Tích phân các hàm số cơ bản

+) $\int\limits_1^2 {{e^x}dx} = \left. {{e^x}} \right|_1^2 = {e^2} - e$

+) \(\int\limits_0^1 {2dx}  = \left. {2x} \right|_0^1 = 2\),

+) \(\int\limits_0^1 {xdx}  = \left. {\dfrac{{{x^2}}}{2}} \right|_0^1 = \dfrac{1}{2}\)

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

Vậy chỉ có đáp án B là có tích phân bằng \(2\).

\(I = \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{{{\cos }^2}x{{\sin }^2}x}}}  = \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\dfrac{{4dx}}{{{{\sin }^2}2x}}}  = \left. { - 2\cot 2x} \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} =  - 2\left( {0 - \dfrac{1}{{\sqrt 3 }}} \right) = \dfrac{2}{{\sqrt 3 }} = \dfrac{{2\sqrt 3 }}{3} \Rightarrow \left\{ \begin{array}{l}a = 0\\b = \dfrac{2}{3}\end{array} \right. \Rightarrow a - b =  - \dfrac{2}{3}\)

\(\begin{array}{l}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos 2x}}{{1 + \cos x}}dx}  = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{2{{\cos }^2}x - 1}}{{1 + \cos x}}dx}  = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\left( {2\cos x - 2} \right) + \dfrac{1}{{1 + \cos x}}} \right)dx}  = \left. {\left( {2\sin x - 2x} \right)} \right|_0^{\dfrac{\pi }{2}} + \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{1 + \cos x}}dx} \\ = 2 - \pi  + \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{1 + \cos x}}dx} \\ = 2 - \pi  + \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{2{{\cos }^2}\dfrac{x}{2}}}dx} \\ = 2 - \pi  + \dfrac{1}{2}.2.\left. {\tan \dfrac{x}{2}} \right|_0^{\dfrac{\pi }{2}}\\ = 2 - \pi  + 1 = 3 - \pi \\ \Rightarrow \left\{ \begin{array}{l}a = 0\\b =  - 1\\c = 3\end{array} \right. \Rightarrow A = ab + bc + ca =  - 3\end{array}\)

Cách 1:

\(\int\limits_{ - 1}^1 {\dfrac{x}{{{x^2} - 5\left| x \right| + 6}}dx}  = \int\limits_{ - 1}^0 {\dfrac{x}{{{x^2} + 5x + 6}}dx}  + \int\limits_0^1 {\dfrac{x}{{{x^2} - 5x + 6}}dx}  = {I_1} + {I_2}\)

Xét tích phân \({I_1} = \int\limits_{ - 1}^0 {\dfrac{x}{{{x^2} + 5x + 6}}dx} \) ta có:

$\begin{array}{l}\dfrac{x}{{{x^2} + 5x + 6}} = \dfrac{x}{{\left( {x + 2} \right)\left( {x + 3} \right)}} = \dfrac{A}{{x + 2}} + \dfrac{B}{{x + 3}} = \dfrac{{Ax + 3A + Bx + 2B}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\\ \Rightarrow \left\{ \begin{array}{l}A + B = 1\\3A + 2B = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A =  - 2\\B = 3\end{array} \right. \Rightarrow \dfrac{x}{{{x^2} + 5x + 6}} = \dfrac{{ - 2}}{{x + 2}} + \dfrac{3}{{x + 3}}\\ \Rightarrow {I_1} = \int\limits_{ - 1}^0 {\left( {\dfrac{{ - 2}}{{x + 2}} + \dfrac{3}{{x + 3}}} \right)dx}  = \left. {\left( { - 2\ln \left| {x + 2} \right| + 3\ln \left| {x + 3} \right|} \right)} \right|_{ - 1}^0\\ =  - 2\ln 2 + 3\ln 3 - 3\ln 2 = 3\ln 3 - 5\ln 2\end{array}$ 

Xét tích phân \({I_2} = \int\limits_0^1 {\dfrac{x}{{{x^2} - 5x + 6}}dx} \) ta có :

