Tích phân (Khái niệm và tính chất)

$\int\limits_2^7 {f\left( z \right)dz} {\rm{\;}} = \int\limits_2^7 {f\left( x \right)dx} {\rm{\;}} = \int\limits_2^{ - 1} {f\left( x \right)dx} {\rm{\;}} + \int\limits_{ - 1}^7 {f\left( x \right)dx} {\rm{\;}} = {\rm{\;}} - \int\limits_{ - 1}^2 {f\left( x \right)dx} {\rm{\;}} + \int\limits_{ - 1}^7 {f\left( t \right)dt} {\rm{\;}} = {\rm{\;}} - 2 + 9 = 7$

$\begin{array}{l}{\rm{\;}}I = \int\limits_0^1 {\left( {2x - {m^2}} \right)dx}  = \left. {\left( {{x^2} - {m^2}x} \right)} \right|_0^1 = 1 - {m^2}\\{\rm{ \;}}I + 3 \ge 0 \Leftrightarrow 1 - {m^2} + 3 \ge 0 \Leftrightarrow {m^2} \le 4 \Leftrightarrow m \in \left[ { - 2;2} \right]\end{array}$

$m$ là số nguyên dương $ \Rightarrow m \in \left\{ {1;2} \right\}$.

Ta có: $\int\limits_0^2 {f\left( x \right)dx = \dfrac{8}{3}}  \Leftrightarrow \int\limits_0^2 {\left( {A\sin \left( {\pi x} \right) + B{x^2}} \right)dx = \dfrac{8}{3}} $

$\begin{array}{*{20}{l}}{\rm{\;}}&{ \Leftrightarrow \left. {\left( {-\dfrac{A}{\pi }\cos \left( {\pi x} \right) + \dfrac{{B{x^3}}}{3}} \right)} \right|_0^2 = \dfrac{8}{3}}\\{}&{ \Leftrightarrow -\dfrac{A}{\pi } + \dfrac{{8B}}{3} + \dfrac{A}{\pi } = \dfrac{8}{3}}\\{}&{ \Leftrightarrow B = 1.}\end{array}$

Ta có $\int\limits_a^b {\left( {2x - 1} \right){\mkern 1mu} {\rm{d}}x}  = \left. {\left( {{x^2} - x} \right)} \right|_a^b = \left( {{b^2} - {a^2}} \right) - \left( {b - a} \right) = 1 \Leftrightarrow {b^2} - {a^2} = b - a + 1.$

Ta có: $\int\limits_0^1 {x\left( {{x^2} + 3} \right)dx}  = \int\limits_0^1 {\left( {{x^3} + 3x} \right)dx = \left. {\left( {\dfrac{{{x^4}}}{4} + \dfrac{{3{x^2}}}{2}} \right)} \right|_0^1 = \dfrac{1}{4} + \dfrac{3}{2} = \dfrac{7}{4}.} $

\(I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\left| {\sin x} \right|dx}  = \int\limits_{ - \dfrac{\pi }{2}}^0 {\left| {\sin x} \right|dx}  + \int\limits_0^{\dfrac{\pi }{2}} {\left| {\sin x} \right|dx}  \)\(=  - \int\limits_{ - \dfrac{\pi }{2}}^0 {\sin xdx}  + \int\limits_0^{\dfrac{\pi }{2}} {\sin xdx} \)\( = \left. {\cos x} \right|_{ - \dfrac{\pi }{2}}^0 - \left. {\cos x} \right|_0^{\dfrac{\pi }{2}} = 1 - \left( {0 - 1} \right) = 2\)

$\int\limits_0^2 {\dfrac{2}{{2x + 1}}dx}  = 2\int\limits_0^2 {\dfrac{1}{{2x + 1}}dx}  = \left. {\dfrac{2}{2}\ln \left| {2x + 1} \right|} \right|_0^2 = \ln 5 - \ln 1 = \ln 5$

$I = \int\limits_0^1 {\dfrac{{dx}}{{x + 1}}}  = \left. {\ln \left| {x + 1} \right|} \right|_0^1 = \ln 2 - \ln 1 = \ln 2$

Ta có $\int\limits_1^2 {\dfrac{{{\rm{d}}x}}{{x + 1}}}  = \left. {\ln \left| {x + 1} \right|} \right|_1^2 = \ln 3 - \ln 2 = \ln \dfrac{3}{2}.$

Ta có $\int\limits_0^1 {\dfrac{{2{x^2} + 3x + 3}}{{{x^2} + 2x + 1}}{\rm{d}}x}  = \int\limits_0^1 {\dfrac{{2\left( {{x^2} + 2x + 1} \right) - \left( {x + 1} \right) + 2}}{{{x^2} + 2x + 1}}{\rm{d}}x}  = \int\limits_0^1 {\left( {2 - \dfrac{1}{{x + 1}} + \dfrac{2}{{{{\left( {x + 1} \right)}^2}}}} \right){\mkern 1mu} {\rm{d}}x} $

${\mkern 1mu}  = \left. {\left( {2x - \ln \left| {x + 1} \right| - \dfrac{2}{{x + 1}}} \right)} \right|\begin{array}{*{20}{c}}1\\0\end{array} = \left( {2 - \ln 2 - 1} \right) + 2 = 3 - \ln 2 \Rightarrow {\mkern 1mu} {\mkern 1mu} \left\{ {\begin{array}{*{20}{l}}{a = 3}\\{b = 2}\end{array}} \right..$

Vậy $P = {a^2} + {b^2} = 13.$

Các mệnh đề A, B, C đều đúng. Mệnh đề D sai.

