Question

1- Seja f(x) uma função derivável. Sabendo que f(1) = 1 e f'(x) = 1 + ㏑(x) , é correto afirmar que:

2- Considere a função f(x) = 1/x + √x5. Determine a integral definida de f(x) para x variando entre 1 e e

Answer 1

Resposta:

1- Integre a derivada para encontrar a f(x).

$\sf f'(x)=1+ln(x)$

$\sf f(x)=\int\sf(1+ln(x))dx$

$\sf f(x)=\int\sf1dx+\int\sf ln(x)dx$

$\sf f(x)=x+\int\sf ln(x)dx$

Integrando por partes, denote respectivamente, por u e dv:

• $\sf u=ln(x)\implies du=\frac{1}{x}dx$

• $\sf dv=dx\implies v=x$

$\sf f(x)=x+u\cdot v-\int\sf v\cdot du$

$\sf f(x)=x+ln(x)\cdot x-\int\sf x\cdot\frac{1}{x}dx$

$\sf f(x)=x+ln(x)\cdot x-\int\sf dx$

$\sf f(x)=x+ln(x)\cdot x-x+\mathnormal{C}$

$\sf f(x)=ln(x)\cdot x+\mathnormal{C}$

Dado que f(1) = 1:

$\sf f(1)=ln(1)\cdot 1+\mathnormal{C}=1$

$\sf ln(1)\cdot 1+\mathnormal{C}=1$

$\sf 0+\mathnormal{C}=1$

$\sf \mathnormal{C}=1$

Portanto:

$\sf f(x)=ln(x)\cdot x+1$

Letra D

2- Calcule a integral indefinida primeiro e aplique o teorema fundamental do cálculo depois pra calcular a integral definida.

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\int^{\sf e}_{\sf1}\sf\big(\frac{1}{x}+\sqrt{x^5}\big)dx$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\int\sf\big(\frac{1}{x}+\sqrt{x^5}dx\big)\big|^{\sf e}_{\sf1}$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\int\sf\frac{1}{x}dx+\int\sf\sqrt{x^5}dx\,\big|^{\sf e}_{\sf1}$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\int\sf\frac{1}{x}dx+\int\sf (x^5)^{\frac{1}{2}}dx\,\big|^{\sf e}_{\sf1}$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\int\sf \frac{1}{x}dx+\int\sf x^{\frac{5}{2}}dx\,\big|^{\sf e}_{\sf1}$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=ln|x|+\frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1}\,\big|^{\sf e}_{\sf1}$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=ln|x|+\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\,\big|^{\sf e}_{\sf1}$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=ln|x|+\frac{2x^{\frac{7}{2}}}{7}\,\big|^{\sf e}_{\sf1}$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=ln|e|+\frac{2(e)^{\frac{7}{2}}}{7}-(ln|1|+\frac{2(1)^{\frac{7}{2}}}{7})$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=1+\frac{2e^{\frac{7}{2}}}{7}-(0+\frac{2(1)}{7})$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=1+\frac{2\sqrt{e^7}}{7}-\frac{2}{7}$

$\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\frac{7\,+\,2\sqrt{e^7}\,-\,2}{7}$

$\red{\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\frac{5\,+\,2\sqrt{e^7}}{7}}$

Letra E