\(\begin{array}{l}\dfrac{x}{{{x^2} - 5x + 6}} = \dfrac{x}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{A}{{x - 2}} + \dfrac{B}{{x - 3}} = \dfrac{{Ax - 3A + Bx - 2B}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\ \Rightarrow \left\{ \begin{array}{l}A + B = 1\\ - 3A - 2B = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A =  - 2\\B = 3\end{array} \right. \Rightarrow \dfrac{x}{{{x^2} - 5x + 6}} = \dfrac{{ - 2}}{{x - 2}} + \dfrac{3}{{x - 3}}\\ \Rightarrow {I_2} = \int\limits_0^1 {\left( {\dfrac{{ - 2}}{{x - 2}} + \dfrac{3}{{x - 3}}} \right)dx}  = \left. {\left( { - 2\ln \left| {x - 2} \right| + 3\ln \left| {x - 3} \right|} \right)} \right|_0^1\\ = 3\ln 2 + 2\ln 2 - 3\ln 3 = 5\ln 2 - 3\ln 3\end{array}\)

\( \Rightarrow I = {I_1} + {I_2} = 0\)

Ta có:

\(\begin{array}{l}I = \int\limits_{ - 1}^0 {\dfrac{{3{x^2} + 5x - 1}}{{x - 2}}dx}  = \int\limits_{ - 1}^0 {\left( {3x + 11 + \dfrac{{21}}{{x - 2}}} \right)dx}  = \left. {\left( {\dfrac{{3{x^2}}}{2} + 11x + 21\ln \left| {x - 2} \right|} \right)} \right|_{ - 1}^0 = 21\ln 2 + \dfrac{{19}}{2} - 21\ln 3 = 21\ln \dfrac{2}{3} + \dfrac{{19}}{2}\\ \Rightarrow \left\{ \begin{array}{l}a = 21\\b = \dfrac{{19}}{2}\end{array} \right. \Rightarrow a + 2b = 21 + 19 = 40\end{array}\)

$\begin{array}{l}\tan \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{\tan x - 1}}{{1 + \tan x}} = \dfrac{{\dfrac{{\sin x}}{{\cos x}} - 1}}{{1 + \dfrac{{\sin x}}{{\cos x}}}} = \dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}\\\cos 2x = {\cos ^2}x - {\sin ^2}x =  - \left( {\sin x + \cos x} \right)\left( {\sin x - \cos x} \right)\\ \Rightarrow \dfrac{{\tan \left( {x - \dfrac{\pi }{4}} \right)}}{{\cos 2x}} = \dfrac{{ - 1}}{{{{\left( {\sin x + \cos x} \right)}^2}}} = \dfrac{{ - 1}}{{2{{\sin }^2}\left( {x + \dfrac{\pi }{4}} \right)}}\\ \Rightarrow I =  - \int\limits_0^{\dfrac{\pi }{6}} {\dfrac{{dx}}{{2{{\sin }^2}\left( {x + \dfrac{\pi }{4}} \right)}}}  = \left. {\dfrac{1}{2}\cot \left( {x + \dfrac{\pi }{4}} \right)} \right|_0^{\dfrac{\pi }{6}} = \dfrac{1}{2}\left( {2 - \sqrt 3  - 1} \right) = \dfrac{1}{2}\left( {1 - \sqrt 3 } \right)\\ \Rightarrow T = 2I + \sqrt 3  = 1 - \sqrt 3  + \sqrt 3  = 1.\end{array}$

Đặt $F\left( x \right) = \int\limits_0^{{x^2}} {f\left( t \right)dt}  = x\sin \left( {\pi x} \right) \Leftrightarrow F\left( {{x^2}} \right) - F\left( 0 \right) = x\sin \left( {\pi x} \right) \Leftrightarrow F\left( {{x^2}} \right) = F\left( 0 \right) + x\sin \left( {\pi x} \right)$

Lấy đạo hàm hai vế ta có

$\begin{array}{l}{\rm{\;}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\left( {F\left( {{x^2}} \right)} \right)^\prime } = \sin \left( {\pi x} \right) + \pi x\cos \left( {\pi x} \right) + {\left( {F\left( 0 \right)} \right)^\prime }\\{\rm{ \;}} \Leftrightarrow 2xf\left( {{x^2}} \right) = \sin \left( {\pi x} \right) + \pi x\cos \left( {\pi x} \right)\end{array}$

