Integrais definidas

aIntegrais definidasGilson Würz. Licenciado em FísicaInstituto Federal de Educação Científica e TecnológicaCampus Jaraguá do SulMath Calcule ∫2x-1

(x-2)(x+1)dx${\displaystyle \underset{}{\overset{}{\int }}\frac{2x-1}{(x-2)(x+1)}dx}$SoluçãoExpandindo f(x)=(x-2)(x+1)=x2-x-2→df

dx=2x-1${\displaystyle f(x)=(x-2)(x+1)=x^2-x-2\to \frac{df}{dx}=2x-1}$∫2x-1

(x-2)(x+1)dx=∫1

f(x)df(x)

dxdx=∫df(x)

f(x)=lnf(x)+c=ln|(x-2)(x+1)|+c${\displaystyle \underset{}{\overset{}{\int }}\frac{2x-1}{(x-2)(x+1)}dx=\int \frac{1}{f(x)}\frac{df(x)}{dx}dx=\int \frac{df(x)}{f(x)}=\ln f(x)+c=\ln |(x-2)(x+1)|+c}$ Calcule ∫1-x2

x(x2+1)dx${\displaystyle \int \frac{1-x^2}{x(x^2+1)}dx}$, para x>0${\displaystyle x>0}$.SoluçãoArranjando a razão polinomial como duas frações por partes, temos1-x2

x(x2+1)≡A

x+Bx+C

x2+1→1-x2

x(x2+1)≡(x2+1)

(x2+1)A

x+(Bx+C)x

(x2+1)x${\displaystyle \frac{1-x^2}{x(x^2+1)}\equiv \frac{A}{x}+\frac{Bx+C}{x^2+1}\to \frac{1-x^2}{x(x^2+1)}\equiv \frac{(x^2+1)}{(x^2+1)}\frac{A}{x}+\frac{(Bx+C)x}{(x^2+1)x}}$Após isso, devemos igualar os coeficientes para determinar seus respectivos valores.1-x2

x(x2+1)≡(x2+1)

(x2+1)A

x+(Bx+C)x

(x2+1)x→1-x2≡A(x2+1)+(Bx+C)x${\displaystyle \frac{1-x^2}{x(x^2+1)}\equiv \frac{(x^2+1)}{(x^2+1)}\frac{A}{x}+\frac{(Bx+C)x}{(x^2+1)x}\to 1-x^2\equiv A(x^2+1)+(Bx+C)x}$1-x2≡Ax2+A+Bx2+Cx${\displaystyle 1-x^2\equiv A{x}^2+A+B{x}^2+Cx}$1-x2≡A+Cx+(A+B)x2${\displaystyle 1-x^2\equiv A+Cx+(A+B)x^2}$Logo, A=1${\displaystyle A=1}$, C=0${\displaystyle C=0}$, A+B=-1→1+B=-1→B=-2${\displaystyle A+B=-1\to 1+B=-1\to B=-2}$Assim, substituindo esses coeficientes na integral abaixo, temos:∫1-x2

x(x2+1)dx=∫(A

x+Bx+C

x2+1)dx=∫dx

x+∫-2x

x2+1dx${\displaystyle \int \frac{1-x^2}{x(x^2+1)}dx=\int (\frac{A}{x}+\frac{Bx+C}{x^2+1})dx=\int \frac{dx}{x}+\int \frac{-2x}{x^2+1}dx}$Fazendo u=x2+1→du=2x⋅dx${\displaystyle u=x^2+1\to du=2x\cdot dx}$∫1-x2

x(x2+1)dx=∫(A

x+Bx+C

x2+1)dx=∫dx

x-2∫du

2

u${\displaystyle \int \frac{1-x^2}{x(x^2+1)}dx=\int (\frac{A}{x}+\frac{Bx+C}{x^2+1})dx=\int \frac{dx}{x}-2\int \frac{\frac{du}{2}}{u}}$Fazendo a simplificação devida, no intuito de cancelar o número 2 da frente da integral com o 2 no denominador de dentro do integrando, temos:∫1-x2

x(x2+1)dx=∫(A

x+Bx+C

x2+1)dx=ln|x|-ln|u|+c${\displaystyle \int \frac{1-x^2}{x(x^2+1)}dx=\int (\frac{A}{x}+\frac{Bx+C}{x^2+1})dx=\ln |x|-\ln |u|+c}$O fator constante deve ser adicionado pois temos uma integral indefinida e substituindo u${\displaystyle u}$ por x2+1${\displaystyle x^2+1}$, temos:∫1-x2

x(x2+1)dx=∫(A

x+Bx+C

x2+1)dx=ln|x|-ln|x2+1|+c${\displaystyle \int \frac{1-x^2}{x(x^2+1)}dx=\int (\frac{A}{x}+\frac{Bx+C}{x^2+1})dx=\ln |x|-\ln |x^2+1|+c}$Da propriedade lnA-lnB=lnA

