cos(x)\,dx=\left.\operatorname{sen}(x)\right|_{% -\pi}^{\pi}$ (4.193)

$\displaystyle\text{}\qquad=\operatorname{sen}(\pi)-\operatorname{sen}(-\pi)$ (4.194)

$\displaystyle\text{}\qquad=0.$ (4.195)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(sym.cos(x), (x, -sym.pi, sym.pi))
```
0
```

4.3.7 Tabela de Integrais

	$\displaystyle\int k\cdot f(x)\,dx=k\cdot\int f(x)\,dx$		(4.196)
	$\displaystyle\int\left[f(x)\pm g(x)\right]\,dx=\int f(x)\,dx\pm\int g(x)\,dx$		(4.197)
	$\displaystyle\int x^{r}\,dx=\frac{x^{r+1}}{r+1}+C,\quad r\neq-1$		(4.198)
	$\displaystyle\int\frac{1}{x}\,dx=\ln x+C$		(4.199)
	$\displaystyle\int e^{x}\,dx=e^{x}+C$		(4.200)
	$\displaystyle\int\operatorname{sen}(x)\,dx=-\cos(x)+C$		(4.201)
	$\displaystyle\int\cos(x)\,dx=\operatorname{sen}(x)+C$		(4.202)

4.3.8 Exercícios resolvidos

ER 4.3.1.

Calcule

\int\frac{x^{2}+2x}{\sqrt{x}}\,dx

(4.203)

Solução.

	$\displaystyle\int\frac{x^{2}+2x}{\sqrt{x}}\,dx=\int\frac{x^{2}}{\sqrt{x}}+2% \frac{x}{\sqrt{x}}\,dx$		(4.204)
	$\displaystyle\text{}\qquad=\int\frac{x^{2}}{x^{\frac{1}{2}}}+2\frac{x}{x^{% \frac{1}{2}}}\,dx$		(4.205)
	$\displaystyle\text{}\qquad=\int x^{2-\frac{1}{2}}+2x^{1-\frac{1}{2}}\,dx$		(4.206)
	$\displaystyle\text{}\qquad=\int x^{\frac{3}{2}}\,dx+2\int x^{\frac{1}{2}}\,dx.$		(4.207)

Agora, usando a regra da função potência (4.117), obtemos

	$\displaystyle\int\frac{x^{2}+2x}{\sqrt{x}}\,dx=\frac{x^{\frac{3}{2}+1}}{\frac{% 3}{2}+1}+2\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C$		(4.208)
	$\displaystyle\text{}\qquad=\frac{2}{5}x^{\frac{5}{2}}+\frac{4}{3}x^{\frac{3}{2% }}+C.$		(4.209)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate((x**2+2*x)/(sqrt(x)))

2*x**(5/2)/5 + 4*x**(3/2)/3

ER 4.3.2.

Calcule

\int_{1}^{e}\frac{1}{2x}\,dx.

(4.210)

Solução.

Das regras básicas de integração, temos

	$\displaystyle\int\frac{1}{2x}\,dx=\int\frac{1}{2}\cdot\frac{1}{x}\,dx$		(4.211)
	$\displaystyle\text{}\qquad=\frac{1}{2}\int\frac{1}{x}\,dx$		(4.212)
	$\displaystyle\text{}\qquad=\frac{1}{2}\ln(x)+C$		(4.213)
	$\displaystyle\text{}\qquad=\ln\sqrt{x}+C.$		(4.214)

Então, do teorema fundamental do cálculo, temos

	$\displaystyle\int_{1}^{e}\frac{1}{2x}\,dx=\left.\ln\sqrt{x}\right\|_{1}^{e}$		(4.215)
	$\displaystyle\text{}\qquad=\ln(\sqrt{e})-\ln(1)$		(4.216)
	$\displaystyle\text{}\qquad=\frac{1}{2}.$		(4.217)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(1/(2*x), (x, 1, sym.E))

1/2

4.3.9 Exercícios

E. 4.3.1.

Calcule

a)

$\displaystyle\int\,dx$
b)

$\displaystyle\int x^{-2}\,dx$
c)

$\displaystyle\int\sqrt{x}\,dx$
d)

$\displaystyle\int\frac{1}{\sqrt{x}}\,dx$

Resposta.