Thay $x = 2$ ta có $2.2.f\left( 4 \right) = \sin \left( {2\pi {\rm{\;}}} \right) + 2\pi \cos \left( {2\pi {\rm{\;}}} \right) \Leftrightarrow 4f\left( 4 \right) = 2\pi  \Leftrightarrow f\left( 4 \right) = \dfrac{\pi }{2}$

Ta có: \(F\left( x \right) = \int {f\left( x \right)} dx = {x^2}\)

\(\begin{array}{l} \Rightarrow \int\limits_1^3 {\left[ {1 + f\left( x \right)} \right]dx}  = \int\limits_1^3 {dx}  + \int\limits_1^3 {f\left( x \right)dx} \\ = \left. x \right|_1^3 + \left. {{x^2}} \right|_1^3 = 3 - 1 + {3^2} - 1 = 10.\end{array}\)

Ta có \(\int\limits_{ - 1}^a {{e^{x + 1}}dx}  = \left. {{e^{x + 1}}} \right|_{ - 1}^a = {e^{a + 1}} - 1\).

Vậy yêu cầu bài toán trở thành \({e^{a + 1}} - 1 = {e^2} - 1{\rm{  }}\Leftrightarrow a+1=2 \Leftrightarrow {\rm{  }}a = 1\).

Tính tích phân cho từng hàm số trong các đáp án:

+) $\int\limits_0^\pi  {\cos 3xdx}  = \left. {\dfrac{1}{3}\sin 3x} \right|_0^\pi  = 0$,

+) $\int\limits_0^\pi  {\sin 3xdx}  =  - \left. {\dfrac{1}{3}\cos 3x} \right|_0^\pi  = \dfrac{2}{3}$,

+) $\int\limits_0^\pi  {\cos \left( {\dfrac{x}{4} + \dfrac{\pi }{2}} \right)dx}  = \left. {4\sin \left( {\dfrac{x}{4} + \dfrac{\pi }{2}} \right)} \right|_0^\pi  = 2\left( {\sqrt 2  - 2} \right)$,

+) $\int\limits_0^\pi  {\sin \left( {\dfrac{x}{4} + \dfrac{\pi }{2}} \right)dx}  = \left. { - 4\cos \left( {\dfrac{x}{4} + \dfrac{\pi }{2}} \right)} \right|_0^\pi  = 2\sqrt 2 $.

Vậy chọn \(f(x) = \cos 3x\).

+) \(\int\limits_0^1 {2dx}  = \left. {2x} \right|_0^1 = 2\),

+) \(\int\limits_0^2 {xdx}  = \left. {\dfrac{{{x^2}}}{2}} \right|_0^2 = 2\)

+) \(\int\limits_0^\pi  {\sin xdx}  = \left. { - \cos x} \right|_0^\pi  = 2\)

Do đó ta dự đoán chỉ có đáp án A là kết quả khác \(2\).

\(I = \int\limits_2^5 {\dfrac{{dx}}{x}}  = \left. {\ln \left| x \right|} \right|_2^5 = \ln 5 - \ln 2 = \ln \dfrac{5}{2}\)

Cách 1:

\(\begin{array}{l}I = \int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\dfrac{{dx}}{{\sin x}}} \\= \int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\dfrac{{\left( {{{\cos }^2}\dfrac{x}{2} + {{\sin }^2}\dfrac{x}{2}} \right)}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}dx} \\ = \dfrac{1}{2}\int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\left( {\cot \dfrac{x}{2} + \tan \dfrac{x}{2}} \right)dx} \\ = \left. {\left[ {\ln \left| {\sin \dfrac{x}{2}} \right| - \ln \left| {\cos \dfrac{x}{2}} \right|} \right]} \right|_{\frac{\pi }{3}}^{\frac{\pi }{2}}\\ = \left[ {\ln \dfrac{{\sqrt 2 }}{2} - \ln \dfrac{{\sqrt 2 }}{2}} \right] - \left[ {\ln \dfrac{1}{2} - \ln \dfrac{{\sqrt 3 }}{2}} \right]\\ = \ln \sqrt 3 .\end{array}\)

Cách 2:

Bước 1: Dùng máy tính như hình dưới, thu được giá trị \(0,549306...\)

Bước 2: Lấy \({e^{0,549306...}}\) cho kết quả \(1,732050808... \approx \sqrt 3 \). Chọn \(\dfrac{1}{2}\ln 3\).