B${\displaystyle \ln A-\ln B=\ln \frac{A}{B}}$, segue que:∫1-x2

x(x2+1)dx=∫(A

x+Bx+C

x2+1)dx=lnax

x2+1a+c${\displaystyle \int \frac{1-x^2}{x(x^2+1)}dx=\int (\frac{A}{x}+\frac{Bx+C}{x^2+1})dx=\ln \left|\frac{x}{x^2+1}\right|+c}$ Calcular os integrais definidos utilizando a fórmula de Barrow. 1) 1∫01+x dx${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x}dx}$Solução1∫01+x dx=1∫01+x d(1+x)=1∫0(1+x)1/2d(1+x)=(1+x)1/2+1

1/2+1=2(1+x)3/2

3a10${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x}dx=\underset{0}{\overset{1}{\int }}\sqrt{1+x}d(1+x)=\underset{0}{\overset{1}{\int }}(1+x{)}^{1/2}d(1+x)={\displaystyle \frac{(1+x{)}^{1/2+1}}{1/2+1}}={\displaystyle \frac{2(1+x{)}^{3/2}}{3}}{|}_0^1}$1∫01+x dx=2

3(1+x)3/2=2

3[(1+1)3/2-(1+0)3/2]=2

3(23/2-13/2)=2

3(223-1)=2

3( 4⋅2-1)${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x}dx={\displaystyle \frac{2}{3}}(1+x{)}^{3/2}={\displaystyle \frac{2}{3}}[(1+1{)}^{3/2}-(1+0{)}^{3/2}]={\displaystyle \frac{2}{3}}(2^{3/2}-1^{3/2})={\displaystyle \frac{2}{3}}(\sqrt[2]{2^3}-1)={\displaystyle \frac{2}{3}}(\sqrt[]{4\cdot 2}-1)}$1∫01+x dx=2

3(22-1)${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x}dx={\displaystyle \frac{2}{3}}(2\sqrt{2}-1)}$

y=1+x2) 1∫-1xdx

(x2+1)2${\displaystyle \underset{-1}{\overset{1}{\int }}{\displaystyle \frac{xdx}{(x^2+1{)}^2}}}$Solução1∫-1xdx

(x2+1)2=1∫-11

2d(x2)

(x2+1)2=1

21∫-1d(x2)

(x2+1)2=1

21∫-1d(x2+1)

(x2+1)2=1

21∫-1(x2+1)-2d(x2+1)=1

2(x2+1)-2+1

-2+1a10→${\displaystyle \underset{-1}{\overset{1}{\int }}{\displaystyle \frac{xdx}{(x^2+1{)}^2}}=\underset{-1}{\overset{1}{\int }}{\displaystyle \frac{{\displaystyle \frac{1}{2}}d\left(x^2\right)}{(x^2+1{)}^2}}={\displaystyle \frac{1}{2}}\underset{-1}{\overset{1}{\int }}{\displaystyle \frac{d\left(x^2\right)}{(x^2+1{)}^2}}={\displaystyle \frac{1}{2}}\underset{-1}{\overset{1}{\int }}{\displaystyle \frac{d(x^2+1)}{(x^2+1{)}^2}}={\displaystyle \frac{1}{2}}\underset{-1}{\overset{1}{\int }}(x^2+1{)}^{-2}d(x^2+1)={\displaystyle \frac{1}{2}}{\displaystyle \frac{(x^2+1{)}^{-2+1}}{-2+1}}{|}_0^1\to }$1∫-1xdx

(x2+1)2=-1

2(x2+1)-1a1-1=-1

2(1

x2+1)a1-1=-1

2(1

12+1-1

(-1)2+1)=0${\displaystyle \underset{-1}{\overset{1}{\int }}{\displaystyle \frac{xdx}{(x^2+1{)}^2}}=-{\displaystyle \frac{1}{2}}(x^2+1{)}^{-1}{|}_{-1}^1=-{\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{x^2+1}}){|}_{-1}^1=-{\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{1^2+1}}-{\displaystyle \frac{1}{(-1{)}^2+1}})=0}$

y=x

(x2+1)2 Bibliografias