a) $\displaystyle x+C$ ; b) $\displaystyle-\frac{1}{x}+C$ ; c) $\displaystyle\frac{2}{3}x^{3/2}+C$ ; d) $\displaystyle 2x^{1/2}+C$

E. 4.3.2.

Calcule

a)

$\displaystyle\int 1+x^{-2}\,dx$
b)

$\displaystyle\int x-\frac{1}{x}\,dx$
c)

$\displaystyle\int 2x^{3}-3x^{2}+1\,dx$

Resposta.

a) $\displaystyle x-\frac{1}{x}+C$ ; b) $\displaystyle\frac{x^{2}}{2}-\ln|x|+C$ ; c) $\displaystyle\frac{1}{2}x^{4}-x^{3}+x+C$

E. 4.3.3.

Calcule

a)

$\displaystyle\int 2\cos(x)\,dx$
b)

$\displaystyle\int 1-\operatorname{sen}(x)\,dx$

Resposta.

a) $\displaystyle 2\operatorname{sen}(x)+C$ ; b) $\displaystyle x+\cos(x)+C$

E. 4.3.4.

Calcule

a)

$\displaystyle\int_{-1}^{1}x^{3}\,dx$
b)

$\displaystyle\int_{e}^{2e}x^{-1}\,dx$

Resposta.

a) $0$ ; b) $\ln(2)$ ;

E. 4.3.5.

Calcule

1.

$\displaystyle\int_{0}^{1}x^{2}-2x^{3}\,dx$
2.

$\displaystyle\int_{1}^{2}\frac{x+1}{\sqrt{x}}\,dx$

Resposta.

a) $\displaystyle-\frac{1}{6}$ ; b) $\displaystyle\frac{10}{3}\sqrt{2}-\frac{8}{3}$ ;

E. 4.3.6.

Cálculo

a)

$\displaystyle\int_{0}^{\frac{\pi}{2}}\operatorname{sen}(x)\,dx$
b)

$\displaystyle\int_{0}^{\frac{\pi}{2}}\cos(x)\,dx$
c)

$\displaystyle\int_{0}^{\pi}\cos(x)-\operatorname{sen}(x)\,dx$

Resposta.

a) $1$ ; b) $1$ ; c) $-2$

Envie seu comentário

Aproveito para agradecer a todas/os que de forma assídua ou esporádica contribuem enviando correções, sugestões e críticas!

Este texto é disponibilizado nos termos da Licença Creative Commons Atribuição-CompartilhaIgual 4.0 Internacional. Ícones e elementos gráficos podem estar sujeitos a condições adicionais.

Cálculo I

4 Integração 4.2 Propriedades de integração 4.4 Integração por substituição

Ajude a manter o site livre, gratuito e sem propagandas. Colabore!

4.3 Regras Básicas de Integração

A integral indefinida de uma função $f$ em relação a $x$ é

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int f(x)\,dx=% F(x)+C,

(4.116)

onde $F$ é uma primitiva de $f$ , i.e. $F^{\prime}=f$ , e $C$ é uma constante indeterminada. Na sequência, vamos estudar as regras básicas para o cálculo de integrais.

4.3.1 Integral de Função Potência

Com base na derivada de função potência, podemos afirmar que

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int x^{r}\,dx% =\frac{x^{r+1}}{r+1}+C,\leavevmode\nobreak\ r\neq-1.

(4.117)

De fato, para $r\neq-1$ , temos

	$\displaystyle F(x)=\frac{x^{r+1}}{r+1}+C,$		(4.118)
	$\displaystyle F^{\prime}(x)=(r+1)\frac{x^{r}}{r+1}=x^{r}.$		(4.119)

Exemplo 4.3.1.

Estudamos os seguintes casos:

a)

$\int x\,dx=\frac{x^{2}}{2}+C.$ (4.120)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(x, x)
```
x**2/2
```
b)

$\displaystyle\int\frac{1}{x^{2}}\,dx=\int x^{\color[rgb]{0,0,1}\definecolor[% named]{pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}-2}\,dx$ (4.121)

$\displaystyle\text{}\qquad=\frac{x^{{\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}-2}+1}}{{\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}-2}+1}+C=-\frac{1}{x}+C.$ (4.122)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(1/x**2, x)
```
-1/x
```

Exemplo 4.3.2.

Vamos calcular

\int_{-1}^{1}x^{2}\,dx.

