Hướng dẫn giải (chi tiết)

Ta có: $f’\left( x \right) = \frac{{ - 3a}}{{{{\left( {x + 1} \right)}^4}}} + b{{\rm{e}}^x} + bx{{\rm{e}}^x}$ nên $f’\left( 0 \right) = - 3a + b = - 22$ (1).

Xét $5 = \int\limits_0^1 {f\left( x \right){\rm{d}}x} = \int\limits_0^1 {\left[ {\frac{a}{{{{\left( {x + 1} \right)}^3}}} + bx{{\rm{e}}^x}} \right]{\rm{d}}x} = a\int\limits_0^1 {{{\left( {x + 1} \right)}^{ - 3}}{\rm{d}}\left( {x + 1} \right)} + b\int\limits_0^1 {x{\rm{d}}\left( {{{\rm{e}}^x}} \right)} $

$ = - \frac{a}{{2{{\left( {x + 1} \right)}^2}}}|_0^1 + b\left[ {\left. {x{{\rm{e}}^x}} \right|_0^1 - \int\limits_0^1 {{{\rm{e}}^x}{\rm{d}}x} } \right] = - \frac{a}{2}\left( {\frac{1}{4} - 1} \right) + b\left[ {\left. {{\rm{e}} - {{\rm{e}}^x}} \right|_0^1} \right] = \frac{{3a}}{8} + b\,(2)$

Từ (1) và (2) suy ra $\left\{ \begin{array}{l} - 3a + b = - 22\\ \frac{{3a}}{8} + b = 5 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = 8\\ b = 2 \end{array} \right. \Rightarrow a + b = 10$

Hướng dẫn giải (chi tiết)

Ta có: $\int\limits_1^2 {\frac{x}{{3x + \sqrt {9{x^2} - 1} }}} {\rm{d}}x = \int\limits_1^2 {x\left( {3x + \sqrt {9{x^2} - 1} } \right)} {\rm{d}}x = \int\limits_1^2 {\left( {3{x^2} - x\sqrt {9{x^2} - 1} } \right){\rm{d}}x} $

$ = \int\limits_1^2 {3{x^2}{\rm{d}}x} - \int\limits_1^2 {x\sqrt {9{x^2} - 1} {\rm{d}}x} = \left. {{x^3}} \right|_1^2 + \int\limits_1^2 {x\sqrt {9{x^2} - 1} {\rm{d}}x} = 7 + \int\limits_1^2 {x\sqrt {9{x^2} - 1} {\rm{d}}x} $

Tính $\int\limits_1^2 {x\sqrt {9{x^2} - 1} {\rm{d}}x} $

Đặt $t = \sqrt {9{x^2} - 1} \Rightarrow 9{x^2} - 1 = {t^2} \Rightarrow x{\rm{d}}x = \frac{{t{\rm{d}}t}}{9}$

Đổi cận $x = 1 \Rightarrow t = 2\sqrt 2 ;\,x = 2 \Rightarrow t = \sqrt {35} $

Khi đó: $\int\limits_1^2 {x\sqrt {9{x^2} - 1} {\rm{d}}x} = \int\limits_{2\sqrt 2 }^{\sqrt {35} } {t\frac{{t{\rm{d}}t}}{9} = \left. {\frac{{{t^3}}}{{27}}} \right|_{2\sqrt 2 }^{\sqrt {35} }} = \frac{{35}}{{27}}\sqrt {35} - \frac{{16}}{{27}}\sqrt 2 $

Vậy $\int\limits_1^2 {\frac{x}{{3x + \sqrt {9{x^2} - 1} }}} {\rm{d}}x = 7 - \frac{{35}}{{27}}\sqrt {35} + \frac{{16}}{{27}}\sqrt 2 $

$a = 7;\,b = \frac{{16}}{{27}};\,c = - \frac{{35}}{{27}}$

Vậy $P = a + 2b + c - 7 = 7 + \frac{{32}}{{27}} - \frac{{35}}{{27}} - 7 = - \frac{1}{9}$

Hướng dẫn giải (chi tiết)

Tính $I = \int\limits_0^\pi {xf\left( {\sin x} \right){\rm{d}}x} $

Đặt $t = \pi - x \Rightarrow {\rm{d}}t = - {\rm{d}}x$

Đổi cận $x = 0 \Rightarrow t = \pi ,\,x = \pi \Rightarrow t = 0$

$ \Rightarrow I = \int\limits_\pi ^0 { - \left( {\pi - t} \right)f\left( {\sin \left( {\pi - t} \right)} \right){\rm{d}}t} = \int\limits_0^\pi {\left( {\pi - t} \right)f\left( {\sin t} \right){\rm{d}}t} = \pi \int\limits_0^\pi {f\left( {\sin t} \right){\rm{d}}t} - \int\limits_0^\pi {tf\left( {\sin t} \right){\rm{d}}t} $

