Tích phân (Khái niệm và tính chất)

Ta có $I = \int\limits_0^3 {\dfrac{{{\rm{d}}x}}{{x + 2}}}  = \left. {\ln \left| {x + 2} \right|} \right|_0^3 = \ln 5 - \ln 2 = \ln \dfrac{5}{2}.$

$\begin{array}{l}\int\limits_1^2 {\dfrac{1}{{{x^2} + 5x + 6}}dx}  = \int\limits_1^2 {\dfrac{1}{{(x + 2)(x + 3)}}dx}  = \left. {\left( {\ln \left| {x + 2} \right| - \ln \left| {x + 3} \right|} \right)} \right|_1^2 = \ln 4 - \ln 5 - \ln 3 + \ln 4\\{\rm{\;}} = 4\ln 2 - \ln 3 - \ln 5 = a\ln 2 + b\ln 3 + c\ln 5,\left( {a,b,c \in Z} \right)\\{\rm{\;}} \Rightarrow a = 4;b =  - 1,c =  - 1 \Rightarrow a + b + c = 2\end{array}$

$\begin{array}{*{20}{l}}{\int_0^1 {\dfrac{{2x + 3}}{{2 - x}}dx} {\rm{\;}} = \int_0^1 {\dfrac{{2x - 4 + 7}}{{2 - x}}dx} {\rm{\;}} = \int_0^1 {\left( { - 2 - \dfrac{7}{{x - 2}}} \right)dx} {\rm{\;}} = \left. {\left( { - 2x - 7\ln \left| {x - 2} \right|} \right)} \right|_0^1}\\{ = {\rm{\;}} - 2 + 7\ln 2 = a\ln 2 + b,{\mkern 1mu} {\mkern 1mu} (a,b \in Z)}\\{ \Rightarrow a = 7,{\mkern 1mu} {\mkern 1mu} b = {\rm{\;}} - 2}\end{array}$

Ta có: $f\left( x \right) = {x^2} + 4x + 3 = \left( {x + 1} \right)\left( {x + 3} \right)$

$f\left( x \right) > 0 \Leftrightarrow \left[ \begin{array}{l}x >  - 1\\x <  - 3\end{array} \right.$ và  $f\left( x \right) < 0 \Leftrightarrow  - 3 < x <  - 1$

Khi đó,

$I = \int\limits_{ - 5}^0 {\left| {{x^2} + 4x + 3} \right|dx}$ $= \int\limits_{ - 5}^{ - 3} {\left( {{x^2} + 4x + 3} \right)dx}$  $+ \int\limits_{ - 3}^{ - 1} {\left[ { - \left( {{x^2} + 4x + 3} \right)} \right]dx}$ $+ \int\limits_{ - 1}^0 {\left( {{x^2} + 4x + 3} \right)dx}$

$= \left. {\left( {\dfrac{{{x^3}}}{3} + 2{x^2} + 3x} \right)} \right|_{ - 5}^{ - 3}$ $- \left. {\left( {\dfrac{{{x^3}}}{3} + 2{x^2} + 3x} \right)} \right|_{ - 3}^{ - 1}$ $+ \left. {\left( {\dfrac{{{x^3}}}{3} + 2{x^2} + 3x} \right)} \right|_{ - 1}^0$

$= 0 + \dfrac{{20}}{3} - \left( { - \dfrac{4}{3} - 0} \right) + \left( {0 + \dfrac{4}{3}} \right) = \dfrac{{28}}{3}$

Do đó $a = 28,b = 3$ hay $a - b = 25$.

$\begin{align}  & I=\int\limits_{1}^{2}{\frac{3{{x}^{2}}+5x+4}{{{x}^{2}}+x+1}dx=}\int\limits_{1}^{2}{\frac{\left( 3{{x}^{2}}+3x+3 \right)+\left( 2x+1 \right)}{{{x}^{2}}+x+1}dx=}\int\limits_{1}^{2}{\left( 3+\frac{2x+1}{{{x}^{2}}+x+1} \right)dx=}\int\limits_{1}^{2}{3dx+\int\limits_{1}^{2}{\frac{2x+1}{{{x}^{2}}+x+1}dx}} \\ & =\left. 3x \right|_{1}^{2}+\int\limits_{1}^{2}{\frac{d({{x}^{2}}+x+1)}{{{x}^{2}}+x+1}=}\left. 3x \right|_{1}^{2}+\ln \left. \left| {{x}^{2}}+x+1 \right| \right|_{1}^{2}=3+\ln 7-\ln 3 \\\end{align}$