(4.123)

Da regra da potência (4.117), temos

\int x^{2}\,dx=\frac{x^{3}}{3}+C.

(4.124)

Logo, do teorema fundamental do cálculo, temos

	$\displaystyle\int_{-1}^{1}x^{2}\,dx=\left.\frac{x^{3}}{3}\right\|_{-1}^{1}$		(4.125)
	$\displaystyle\text{}\qquad=\frac{1^{3}}{3}-\frac{(-1)^{3}}{3}$		(4.126)
	$\displaystyle\text{}\qquad=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}.$		(4.127)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(x**2, (x, -1, 1))

2/3

4.3.2 Regra da Multiplicação por Constante

Seja $k$ uma constante. Então, temos a seguinte regra da multiplicação por constante

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int k\cdot f(% x)\,dx=k\cdot\int f(x)\,dx

(4.128)

De fato, se $F$ é uma primitiva de $f$ , então pela regra da multiplicação por constante para derivadas, temos

	$\displaystyle(k\cdot F)^{\prime}=k\cdot F^{\prime}$		(4.129)
	$\displaystyle\text{}\qquad=k\cdot f,$		(4.130)

i.e. $k\cdot F$ é primitiva de $k\cdot f$ .

Exemplo 4.3.3.

Estudamos os seguintes casos:

a)

$\displaystyle\int 2x\,dx=2\int x\,dx$ (4.131)

$\displaystyle\text{}\qquad=2\left(\frac{x^{2}}{2}+C\right)$ (4.132)

$\displaystyle\text{}\qquad=x^{2}+2C$ (4.133)

$\displaystyle\text{}\qquad=x^{2}+C.$ (4.134)

Aqui, fizemos um abuso de linguagem ao assumir $2C=C$ . Isso pode ser feito, pois $C$ denota uma constante indeterminada e, multiplicá-la por dois continua sendo indeterminada e constante. Vamos fazer este tipo de simplificação de notação várias vezes ao longo do texto.

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(2*x, x)
```
x**2
```
b)

$\displaystyle\int\frac{1}{3}\sqrt{x}\,dx=\frac{1}{3}\int x^{\frac{1}{2}}\,dx$ (4.135)

$\displaystyle\text{}\qquad=\frac{1}{3}\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C$ (4.136)

$\displaystyle\text{}\qquad=\frac{1}{3}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+C$ (4.137)

$\displaystyle\text{}\qquad=\frac{2}{9}\sqrt{x^{3}}+C.$ (4.138)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(1/3*sym.sqrt(x), x)
```
0.222222222222222*x**(3/2)
```
c)

$\displaystyle\int_{0}^{1}-x^{2}\,dx=-\int_{0}^{1}x^{2}\,dx$ (4.139)

$\displaystyle\text{}\qquad=-\left[\frac{x^{3}}{3}\right]_{0}^{1}$ (4.140)

$\displaystyle\text{}\qquad=-\left(\frac{1}{3}-\frac{0}{3}\right)$ (4.141)

$\displaystyle\text{}\qquad=-\frac{1}{3}.$ (4.142)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(-x**2, (x, 0, 1))
```
-1/3
```

4.3.3 Regra da soma ou subtração

Se $f$ e $g$ são funções integráveis, então vale a seguinte regra da soma/subtração

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int\left[f(x)% \pm g(x)\right]\,dx=\int f(x)\,dx\pm\int g(x)\,dx.

(4.143)

De fato, sejam $F$ uma primitiva de $f$ e $G$ uma primitiva de $g$ . Temos

	$\displaystyle(F\pm G)^{\prime}=F^{\prime}\pm G^{\prime}$		(4.144)
	$\displaystyle\text{}\qquad=f\pm g,$		(4.145)

i.e. $F\pm G$ é primitiva de $f\pm g$ .

Exemplo 4.3.4.

Estudamos os seguintes casos:

a)

$\displaystyle\int{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{% 0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}x}+{% \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}1}\,dx={\color% [rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int x\,dx}+{% \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}\int dx}$ (4.146)

$\displaystyle\text{}\qquad={\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}\frac{x^{2}}{2}+C_{1}}+{\color[rgb]{1,0,0}% \definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\pgfsys@color@rgb@stroke{1}{0}{% 0}\pgfsys@color@rgb@fill{1}{0}{0}x+C_{2}}$ (4.147)

$\displaystyle\text{}\qquad=\frac{x^{2}}{2}+x+C.$ (4.148)

Aqui, $C_{1}$ , $C_{2}$ e $C=C_{1}+C_{2}$ denotam constantes indeterminadas.