$ \Rightarrow I = \pi - \int\limits_0^\pi {t.f\left( {\sin t} \right){\rm{d}}t} \Rightarrow \int\limits_0^\pi {t.f\left( {\sin t} \right){\rm{d}}} t = \frac{\pi }{2}$

Vậy $\int\limits_0^\pi {x.f\left( {\sin x} \right){\rm{d}}x} = \frac{\pi }{2}$

Hướng dẫn giải (chi tiết)

Đặt $t = \frac{1}{x}$. Suy ra ${\rm{d}}t = {\rm{d}}\left( {\frac{1}{x}} \right) = \frac{{ - 1}}{{{x^2}}}{\rm{d}}x \Rightarrow {\rm{d}}x = - \frac{1}{{{t^2}}}{\rm{d}}t$

Đổi cận $x = \frac{1}{2} \Rightarrow t = 2;\,x = 2 \Rightarrow t = \frac{1}{2}$

Ta có: $I = \int\limits_2^{\frac{1}{2}} {tf\left( {\frac{1}{t}} \right)\left( {\frac{{ - 1}}{{{t^2}}}} \right){\rm{d}}t} = \int\limits_{\frac{1}{2}}^2 {f\left( {\frac{1}{t}} \right)\left( {\frac{1}{t}} \right){\rm{d}}t} = \int\limits_{\frac{1}{2}}^2 {f\left( {\frac{1}{x}} \right)\left( {\frac{1}{x}} \right){\rm{d}}x} $

Suy ra $3I = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( x \right)}}{x}{\rm{d}}x} + 2\int\limits_{\frac{1}{2}}^2 {f\left( {\frac{1}{x}} \right)\left( {\frac{1}{x}} \right){\rm{d}}x} = \int\limits_{\frac{1}{2}}^2 {\frac{1}{x}\left( {f\left( x \right) + 2f\left( {\frac{1}{x}} \right)} \right){\rm{d}}x} = \int\limits_{\frac{1}{2}}^2 {{\rm{3d}}x} = \left. {3x} \right|_{\frac{1}{2}}^2 = \frac{9}{2}$

Vậy $I = \frac{3}{2}$ 

Hướng dẫn giải (chi tiết)

Dựa vào đồ thị hàm số ta có:

+ Đồ thị hàm số đi qua các điểm $\left( { - 1; - 1} \right);\,\left( {0;3} \right);\,\left( {2; - 1} \right);\,\left( {3;3} \right)$. Nên hàm số có dạng:  $y = f\left( x \right) = {x^3} - 3{x^2} + 3$

Ta có:

 $I = \int\limits_1^2 {f’\left( {2x - 1} \right)} \,{\rm{d}}x \\= \frac{1}{2}\int\limits_1^2 {f’\left( {2x - 1} \right)} \,{\rm{d}}\left( {2x - 1} \right) \\= \frac{1}{2}\left. {f\left( {2x - 1} \right)} \right|_1^2 \\= \frac{1}{2}\left[ {f\left( 3 \right) - f\left( 1 \right)} \right] \\= 1$

Hướng dẫn giải (chi tiết)

Ta có: $\int\limits_0^d {f\left( x \right){\rm{d}}x} = \left. {\left( {\frac{a}{3}{x^3} + \frac{b}{2}{x^2} + cx} \right)} \right|_0^d = \frac{a}{3}{d^3} + \frac{b}{2}{d^2} + cd$

Do đó: $\left\{ \begin{array}{l}\n\int\limits_0^1 {f\left( x \right){\rm{d}}x} = - \frac{7}{2} \Leftrightarrow \frac{a}{3} + \frac{b}{2} + c = - \frac{7}{2}\\\n\int\limits_0^2 {f\left( x \right){\rm{d}}x} = - 2 \Leftrightarrow \frac{8}{3}a + 2b + 2c = - 2\\\n\int\limits_0^3 {f\left( x \right){\rm{d}}x} = \frac{{13}}{2} \Leftrightarrow 9a + \frac{9}{2}b + 3c = \frac{{13}}{2}\n\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\na = 1\\\nb = 3\\\nc = - \frac{{16}}{3}\n\end{array} \right.$