$\Rightarrow a=3,\,\,b=1,\,\,c=-1\Rightarrow a+b+c=3$

$\int\limits_1^4 {\left[ {f\left( x \right) - g\left( x \right)} \right]dx}  = \int\limits_1^4 {f\left( x \right)dx}  - \int\limits_1^4 {g\left( x \right)dx}$ $= 5 - \left( { - 4} \right) = 9$.

Ta có $\int\limits_0^3 {3f\left( x \right)dx}  = 3\int\limits_0^3 {f\left( x \right)dx}  = 3.2 = 6$.

$\int\limits_0^1 {\left[ {f\left( x \right) + g\left( x \right)} \right]dx}  = \int\limits_0^1 {f\left( x \right)dx}  + \int\limits_0^1 {g\left( x \right)dx}  = 2 + \left( { - 4} \right) =  - 2$.

Ta có: $\int\limits_1^4 {\left[ {f\left( x \right) - g\left( x \right)} \right]dx}  =$$\int\limits_1^4 {f\left( x \right)dx}  - \int\limits_1^4 {g\left( x \right)dx = 4 - \left( { - 3} \right) = 7}$

Ta có: $\int\limits_0^2 {2f\left( x \right)dx}  = 4$ $\Leftrightarrow 2\int\limits_0^2 {f\left( x \right)dx}  = 4$$\Leftrightarrow \int\limits_0^2 {f\left( x \right)dx}  = 2$

$\Rightarrow I = \int\limits_{ - 2}^2 {f\left( x \right)dx}  = \int\limits_{ - 2}^0 {f\left( x \right)dx}  + \int\limits_0^2 {f\left( x \right)dx}  = 3 + 2 = 5.$

Ta có: $\int\limits_1^2 {2f\left( x \right)dx}  = 2\int\limits_1^2 {f\left( x \right)dx}  = 2.3 = 6.$

Ta có:  $\int\limits_2^6 {\left[ {3f\left( x \right) - g\left( x \right)} \right]dx}$ $= 3\int\limits_2^6 {f\left( x \right)dx}  - \int\limits_2^6 {g\left( x \right)dx}$$= 3.4 - 5 = 7.$

Ta có: $\int\limits_0^2 {\left[ {2f\left( x \right) - 1} \right]dx}  = 2\int\limits_0^2 {f\left( x \right)dx}  - \int\limits_0^2 {1dx}$ $= 2.6 - \left. x \right|_0^2 = 12 - \left( {2 - 0} \right) = 10$.

Ta có: $\int\limits_0^2 {f\left( x \right)dx}$ $= \int\limits_0^1 {f\left( x \right)dx}  + \int\limits_1^2 {f\left( x \right)dx}$$= 2 + 6 = 8.$

Cho hàm số $F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)$ trên đoạn $\left[ {a;\,\,b} \right].$

Khi đó ta có: $\int\limits_a^b {f\left( x \right)dx}  = \left. {F\left( x \right)} \right|_a^b = F\left( b \right) - F\left( a \right).$

Cách 1: Không đi tìm hàm $F\left( x \right)$.

Ta có:

$\begin{array}{l}P = F\left( { - 1} \right) + 2F\left( 2 \right)\\\,\,\,\,\, = \left[ {F\left( { - 1} \right) - F\left( 0 \right)} \right] + 2\left[ {F\left( 2 \right) - F\left( 0 \right)} \right] + 3F\left( 0 \right)\\\,\,\,\,\,\, = \int\limits_0^{ - 1} {f\left( x \right)dx}  + 2\int\limits_0^2 {f\left( x \right)dx}  + 3F\left( 0 \right)\end{array}$

(Hàm số $F\left( x \right)$ là hàm số thay đổi công thức tại $x = 1$, nhưng liên tục tại $x = 1$, nên việc ta khẳng định $\int\limits_0^2 {f\left( x \right)dx}  = F\left( 2 \right) - F\left( 0 \right)$ là hoàn toàn chặt chẽ bản chất và việc phân đoạn tích phân vẫn đúng).