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(x+1, x)
```
x**2/2 + x
```
b)

$\displaystyle\int{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{% 0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\sqrt{x}% }-{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}x}\,dx={\color% [rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int x^{\frac{% 1}{2}}\,dx}-{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}\int x\,dx}$ (4.149)

$\displaystyle\text{}\qquad={\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}\frac{2}{3}x^{\frac{3}{2}}}-{\color[rgb]{1,0,0}% \definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\pgfsys@color@rgb@stroke{1}{0}{% 0}\pgfsys@color@rgb@fill{1}{0}{0}\frac{x^{2}}{2}}+C.$ (4.150)

⬇

1import sympy as sym

2from sympy.abc import x

3f = sym.sqrt(x) - x

4sym.integrate(f, x)
```
2*x**(3/2)/3 - x**2/2
```

	$\displaystyle\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int(2x^{2}+3x% -1)\,dx=\int\left[2x^{2}+(3x-1)\right]\,dx$		(4.151)
	$\displaystyle\text{}\quad=\int 2x^{2}\,dx+\int 3x-1\,dx$		(4.152)
	$\displaystyle\text{}\quad=\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor% }{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}% \int 2x^{2}\,dx-\int 3x\,dx-\int\,dx$		(4.153)
	$\displaystyle\text{}\quad=2\int x^{2}\,dx+3\int x\,dx-\int\,dx$		(4.154)
	$\displaystyle\text{}\quad=\frac{2}{3}x^{3}+\frac{3}{2}x^{2}-x+C.$		(4.155)

⬇

1import sympy as sym

2from sympy.abc import x

3f = 2*x**2 + 3*x - 1

4sym.integrate(f, x)

2*x**3/3 + 3*x**2/2 - x

Exemplo 4.3.5.

Vamos calcular

\int_{0}^{1}x^{2}+1\,dx.

(4.156)

Temos

	$\displaystyle\int x^{2}+1\,dx=\int x^{2}\,dx+\int\,dx$		(4.157)
	$\displaystyle\text{}\qquad=\frac{x^{3}}{3}+x+C.$		(4.158)

Agora, do teorema fundamental do cálculo, temos

	$\displaystyle\int_{0}^{1}x^{2}+1\,dx=\left.\frac{x^{3}}{3}+x\right\|_{0}^{1}$		(4.159)
	$\displaystyle\text{}\qquad=\left(\frac{1}{3}+1\right)-\left(\frac{0^{3}}{3}+0\right)$		(4.160)
	$\displaystyle\text{}\qquad=\frac{4}{3}.$		(4.161)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(x**2 + 1, (x, 0, 1))

4/3

4.3.4 Integral de $x^{-1}$

Começamos lembrando que

\frac{d}{dx}\ln x=\frac{1}{x},\leavevmode\nobreak\ x>0.

(4.162)

Para $x<0$ , usamos a regra da cadeia

	$\displaystyle\frac{d}{dx}\ln(-x)=\frac{1}{-x}\cdot(-x)^{\prime}$		(4.163)
	$\displaystyle\text{}\qquad=-\frac{1}{x}\cdot(-1)$		(4.164)
	$\displaystyle\text{}\qquad=\frac{1}{x}.$		(4.165)

Ou seja, temos que

\frac{d}{dx}\ln|x|=\frac{1}{x},

(4.166)

donde concluímos que a integral de $x^{-1}$ é

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int\frac{1}{x% }\,dx=\ln|x|+C.

(4.167)

Exemplo 4.3.6.

	$\displaystyle\int_{1}^{e}x^{-1}\,dx=\int_{1}^{e}\frac{1}{x}\,dx$		(4.168)
	$\displaystyle\text{}\qquad=\left[\ln\|x\|\right]_{1}^{e}$		(4.169)
	$\displaystyle\text{}\qquad=\ln\|e\|-\ln\|1\|$		(4.170)
	$\displaystyle\text{}\qquad=1-0=1.$		(4.171)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(1/x, (x, 1, sym.E))

4.3.5 Integral da Função Exponencial Natural

Da derivada da função exponencial natural, temos

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int e^{x}\,dx% =e^{x}+C.

