Question

Qual a medida da região hachurada.

Answer 1

Resposta:

Explicação passo a passo:

$A=\displaystyle\int_{1}^{\frac{\pi }{2} }(senx-lnx)dx$

Vamos determinar, por questões didáticas, primeiramente:

$\displaystyle\int lnx~dx=\\Sejam ~lnx =U ~e~dx=V \implies\displaystyle\int dx=\displaystyle\int dV \implies x=V \\\\x=e^{U} \\\\\frac{1}{x} ~dx=dU \implies \ dx=xdU\implies dx=e^{U}dU \\\\\displaystyle\int lnxdx=UV-\displaystyle\int VdU\\\\\displaystyle\int lnx~dx=lnx*x-\displaystyle\int x*\frac{dx}{x} +c=xlnx-\displaystyle\int dx+c=xlnx-x +c$

Na integral definida dispensamos a constante de integração c.

$\displaystyle\int_{1}^{\frac{\pi }{2} }(senx-lnx)dx=[-cosx-(xlnx-x)\left ] {{\frac{\pi }{2} } \atop {1}} \right. =(x-xlnx-cosx\left ] {{\frac{\pi }{2} } \atop {1}} \right. =\frac{\pi }{2}-\frac{\pi }{2} ln\frac{\pi }{2} -cos\frac{\pi }{2}- (1-1.ln1-cox1)=\frac{\pi }{2} -\frac{\pi }{2} ln\frac{\pi }{2} -0-1+0+cos1=\frac{\pi -\pi ln\frac{\pi }{2} }{2} -1+cos1$