$\begin{array}{l} \Rightarrow P = \int\limits_0^{ - 1} {f\left( x \right)dx}  + 2\left[ {\int\limits_0^1 {f\left( x \right)dx}  + \int\limits_1^2 {f\left( x \right)dx} } \right] + 3.2\\\,\,\,\,\,\,\,\,\,\,\, = \int\limits_0^{ - 1} {\left( {3{x^2} + 2} \right)dx}  + 2\left[ {\int\limits_0^1 {\left( {3{x^2} + 2} \right)dx}  + \int\limits_1^2 {\left( {2x + 3} \right)dx} } \right] + 6\\\,\,\,\,\,\,\,\,\,\,\, = 21\end{array}$

Cách 2: Tìm hàm $F\left( x \right)$.

$f\left( x \right) = \left\{ \begin{array}{l}2x + 3\\3{x^2} + 2\end{array} \right. \Rightarrow F\left( x \right) = \int {f\left( x \right)dx}  = \left\{ \begin{array}{l}{x^2} + 3x + {C_1}\,\,khi\,\,x \ge 1\\{x^3} + 2x + {C_2}\,\,khi\,\,x < 1\end{array} \right.$.

+ Vì $F\left( 0 \right) = 2 \Rightarrow {0^3} + 2.0 + {C_2} = 2 \Leftrightarrow {C_2} = 2$.

+ Theo giả thiết, $F\left( x \right)$ là hàm số tồn tại đạo hàm trên $\mathbb{R}$.

$\Rightarrow F\left( x \right)$ tồn tại đạo hàm tại $x = 1 \Rightarrow F\left( x \right)$ liên tục tại $x = 1$.

$\Rightarrow F\left( {{1^ + }} \right) = F\left( {{1^ - }} \right) = F\left( 1 \right) \Rightarrow 1 + 3 + {C_1} = 1 + 2 + {C_2}$ $\Rightarrow {C_1} =  - 1 + {C_2} = 1$.

$\begin{array}{l} \Rightarrow F\left( x \right) = \left\{ \begin{array}{l}{x^2} + 3x + 1\,\,khi\,\,x \ge 1\\{x^3} + 2x + 2\,\,khi\,\,x < 1\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}F\left( { - 1} \right) = {\left( { - 1} \right)^3} + 2.\left( { - 1} \right) + 2 =  - 1\\F\left( 2 \right) = {2^2} + 3.2 + 1 = 11\end{array} \right.\\ \Rightarrow P = F\left( { - 1} \right) + 2F\left( 2 \right) =  - 1 + 2.11 = 21\end{array}$

Ta có: $\int\limits_1^3 {\left[ {f\left( x \right) + 1} \right]} dx$ $= \int\limits_1^3 {f\left( x \right)dx}  + \int\limits_1^3 {dx}$$= 4 + \left. x \right|_1^3$$= 4 + 3 - 1 = 6.$

Ta có: $\int\limits_0^1 {f'\left( x \right)dx}  = 5$ $\Leftrightarrow \left. {f\left( x \right)} \right|_0^1 = 5$ $\Leftrightarrow f\left( 1 \right) - f\left( 0 \right) = 5$$\Leftrightarrow f\left( 1 \right) = 5 + f\left( 0 \right) = 5 + 2 = 7$

Ta có: $\int\limits_0^3 {4f\left( x \right)dx}  = 4\int\limits_0^3 {f\left( x \right)dx = 4.3 = 12}$

Ta có: $\int\limits_1^2 {\left[ {2f\left( x \right) + g\left( x \right)} \right]} dx = 13$

$\begin{array}{l} \Leftrightarrow 2\int\limits_1^2 {f\left( x \right)dx}  + \int\limits_1^2 {g\left( x \right)dx}  = 13\\ \Leftrightarrow 2.5 + \int\limits_1^2 {g\left( x \right)dx}  = 13\\ \Leftrightarrow 10 + \int\limits_1^2 {g\left( x \right)dx}  = 13\\ \Leftrightarrow \int\limits_1^2 {g\left( x \right)dx}  = 13 - 10 = 3.\end{array}$