(4.172)

Exemplo 4.3.7.

Vamos estudar os seguintes casos:

a)

$\displaystyle\int e^{2+x}\,dx=\int e^{2}e^{x}\,dx$ (4.173)

$\displaystyle\text{}\qquad=e^{2}\int e^{x}\,dx$ (4.174)

$\displaystyle\text{}\qquad=e^{2}e^{x}+C$ (4.175)

$\displaystyle\text{}\qquad=e^{2+x}+C.$ (4.176)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(sym.exp(2+x), x)
```
exp(x + 2)
```
b)

$\displaystyle\int_{0}^{\ln 2}e^{x}\,dx=\left.e^{x}\right|_{0}^{\ln 2}$ (4.177)

$\displaystyle\text{}\qquad=e^{\ln 2}-e^{0}$ (4.178)

$\displaystyle\text{}\qquad=2-1$ (4.179)

$\displaystyle\text{}\qquad=1.$ (4.180)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(sym.exp(x), (x, 0, sym.log(2)))
```
1
```

4.3.6 Integrais de Funções Trigonométricas

No que lembramos que

\frac{d}{dx}\cos(x)=-\operatorname{sen}(x)

(4.181)

temos que a integral da função seno é

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int% \operatorname{sen}(x)\,dx=-\cos(x)+C.

(4.182)

Exemplo 4.3.8.

Estudamos os seguintes casos:

a)

$\displaystyle\int 2\operatorname{sen}(x)\,dx=2\int\operatorname{sen}(x)dx$ (4.183)

$\displaystyle\text{}\qquad=-2\cos(x)+C$ (4.184)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(2*sym.sin(x), x)
```
-2*cos(x)
```
b)

$\displaystyle\int_{-\pi}^{\pi}\operatorname{sen}(x)\,dx=\left.-\cos(x)\right|_% {-\pi}^{\pi}$ (4.185)

$\displaystyle\text{}\qquad=-\cos(\pi)-[-\cos(-\pi)]$ (4.186)

$\displaystyle\text{}\qquad=1-1=0$ (4.187)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(sym.sin(x), (x, -sym.pi, sym.pi))
```
0
```

Também, lembramos que

\frac{d}{dx}\operatorname{sen}(x)=\cos(x),

(4.188)

donde temos que a integral da função cosseno

\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int\cos(x)\,% dx=\operatorname{sen}(x)+C.

(4.189)

Exemplo 4.3.9.

Estudamos os seguintes casos:

a)

$\displaystyle\int\frac{1}{2}\cos(x)\,dx=\frac{1}{2}\int\cos(x)\,dx$ (4.190)

$\displaystyle\text{}\qquad=\frac{1}{2}\operatorname{sen}(x)+C.$ (4.191)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(1/2*sym.cos(x), x)
```
0.5*sin(x)
```
b)

$\displaystyle\int_{-\pi}^{\pi}\cos(x)\,dx=\left.\operatorname{sen}(x)\right|_{% -\pi}^{\pi}$ (4.193)

$\displaystyle\text{}\qquad=\operatorname{sen}(\pi)-\operatorname{sen}(-\pi)$ (4.194)

$\displaystyle\text{}\qquad=0.$ (4.195)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(sym.cos(x), (x, -sym.pi, sym.pi))
```
0
```

4.3.7 Tabela de Integrais

	$\displaystyle\int k\cdot f(x)\,dx=k\cdot\int f(x)\,dx$		(4.196)
	$\displaystyle\int\left[f(x)\pm g(x)\right]\,dx=\int f(x)\,dx\pm\int g(x)\,dx$		(4.197)
	$\displaystyle\int x^{r}\,dx=\frac{x^{r+1}}{r+1}+C,\quad r\neq-1$		(4.198)
	$\displaystyle\int\frac{1}{x}\,dx=\ln x+C$		(4.199)
	$\displaystyle\int e^{x}\,dx=e^{x}+C$		(4.200)
	$\displaystyle\int\operatorname{sen}(x)\,dx=-\cos(x)+C$		(4.201)
	$\displaystyle\int\cos(x)\,dx=\operatorname{sen}(x)+C$		(4.202)

4.3.8 Exercícios resolvidos

ER 4.3.1.

Calcule

\int\frac{x^{2}+2x}{\sqrt{x}}\,dx

(4.203)

Solução.