Dễ thấy các công thức a) đúng vì tích phân có hai cận bằng nhau thì có giá trị $0$.

Công thức c) là đúng theo tính chất tích phân.

Công thức b) sai vì \(\int\limits_a^b {f\left( x \right)dx}  =  - \int\limits_b^a {f\left( x \right)dx} \)

$I = \int\limits_1^4 {f'(x)dx}  = f\left. {(x)} \right|_1^4 = f(4) - f(1) = 10 - 2 = 8$

Ta có $\int\limits_0^1 {\left[ {3 - 2f\left( x \right)} \right]{\rm{d}}x}  = \int\limits_0^1 {{\rm{3d}}x}  - 2\int\limits_0^1 {f\left( x \right){\rm{d}}x}  = \left. {3x} \right|_0^1 - 2\int\limits_0^1 {f\left( x \right){\rm{d}}x}  = 3 - 2\int\limits_0^1 {f\left( x \right){\rm{d}}x} $

Mặt khác $\int\limits_0^1 {\left[ {3 - 2f\left( x \right)} \right]{\rm{d}}x}  = 5 \Rightarrow 3 - 2\,\int\limits_0^1 {f\left( x \right){\rm{d}}x}  = 5 \Leftrightarrow \int\limits_0^1 {f\left( x \right){\rm{d}}x}  =  - \,1.$

Ta có: \(F\left( x \right) = \int\limits_1^x {tdt}  = \left. {\dfrac{{{t^2}}}{2}} \right|_1^x = \dfrac{{{x^2}}}{2} - \dfrac{1}{2} \Rightarrow F'\left( x \right) = x\)

Ta có: \(F\left( x \right) = \int\limits_1^x {\left( {t + 1} \right)dt}  = \left. {\left( {\dfrac{{{t^2}}}{2} + t} \right)} \right|_1^x = \dfrac{{{x^2}}}{2} + x - \dfrac{1}{2} - 1 = \dfrac{{{x^2}}}{2} + x - \dfrac{3}{2}\)

Hàm số \(y = F\left( x \right)\) là hàm số bậc hai, hệ số \(a > 0\) nên nó đạt GTNN tại \(x =  -1 \in \left[ { - 1;1} \right]\).

Khi đó $F(-1)=\dfrac{1}{2}+(-1)-\dfrac{3}{2}=-2$

Vì \(f\left( x \right) = {x^2} \ge 0,\forall x \in \left[ {0;1} \right]\) nên \(\int\limits_0^1 {f\left( x \right)dx}  \ge 0\). Do đó A đúng, D sai.

Vì \(g\left( x \right) = {x^3} \ge 0,\forall x \in \left[ {0;1} \right]\) nên \(\int\limits_0^1 {g\left( x \right)dx}  \ge 0\). Do đó B sai.

Vì \({x^2} \ge {x^3}\) trên \(\left[ {0;1} \right]\) nên \(\int\limits_0^1 {f\left( x \right)dx}  \ge \int\limits_0^1 {g\left( x \right)dx} \). Do đó C sai.

Dựa vào các đáp án ta có nhận xét sau:

$\int\limits_a^c {f(x)dx = } \int\limits_a^b {f(x)dx}  + \int\limits_b^c {f(x)dx} $ => A đúng

$\int\limits_a^b {f(x)dx = } \int\limits_a^c {f(x)dx}  - \int\limits_b^c {f(x)dx} $ B đúng

$\int\limits_a^b {f(x)dx = } \int\limits_b^a {f(x)dx}  + \int\limits_a^c {f(x)dx} $ C sai

$\int\limits_a^b {cf(x)dx =  - c} \int\limits_b^a {f(x)dx} $  D đúng.

Ta có: \(\int\limits_1^4 {f'\left( x \right)dx}  = 17 \Rightarrow \left. {f\left( x \right)} \right|_1^4 = 17 \Rightarrow f\left( 4 \right) - f\left( 1 \right) = 17 \Rightarrow f\left( 4 \right) - 12 = 17 \Rightarrow f\left( 4 \right) = 29\)

Ta có:

\(\int\limits_5^2 {\left[ {2 - 4f\left( x \right)} \right]dx}  = \int\limits_5^2 {2dx}  - 4\int\limits_5^2 {f\left( x \right)dx}  = \left. {2x} \right|_5^2 + 4\int\limits_2^5 {f\left( x \right)dx}  = 2.2 - 2.5 + 4.10 = 34\)