Vậy $P = a + b + c = - \frac{4}{3}$

Hướng dẫn giải (chi tiết)

Ta có: $f\left( x \right).f\left( {1 - x} \right) + f\left( x \right) = 1 + f\left( x \right) \Rightarrow \frac{1}{{f\left( {1 - x} \right) + 1}} = \frac{{f\left( x \right)}}{{1 + f\left( x \right)}}{\mkern 1mu} $ 

Xét $I = \int\limits_0^1 {\frac{{{\rm{d}}x}}{{1 + f\left( x \right)}}} $

Đặt $t = 1 - x \Leftrightarrow x = 1 - t \Rightarrow {\rm{d}}x = - {\rm{d}}t$ 

Đổi cận $x = 0 \Rightarrow t = 1;\,x = 1 \Rightarrow t = 0$

Khi đó $I = - \int\limits_1^0 {\frac{{{\rm{d}}t}}{{1 + f\left( {1 - t} \right)}}} = \int\limits_0^1 {\frac{{{\rm{d}}t}}{{1 + f\left( {1 - t} \right)}}} = \int\limits_0^1 {\frac{{{\rm{d}}x}}{{1 + f\left( {1 - x} \right)}}} = \int\limits_0^1 {\frac{{f\left( x \right){\rm{d}}x}}{{1 + f\left( x \right)}}} $

Mặt khác $\int\limits_0^1 {\frac{{{\rm{d}}x}}{{1 + f\left( x \right)}}} + \int\limits_0^1 {\frac{{f\left( x \right){\rm{d}}x}}{{1 + f\left( x \right)}}} = \int\limits_0^1 {\frac{{1 + f\left( x \right)}}{{1 + f(t)}}{\rm{d}}x} = \int\limits_0^1 {{\rm{d}}x} = 1$ hay $2I = 1$. Vậy $I = \frac{1}{2}$ 

Hướng dẫn giải (chi tiết)

Xét $I = \int\limits_0^{\frac{\pi }{4}} {f(\tan x)dx} = \int\limits_0^{\frac{\pi }{4}} {\frac{{f(\tan x)}}{{1 + {{\tan }^2}x}}\left( {1 + {{\tan }^2}x} \right)dx} $

Đặt $u = \tan x \Rightarrow du = \left( {1 + {{\tan }^2}x} \right)dx$

Đổi cận $x = 0 \Rightarrow u = 0;\,x = \frac{\pi }{4} \Rightarrow u = 1$

Nên $I = \int\limits_0^1 {\frac{{f(u)}}{{1 + {u^2}}}du} = \int\limits_0^1 {\frac{{f(x)}}{{1 + {x^2}}}dx} $. Suy ra $\int\limits_0^1 {\frac{{f(x)}}{{1 + {x^2}}}dx} = 4$

Mặt khác $\int\limits_0^1 {\frac{{{x^2}f(x)}}{{{x^2} + 1}}dx} = \int\limits_0^1 {\frac{{\left[ {\left( {{x^2} + 1} \right) - 1} \right]f(x)}}{{{x^2} + 1}}dx} = \int\limits_0^1 {f\left( x \right)dx} - \int\limits_0^1 {\frac{{f(x)}}{{1 + {x^2}}}dx} $

Do đó $2 = \int\limits_0^1 {f\left( x \right)dx} - 4 \Leftrightarrow \int\limits_0^1 {f\left( x \right)dx} = 6$ 

Hướng dẫn giải (chi tiết)

Đặt $t = 5 - 3x \Rightarrow {\rm{d}}x = - \frac{{{\rm{d}}t}}{3}$ 

Đồi cận $\left\{ {\begin{array}{*{20}{l}}\n{x = 0 \Rightarrow t = 5}\\\n{x = 2 \Rightarrow t = - 1}\n\end{array}} \right.$

Tích phân ban đầu trở thành: $P = \int\limits_5^{ - 1} {\left[ {f\left( t \right) + 7} \right]} \left( { - \frac{{dt}}{3}} \right) = \frac{1}{3}\int\limits_{ - 1}^5 {\left[ {f\left( t \right) + 7} \right]dt} = \frac{1}{3}\left( {\int\limits_{ - 1}^5 {f\left( t \right){\rm{d}}t} + 7\int\limits_{ - 1}^5 {{\rm{d}}t} } \right) = \frac{1}{3}.15 + \frac{1}{3}.7.6 = 19$