	$\displaystyle\int\frac{x^{2}+2x}{\sqrt{x}}\,dx=\int\frac{x^{2}}{\sqrt{x}}+2% \frac{x}{\sqrt{x}}\,dx$		(4.204)
	$\displaystyle\text{}\qquad=\int\frac{x^{2}}{x^{\frac{1}{2}}}+2\frac{x}{x^{% \frac{1}{2}}}\,dx$		(4.205)
	$\displaystyle\text{}\qquad=\int x^{2-\frac{1}{2}}+2x^{1-\frac{1}{2}}\,dx$		(4.206)
	$\displaystyle\text{}\qquad=\int x^{\frac{3}{2}}\,dx+2\int x^{\frac{1}{2}}\,dx.$		(4.207)

Agora, usando a regra da função potência (4.117), obtemos

	$\displaystyle\int\frac{x^{2}+2x}{\sqrt{x}}\,dx=\frac{x^{\frac{3}{2}+1}}{\frac{% 3}{2}+1}+2\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C$		(4.208)
	$\displaystyle\text{}\qquad=\frac{2}{5}x^{\frac{5}{2}}+\frac{4}{3}x^{\frac{3}{2% }}+C.$		(4.209)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate((x**2+2*x)/(sqrt(x)))

2*x**(5/2)/5 + 4*x**(3/2)/3

ER 4.3.2.

Calcule

\int_{1}^{e}\frac{1}{2x}\,dx.

(4.210)

Solução.

Das regras básicas de integração, temos

	$\displaystyle\int\frac{1}{2x}\,dx=\int\frac{1}{2}\cdot\frac{1}{x}\,dx$		(4.211)
	$\displaystyle\text{}\qquad=\frac{1}{2}\int\frac{1}{x}\,dx$		(4.212)
	$\displaystyle\text{}\qquad=\frac{1}{2}\ln(x)+C$		(4.213)
	$\displaystyle\text{}\qquad=\ln\sqrt{x}+C.$		(4.214)

Então, do teorema fundamental do cálculo, temos

	$\displaystyle\int_{1}^{e}\frac{1}{2x}\,dx=\left.\ln\sqrt{x}\right\|_{1}^{e}$		(4.215)
	$\displaystyle\text{}\qquad=\ln(\sqrt{e})-\ln(1)$		(4.216)
	$\displaystyle\text{}\qquad=\frac{1}{2}.$		(4.217)

⬇

1import sympy as sym

2from sympy.abc import x

3sym.integrate(1/(2*x), (x, 1, sym.E))

1/2

4.3.9 Exercícios

E. 4.3.1.

Calcule

a)

$\displaystyle\int\,dx$
b)

$\displaystyle\int x^{-2}\,dx$
c)

$\displaystyle\int\sqrt{x}\,dx$
d)

$\displaystyle\int\frac{1}{\sqrt{x}}\,dx$

Resposta.

a) $\displaystyle x+C$ ; b) $\displaystyle-\frac{1}{x}+C$ ; c) $\displaystyle\frac{2}{3}x^{3/2}+C$ ; d) $\displaystyle 2x^{1/2}+C$

E. 4.3.2.

Calcule

a)

$\displaystyle\int 1+x^{-2}\,dx$
b)

$\displaystyle\int x-\frac{1}{x}\,dx$
c)

$\displaystyle\int 2x^{3}-3x^{2}+1\,dx$

Resposta.

a) $\displaystyle x-\frac{1}{x}+C$ ; b) $\displaystyle\frac{x^{2}}{2}-\ln|x|+C$ ; c) $\displaystyle\frac{1}{2}x^{4}-x^{3}+x+C$

E. 4.3.3.

Calcule

a)

$\displaystyle\int 2\cos(x)\,dx$
b)

$\displaystyle\int 1-\operatorname{sen}(x)\,dx$

Resposta.

a) $\displaystyle 2\operatorname{sen}(x)+C$ ; b) $\displaystyle x+\cos(x)+C$

E. 4.3.4.

Calcule

a)

$\displaystyle\int_{-1}^{1}x^{3}\,dx$
b)

$\displaystyle\int_{e}^{2e}x^{-1}\,dx$

Resposta.

a) $0$ ; b) $\ln(2)$ ;

E. 4.3.5.

Calcule

1.

$\displaystyle\int_{0}^{1}x^{2}-2x^{3}\,dx$
2.

