ANPEC 2003 - Q9

por temujin Sáb Abr 13, 2013 5:11 pm

(0) $\small \int_{-1}^0\frac{x^2}{\sqrt{1-x^3}}dx \leq \frac{1}{3}$

V. Substituição:

$\small \\ t=1-x^3 \ ; \ dt=3x^2dx \Rightarrow x^2dx=-\frac{1}{3}dt \\ \\ x=-1 \rightarrow t=2 \ ; \ x=0 \rightarrow t=1 \\ \\ \int_{-1}^0\frac{x^2}{\sqrt{1-x^3}}dx = \int_2^1-\frac{1}{3}.\frac{1}{\sqrt{t}}dt = \int_1^2\frac{1}{3}.t^{-1/2}dt = \frac{1}{3}.\left [ \frac{t^{1/2}}{1/2} \right ]_1^2 = \frac{2}{3}.\left [ t^{1/2} \right ]_1^2=\\ \\ \frac{2}{3}.\left [ \sqrt{2}-1 \right ] \approx \frac{0,82}{3}<\frac{1}{3}$

(1) $\small \int_1^2 x.ln(x)dx=-\frac{3}{4}+2ln(2)$

V. Por partes:

$\small \\ u=ln(x) \ ; \ u'(x)=\frac{1}{x} \ ; \ v'=xdx \ ; \ v=\frac{x^2}{2}\\ \\ \int_1^2 xln(x)dx = \frac{x^2}{2}ln(x)-\int_1^2 \frac{1}{x}.\frac{x^2}{2}dx = \frac{x^2}{2}ln(x)-\int_1^2 \frac{x}{2}dx = \left [ \frac{x^2}{2}ln(x)-\frac{x^2}{4} \right ]_1^2=2ln(2)-1+\frac{1}{4}=-\frac{3}{4}+2ln(2)$

(2) $\small \int_x^{2x} \frac{1}{t}dt >1$ , para todo x > 0.

F.

$\small \int_x^{2x} \frac{1}{t}dt = \left | ln t \right |_x^{2x} = ln(2x)-ln(x) =ln\left(\frac{2x}{x}\right)=ln(2)<1$

(3) $\small \int_{-1}^0 e^{-\frac{x^2}{2}}dx=-\int_0^1 e^{-\frac{(x-1)^2}{2}}dx$

F. Substituição na integral do lado direito:

$\small \\ t=x-1 \ ; \ dt = dx \ ; \ x=0 \rightarrow t=-1 \ ; \ x = 1 \rightarrow t=0\\ \\ -\int_{0}^1 e^{-\frac{(x-1)^2}{2}}dx=-\int_{-1}^0 e^{\frac{-t^2}{2}}dt \neq \int_{-1}^0 e^{\frac{-x^2}{2}}dx$

(4) $\small \frac{15}{16}.\int_{-1}^3 x.\sqrt{1+x}dx=7$

V. Por partes:

$\small \\ u=x \ ; \ u'=dx \ ; \ v'=(1+x)^{1/2} \ ;91 \ v = \frac{2(1+x)^{3/2}}{3} \\ \\ \frac{15}{16}.\int_{-1}^3 x.\sqrt{1+x}dx=\frac{15}{16}.\left [x.\frac{2(1+x)^{3/2}}{3}-\int_{-1}^3 \frac{2(1+x)^{3/2}}{3}dx \right ]=\frac{5}{8}.\left [ x(1+x)^{3/2}-\frac{2(1+x)^{5/2}}{5} \right ]_{-1}^3=\frac{5}{8}.\left [ \left ( 3.4^{3/2}-\frac{2.4^{5/2}}{5} \right )-\left ( -1.0^{3/2}-\frac{2.0^{5/2}}{5} \right ) \right ]=\frac{5}{2^3}.\left [ 3.2^3-\frac{2.2^5}{5} \right ]=15-8=7$