Tích phân các hàm số cơ bản

$\dfrac{{4{{\sin }^3}x}}{{1 + \cos x}} = \dfrac{{4{{\sin }^3}x(1 - \cos x)}}{{{{\sin }^2}x}} = 4\sin x - 4\sin x\cos x = 4\sin x - 2\sin 2x$

$ \Rightarrow I = \int_0^{\dfrac{\pi }{2}} {(4\sin x - 2\sin 2x)} dx = 2.$

Phương pháp tự luận

$\begin{array}{c}I = \int\limits_0^{2\pi } {\sqrt {{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}^2}} } dx = \int\limits_0^{2\pi } {\left| {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right|} dx = \sqrt 2 \int\limits_0^{2\pi } {\left| {\sin \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)} \right|} dx\\ = \sqrt 2 \left[ {\int\limits_0^{\dfrac{{3\pi }}{2}} {\sin \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)} dx - \int\limits_{\dfrac{{3\pi }}{2}}^{2\pi } {\sin \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)dx} } \right] = 4\sqrt 2 \end{array}$

$\begin{array}{c}\int\limits_{ - 1}^5 {\left| {{x^2} - 2x - 3} \right|dx}  = \int\limits_{ - 1}^5 {\left| {(x - 3)(x + 1)} \right|dx}  =  - \int\limits_{ - 1}^3 {\left( {{x^2} - 2x - 3} \right)dx}  + \int\limits_3^5 {\left( {{x^2} - 2x - 3} \right)dx} \\ =  - \left. {\left( {\dfrac{{{x^3}}}{3} - {x^2} - 3x} \right)} \right|_{ - 1}^3 + \left. {\left( {\dfrac{{{x^3}}}{3} - {x^2} - 3x} \right)} \right|_3^5 = \dfrac{{64}}{3}.\end{array}$

$\int\limits_2^3 {\dfrac{{{x^2} - x + 4}}{{x + 1}}dx}  = \int\limits_2^3 {\left( {x - 2 + \dfrac{6}{{x + 1}}} \right)dx}  = \left. {\left( {\dfrac{{{x^2}}}{2} - 2x + 6\ln \left| {x + 1} \right|} \right)} \right|_2^3 = \dfrac{1}{2} + 6\ln \dfrac{4}{3}$.

Ta có:

$\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$

\( = \left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right]\)\(\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x{{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]\)

\( = \left( {1 - \dfrac{1}{2}{{\sin }^2}2x} \right)\left( {1 - \dfrac{3}{4}{{\sin }^2}2x} \right)\)\( = 1 - \dfrac{5}{4}{\sin ^2}2x + \dfrac{3}{8}{\left( {{{\sin }^2}2x} \right)^2}\)\( = 1 - \dfrac{5}{4}.\dfrac{{1 - \cos 4x}}{2} + \dfrac{3}{8}.{\left( {\dfrac{{1 - \cos 4x}}{2}} \right)^2}\)\( = \dfrac{3}{8} + \dfrac{5}{8}\cos 4x + \dfrac{3}{{32}}\left( {1 - 2\cos 4x + {{\cos }^2}4x} \right)\)\( = \dfrac{{15}}{{32}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{32}}{\cos ^2}4x\)\( = \dfrac{{15}}{{32}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{32}}.\dfrac{{1 + \cos 8x}}{2}\)\( = \dfrac{{33}}{{64}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{64}}\cos 8x\)

Do đó $I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{33}}{{64}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{64}}\cos 8x} \right)dx} $$ = \left. {\dfrac{{33}}{{64}}x} \right|_0^{\dfrac{\pi }{2}} + \left. {\dfrac{7}{{64}}\sin 4x} \right|_0^{\dfrac{\pi }{2}} + \left. {\dfrac{3}{{512}}\sin 8x} \right|_0^{\dfrac{\pi }{2}}$$ = \dfrac{{33}}{{128}}\pi $

$\begin{array}{l}f(x) = A\sin\pi x + B \Rightarrow f'(x) = A\pi \cos \pi x\\f'(1) = 2 \Rightarrow A\pi \cos \pi = 2 \Rightarrow A = - \dfrac{2}{\pi }\end{array}$
\(\int\limits_0^2 {f(x)dx = 4} \Rightarrow \int\limits_0^2 {(A\sin\pi x + B)dx = 4} \) \(\Rightarrow - \dfrac{A}{\pi }\cos 2\pi + 2B + \dfrac{A}{\pi }\cos 0 = 4 \Rightarrow B = 2\)

Ta có: \(\int\limits_2^4 {2xdx}  = \left. {{x^2}} \right|_2^4 = 12\)

\(\int\limits_1^2 {\left[ {{a^2} + (4 - 4a)x + 4{x^3}} \right]dx = } \)\( = \left. {\left[ {{a^2}x + (2 - 2a){x^2} + {x^4}} \right]} \right|_1^2 = {a^2} - 6a + 21\)

\( \Rightarrow {a^2} - 6a + 21 = 12 \Leftrightarrow a = 3.\)

Xét hiệu số \(1 - {x^2}\) trên đoạn ${\rm{[}}0;2]$ để tìm $\min \left\{ {1,{x^2}} \right\}$.

Ta có: $1 - {x^2} \ge 0 \Leftrightarrow 0 \le x \le 1$ nên với $0 \le x \le 1$ thì $1 \ge {x^2} \Rightarrow \min \left\{ {1;{x^2}} \right\} = {x^2}$.

Với $1 \le x \le 2$ thì $1 \le {x^2} \Rightarrow \min \left\{ {1;{x^2}} \right\} = 1$.