$\displaystyle\int_{1}^{2}\frac{x+1}{\sqrt{x}}\,dx$

Resposta.

a) $\displaystyle-\frac{1}{6}$ ; b) $\displaystyle\frac{10}{3}\sqrt{2}-\frac{8}{3}$ ;

E. 4.3.6.

Cálculo

a)

$\displaystyle\int_{0}^{\frac{\pi}{2}}\operatorname{sen}(x)\,dx$
b)

$\displaystyle\int_{0}^{\frac{\pi}{2}}\cos(x)\,dx$
c)

$\displaystyle\int_{0}^{\pi}\cos(x)-\operatorname{sen}(x)\,dx$

Resposta.

a) $1$ ; b) $1$ ; c) $-2$

Envie seu comentário

Aproveito para agradecer a todas/os que de forma assídua ou esporádica contribuem enviando correções, sugestões e críticas!

Este texto é disponibilizado nos termos da Licença Creative Commons Atribuição-CompartilhaIgual 4.0 Internacional. Ícones e elementos gráficos podem estar sujeitos a condições adicionais.

Pedro H A Konzen

| | | |

	$\displaystyle\int\frac{1}{x^{2}}\,dx=\int x^{\color[rgb]{0,0,1}\definecolor[% named]{pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}-2}\,dx$		(4.121)
	$\displaystyle\text{}\qquad=\frac{x^{{\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}-2}+1}}{{\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}-2}+1}+C=-\frac{1}{x}+C.$		(4.122)

	$\displaystyle\int 2x\,dx=2\int x\,dx$		(4.131)
	$\displaystyle\text{}\qquad=2\left(\frac{x^{2}}{2}+C\right)$		(4.132)
	$\displaystyle\text{}\qquad=x^{2}+2C$		(4.133)
	$\displaystyle\text{}\qquad=x^{2}+C.$		(4.134)

	$\displaystyle\int\frac{1}{3}\sqrt{x}\,dx=\frac{1}{3}\int x^{\frac{1}{2}}\,dx$		(4.135)
	$\displaystyle\text{}\qquad=\frac{1}{3}\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C$		(4.136)
	$\displaystyle\text{}\qquad=\frac{1}{3}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+C$		(4.137)
	$\displaystyle\text{}\qquad=\frac{2}{9}\sqrt{x^{3}}+C.$		(4.138)

	$\displaystyle\int_{0}^{1}-x^{2}\,dx=-\int_{0}^{1}x^{2}\,dx$		(4.139)
	$\displaystyle\text{}\qquad=-\left[\frac{x^{3}}{3}\right]_{0}^{1}$		(4.140)
	$\displaystyle\text{}\qquad=-\left(\frac{1}{3}-\frac{0}{3}\right)$		(4.141)
	$\displaystyle\text{}\qquad=-\frac{1}{3}.$		(4.142)

	$\displaystyle\int{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{% 0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}x}+{% \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}1}\,dx={\color% [rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int x\,dx}+{% \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}\int dx}$		(4.146)
	$\displaystyle\text{}\qquad={\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}\frac{x^{2}}{2}+C_{1}}+{\color[rgb]{1,0,0}% \definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\pgfsys@color@rgb@stroke{1}{0}{% 0}\pgfsys@color@rgb@fill{1}{0}{0}x+C_{2}}$		(4.147)
	$\displaystyle\text{}\qquad=\frac{x^{2}}{2}+x+C.$		(4.148)

	$\displaystyle\int{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{% 0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\sqrt{x}% }-{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}x}\,dx={\color% [rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\int x^{\frac{% 1}{2}}\,dx}-{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}% \pgfsys@color@rgb@stroke{1}{0}{0}\pgfsys@color@rgb@fill{1}{0}{0}\int x\,dx}$		(4.149)
	$\displaystyle\text{}\qquad={\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\pgfsys@color@rgb@stroke{0}{0}{1}% \pgfsys@color@rgb@fill{0}{0}{1}\frac{2}{3}x^{\frac{3}{2}}}-{\color[rgb]{1,0,0}% \definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\pgfsys@color@rgb@stroke{1}{0}{% 0}\pgfsys@color@rgb@fill{1}{0}{0}\frac{x^{2}}{2}}+C.$		(4.150)

