Chủ đề này có 3 trả lời

[ChunChun]

1) $$\int_{0}^{-\propto }\dfrac{dx}{e^{x}+1}$$
2)$\int_{3}^{1}\dfrac{dx}{\sqrt{-x^{2}+4x-3}}$

MOD: Bạn nên gõ latex lên tiêu đề

Bài viết đã được chỉnh sửa nội dung bởi xusinst: 25-12-2011 - 20:53

[Crystal]

1) $$\int_{0}^{-\propto }\dfrac{dx}{e^{x}+1}$$

Ta có: $$I = \mathop {\lim }\limits_{a \to - \infty } \int\limits_0^a {\dfrac{{dx}}{{{e^x} + 1}}} $$
$$t = {e^x} \Rightarrow dt = {e^x}dx \Rightarrow dx = \dfrac{{dt}}{{{e^x}}} = \dfrac{{dt}}{t};\,x = 0 \Rightarrow t = 1,\,x = a \Rightarrow t = {e^a}$$
Khi đó: $$I = - \mathop {\lim }\limits_{a \to - \infty } \int\limits_{{e^a}}^1 {\dfrac{{\dfrac{{dt}}{t}}}{{t + 1}} = } - \mathop {\lim }\limits_{a \to - \infty } \int\limits_{{e^a}}^1 {\dfrac{{dt}}{{t\left( {t + 1} \right)}} = } - \mathop {\lim }\limits_{a \to - \infty } \int\limits_{{e^a}}^1 {\left( {\dfrac{1}{t} - \dfrac{1}{{t + 1}}} \right)dt} $$
$$ = - \mathop {\lim }\limits_{a \to - \infty } \left. {\left( {\ln \left| {\dfrac{t}{{t + 1}}} \right|} \right)} \right|_{{e^a}}^1 = - \mathop {\lim }\limits_{a \to - \infty } \left( {\ln \dfrac{1}{2} - \ln \dfrac{{{e^a}}}{{{e^a} + 1}}} \right) = - \ln \dfrac{1}{2} = \ln 2$$
Vậy $\boxed{I = \ln 2}$

[Vũ Sơn]

Mình phang thử bài 2:

2)$\int_{3}^{1}\dfrac{dx}{\sqrt{-x^{2}+4x-3}}$

\[
I = \int_3^1 {\dfrac{{dx}}{{\sqrt { - x^2 + 4x - 3} }}} = \int_3^1 {\dfrac{{dx}}{{\sqrt {\left( {x - 3} \right)\left( {1 - x} \right)} }}}
\]
Điểm bất thường: $x=3$ và $x=1$. Cố định $\alpha ,\beta$, rồi đặt
\[
x = 3\cos ^2 t + \sin ^2 t \Rightarrow dx = ( - 6\cos t.\sin t + 2\sin t\cos t)dt = - 4\cos t\sin tdt
\]
\[x = 3,\,t = 0;\,x = 1,\,t = \dfrac{\pi }{2}
\]
Từ đó:
\[
I = \mathop {\lim }\limits_{\scriptstyle \alpha \to 0^ + \atop
\scriptstyle \beta \to \dfrac{\pi }{2}^ -} \int_\alpha ^\beta {\dfrac{{ - 4\sin t\cos tdt}}{{\sqrt {\left( { - 2\sin ^2 t} \right)\left( { - 2\cos ^2 t} \right)} }}} = - 2\mathop {\lim }\limits_{\scriptstyle \alpha \to 0^ + \atop
\scriptstyle \beta \to \dfrac{\pi }{2}^ -} \int_\alpha ^\beta {dt} = - 2\int_0^{\dfrac{\pi }{2}} {dt} = -\pi
\]

Bài viết đã được chỉnh sửa nội dung bởi Vũ Sơn: 16-11-2011 - 14:20

[Crystal]

2)$$J=\int_{3}^{1}\dfrac{dx}{\sqrt{-x^{2}+4x-3}}$$

Ta có nhận xét: Hàm số $f\left( x \right) = \dfrac{1}{{\sqrt { - {x^2} + 4x - 3} }}$ xác định trên $\left( {1,3} \right)$ và có $\mathop {\lim }\limits_{x \to 1 + 0} f\left( x \right) = \infty ,\,\,\,\mathop {\lim }\limits_{x \to 3 - 0} f\left( x \right) = \infty $.
Do đó: $$J = \int\limits_3^1 {\dfrac{{dx}}{{\sqrt { - {x^2} + 4x - 3} }}} = - \int\limits_1^3 {\dfrac{{dx}}{{\sqrt { - {x^2} + 4x - 3} }}} $$
$$ = - \left( {\int\limits_1^2 {\dfrac{{dx}}{{\sqrt { - {x^2} + 4x - 3} }} + } \int\limits_2^3 {\dfrac{{dx}}{{\sqrt { - {x^2} + 4x - 3} }}} } \right)$$
$$ = - \left( {\mathop {\lim }\limits_{\varepsilon \to 0 + 0} \int\limits_{1 + \varepsilon }^2 {\dfrac{{dx}}{{\sqrt { - {x^2} + 4x - 3} }} + \mathop {\lim }\limits_{\varepsilon \to 0 - 0} \int\limits_2^{3 + \varepsilon } {\dfrac{{dx}}{{\sqrt { - {x^2} + 4x - 3} }}} } } \right)$$
Xét $$\int {\dfrac{{dx}}{{\sqrt { - {x^2} + 4x - 3} }}} = \int {\dfrac{{dx}}{{\sqrt { - \left( {{x^2} - 4x + 4} \right) + 1} }} = } \int {\dfrac{{dx}}{{\sqrt {1 - {{\left( {x - 2} \right)}^2}} }} = arcsin\left( {x - 2} \right) + C} $$
Suy ra $$J = - \left( {\mathop {\lim }\limits_{\varepsilon \to 0 + 0} \left. {arcsin\left( {x - 2} \right)} \right|_{1 + \varepsilon }^2 + \mathop {\lim }\limits_{\varepsilon \to 0 - 0} \left. {arcsin\left( {x - 2} \right)} \right|_2^{3 + \varepsilon }} \right)$$
$$ = - \left( {\mathop {\lim }\limits_{\varepsilon \to 0 + 0} \left( {arcsin0 - arcsin\left( {\varepsilon - 1} \right)} \right) + \mathop {\lim }\limits_{\varepsilon \to 0 - 0} \left( {arcsin\left( {1 + \varepsilon } \right) - arcsin0} \right)} \right)$$
$$ = - \left( {\dfrac{\pi }{2} + \dfrac{\pi }{2}} \right) = - \pi $$
Vậy $\boxed{J = - \pi }$.