Cách 3:

Thực hiện các phép tính sau trên máy tính (đến khi thu được kết quả bằng \(0\) thì ngưng)

Chọn \(\dfrac{1}{2}\ln 3\).

\(K = \int\limits_{ - 2}^0 {\left( {4 - {e^{-\frac{x}{2}}}} \right)dx}  + 2e = \left. {\left( {4x + 2{e^{-\frac{x}{2}}}} \right)} \right|_{ - 2}^0 + 2e = 2 - \left( { - 8 + 2e} \right) + 2e = 10\)

\(\int\limits_0^1 {\dfrac{1}{{{x^2} - x - 2}}dx}  = \int\limits_0^1 {\dfrac{1}{{(x - 2)(x + 1)}}dx}  = \dfrac{1}{3}\int\limits_0^1 {\left[ {\dfrac{1}{{x - 2}} - \dfrac{1}{{x + 1}}} \right]dx}  = \dfrac{1}{3}\left. {\left[ {\ln \left| {x - 2} \right| - \ln \left| {x + 1} \right|} \right]} \right|_0^1 =  - \dfrac{{2\ln 2}}{3}\)

Ta có : \(\int\limits_0^3 {x(x - 1)dx} \)\( = \int\limits_0^3 {\left( {{x^2} - x} \right)dx}  = \left. {\dfrac{{{x^3}}}{3} - \dfrac{{{x^2}}}{2}} \right|_0^3\) \( = 9 - \dfrac{9}{2} = \dfrac{9}{2}\)

+) $\int\limits_0^{\ln \sqrt {10} } {{e^{2x}}dx}  = \left. {\dfrac{{{e^{2x}}}}{2}} \right|_0^{\ln \sqrt {10} } = \dfrac{{{e^{2\ln \sqrt {10} }} - 1}}{2} = \dfrac{9}{2}$,

+) $3\int\limits_0^{3\pi } {\sin xdx}  = \left. { - 3\cos x} \right|_0^{3\pi } = 6$,

+) $\int\limits_0^2 {\left( {{x^2} + x - 3} \right)dx}  = \left. {\left( {\dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2} - 3x} \right)} \right|_0^2 = \dfrac{8}{3} + 2 - 6 =  - \dfrac{4}{3}$,

+) $\int\limits_0^\pi  {\cos (3x + \pi )dx}  = \dfrac{1}{3}\left. {\sin (3x + \pi )} \right|_0^\pi  = \dfrac{1}{3}\left( {\sin 4\pi  - \sin \pi } \right) = 0$.

Vậy chọn $\int\limits_0^{\ln \sqrt {10} } {{e^{2x}}dx} $

Ta có: $I = \int\limits_1^2 {{x^5}} dx = \left. {\dfrac{{{x^6}}}{6}} \right|_1^2 = \dfrac{{21}}{2}$.

Ta có

$\dfrac{x}{{{{(x + 1)}^3}}} = \dfrac{{x + 1 - 1}}{{{{(x + 1)}^3}}} = {(x + 1)^{ - 2}} - {(x + 1)^{ - 3}}$ 

$ \Rightarrow I = \int\limits_0^1 {\left[ {{{(x + 1)}^{ - 2}} - {{(x + 1)}^{ - 3}}} \right]} dx$$ = \left. {\left[ { - {{\left( {x + 1} \right)}^{ - 1}} + \dfrac{1}{2}{{\left( {x + 1} \right)}^{ - 2}}} \right]} \right|_0^1 = \dfrac{1}{8}$

$I = \int\limits_0^2 {{x^3}dx}  = \left. {\dfrac{{{x^4}}}{4}} \right|_0^2 = 4$ và $J = \int\limits_0^2 {xdx}  = \left. {\dfrac{{{x^2}}}{2}} \right|_0^2 = 2$.

Suy ra \(I.J = 8\).

$I = \int\limits_0^2 {{x^3}dx}  = 4$ và $J = \int\limits_0^2 {xdx}  = 2$, suy ra \(I.J = 8\).