	$\displaystyle\int e^{2+x}\,dx=\int e^{2}e^{x}\,dx$		(4.173)
	$\displaystyle\text{}\qquad=e^{2}\int e^{x}\,dx$		(4.174)
	$\displaystyle\text{}\qquad=e^{2}e^{x}+C$		(4.175)
	$\displaystyle\text{}\qquad=e^{2+x}+C.$		(4.176)

	$\displaystyle\int_{0}^{\ln 2}e^{x}\,dx=\left.e^{x}\right\|_{0}^{\ln 2}$		(4.177)
	$\displaystyle\text{}\qquad=e^{\ln 2}-e^{0}$		(4.178)
	$\displaystyle\text{}\qquad=2-1$		(4.179)
	$\displaystyle\text{}\qquad=1.$		(4.180)

	$\displaystyle\int 2\operatorname{sen}(x)\,dx=2\int\operatorname{sen}(x)dx$		(4.183)
	$\displaystyle\text{}\qquad=-2\cos(x)+C$		(4.184)

	$\displaystyle\int_{-\pi}^{\pi}\operatorname{sen}(x)\,dx=\left.-\cos(x)\right\|_% {-\pi}^{\pi}$		(4.185)
	$\displaystyle\text{}\qquad=-\cos(\pi)-[-\cos(-\pi)]$		(4.186)
	$\displaystyle\text{}\qquad=1-1=0$		(4.187)

	$\displaystyle\int\frac{1}{2}\cos(x)\,dx=\frac{1}{2}\int\cos(x)\,dx$		(4.190)
	$\displaystyle\text{}\qquad=\frac{1}{2}\operatorname{sen}(x)+C.$		(4.191)

	$\displaystyle\int_{-\pi}^{\pi}\cos(x)\,dx=\left.\operatorname{sen}(x)\right\|_{% -\pi}^{\pi}$		(4.193)
	$\displaystyle\text{}\qquad=\operatorname{sen}(\pi)-\operatorname{sen}(-\pi)$		(4.194)
	$\displaystyle\text{}\qquad=0.$		(4.195)

Política de uso de dados

Política de uso de dados

Cálculo I

4.3 Regras Básicas de Integração

4.3.1 Integral de Função Potência

Exemplo 4.3.1.

Exemplo 4.3.2.

4.3.2 Regra da Multiplicação por Constante

Exemplo 4.3.3.

4.3.3 Regra da soma ou subtração

Exemplo 4.3.4.

Exemplo 4.3.5.

4.3.4 Integral de x−1

Exemplo 4.3.6.

4.3.5 Integral da Função Exponencial Natural

Exemplo 4.3.7.

4.3.6 Integrais de Funções Trigonométricas

Exemplo 4.3.8.

Exemplo 4.3.9.

4.3.7 Tabela de Integrais

4.3.8 Exercícios resolvidos

ER 4.3.1.

Solução.

ER 4.3.2.

Solução.

4.3.9 Exercícios

E. 4.3.1.

Resposta.

E. 4.3.2.

Resposta.

E. 4.3.3.

Resposta.

E. 4.3.4.

Resposta.

E. 4.3.5.

Resposta.

E. 4.3.6.

Resposta.

Envie seu comentário

Cálculo I

4.3 Regras Básicas de Integração

4.3.1 Integral de Função Potência

Exemplo 4.3.1.

Exemplo 4.3.2.

4.3.2 Regra da Multiplicação por Constante

Exemplo 4.3.3.

4.3.3 Regra da soma ou subtração

Exemplo 4.3.4.

Exemplo 4.3.5.

4.3.4 Integral de x−1

Exemplo 4.3.6.

4.3.5 Integral da Função Exponencial Natural

Exemplo 4.3.7.

4.3.6 Integrais de Funções Trigonométricas

Exemplo 4.3.8.

Exemplo 4.3.9.

4.3.7 Tabela de Integrais

4.3.8 Exercícios resolvidos

ER 4.3.1.

Solução.

ER 4.3.2.

Solução.

4.3.9 Exercícios

E. 4.3.1.

Resposta.

E. 4.3.2.

Resposta.

E. 4.3.3.

Resposta.

E. 4.3.4.

Resposta.

E. 4.3.5.

Resposta.

E. 4.3.6.

Resposta.

Envie seu comentário

4.3.4 Integral de $x^{-1}$

4.3.4 Integral de $x^{-1}$