Vậy $I = \int\limits_0^2 {\min \left\{ {1,{x^2}} \right\}dx}  = \int\limits_0^1 {{x^2}dx}  + \int\limits_1^2 {dx}  = \left. {\frac{{{x^3}}}{3}} \right|_0^1 + \left. x \right|_1^2 = \frac{4}{3}.$

Do hàm số $f(x) = \sqrt {1 - \cos 2x}  = \sqrt 2 \left| {\sin x} \right|$là hàm liên tục và tuần hoàn với chu kì $T = \pi $ nên ta có

\(\begin{array}{l}\int\limits_0^T {f\left( x \right)dx = \int\limits_T^{2T} {f\left( x \right)dx} }  = \int\limits_{2T}^{3T} {f\left( x \right)dx} \\ = ... = \int\limits_{\left( {n - 1} \right)T}^{nT} {f\left( x \right)dx} \\ \Rightarrow \int\limits_0^{nT} {f\left( x \right)dx}  = \int\limits_0^T {f\left( x \right)dx}  \\+ \int\limits_T^{2T} {f\left( x \right)dx}+ \int\limits_{2T}^{3T} {f\left( x \right)dx}   + ... + \\ \int\limits_{\left( {n - 1} \right)T}^{nT} {f\left( x \right)dx}  = n\int\limits_{0}^{T} {f\left( x \right)dx} \\ \Rightarrow \int\limits_0^{2017\pi } {\sqrt {1 - \cos 2x} dx}  \\= 2017\int\limits_0^\pi  {\sqrt {1 - \cos 2x} dx} \\ = 2017\sqrt 2 \int\limits_0^\pi  {\sin x dx = 4034\sqrt 2 } \end{array}\)

Ta có

\(I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\cos 2x}}{{{{\left( {\sin x - \cos x + 3} \right)}^2}}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\sin x - \cos x + 3} \right)}^2}}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{{{\left( {\sin x - \cos x + 3} \right)}^2}}}dx} \)

Đặt \(\sin x - \cos x + 3 = t \Leftrightarrow \left\{ \begin{array}{l}\left( {\cos x + \sin x} \right)dx = dt\\\cos x - \sin x = 3 - t\end{array} \right.\)

Đổi cận \(x = 0 \Rightarrow t = 2;\,x = \dfrac{\pi }{4} \Rightarrow t = 3\)

Suy ra \(I = \int\limits_2^3 {\dfrac{{\left( {3 - t} \right)dt}}{{{t^2}}} = \int\limits_2^3 {\left( {\dfrac{3}{{{t^2}}} - \dfrac{1}{t}} \right)dt}  = \left. {\left( { - \dfrac{3}{t} - \ln \left| t \right|} \right)} \right|_2^3}  = \dfrac{1}{2} + \ln 2 - \ln 3 = \dfrac{1}{2} + \ln \dfrac{2}{3}\)

Hay \(a = \dfrac{1}{2};b = \dfrac{2}{3} \Rightarrow 2a + 3b = 3.\)

Ta có \(g\left( x \right) = 1 + 2\int\limits_0^x {f\left( t \right)dt} \)  suy ra \(\left\{ \begin{array}{l}g\left( x \right) - 1 = 2\int\limits_0^x {f\left( t \right)dt} \\g\left( 0 \right) = 1 + \int\limits_0^0 {f\left( t \right)dt} \end{array} \right. \)\(\Leftrightarrow \left\{ \begin{array}{l}g'\left( x \right) = 2f\left( x \right) \Rightarrow f\left( x \right) = \dfrac{{g'\left( x \right)}}{2}\\g\left( 0 \right) = 1\end{array} \right.\)

Mà

\(g\left( x \right) \ge {\left[ {f\left( x \right)} \right]^3} \Leftrightarrow g\left( x \right) \ge {\left[ {\dfrac{{g'\left( x \right)}}{2}} \right]^3} \)\(\Leftrightarrow \sqrt[3]{{g\left( x \right)}} \ge \dfrac{{g'\left( x \right)}}{2} \Leftrightarrow \dfrac{{g'\left( x \right)}}{{\sqrt[3]{{g\left( x \right)}}}} \le 2\)

Với \(t \in \left[ {0;1} \right]\), Lấy tích phân hai vế ta được

\(\begin{array}{l}\int\limits_0^t {\dfrac{{g'\left( x \right)}}{{\sqrt[3]{{g\left( x \right)}}}}} dx \le \int\limits_0^t {2dx} \\ \Leftrightarrow \int\limits_0^t {{{\left[ {g\left( x \right)} \right]}^{\dfrac{{ - 1}}{3}}}} d\left( {g\left( x \right)} \right) \le 2t\\ \Leftrightarrow 2t \ge \dfrac{3}{2}\left. {{{\left[ {g\left( x \right)} \right]}^{\dfrac{2}{3}}}} \right|_0^t \\\Leftrightarrow \dfrac{4}{3}t \ge \sqrt[3]{{{g^2}\left( t \right)}} - \sqrt[3]{{{g^2}\left( 0 \right)}}\end{array}\)

Mà \(g\left( 0 \right) = 1\) nên \(\sqrt[3]{{{g^2}\left( t \right)}} \le \dfrac{4}{3}t + 1 \Rightarrow \sqrt[3]{{{g^2}\left( x \right)}} \le \dfrac{4}{3}x + 1\)

Từ đó ta có \(\int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \int\limits_0^1 {\left( {\dfrac{4}{3}x + 1} \right)dx} \)\( \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \left. {\left( {\dfrac{2}{3}{x^2} + x} \right)} \right|_0^1\\ \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \dfrac{5}{3}\)

Hay giá trị lớn nhất cần tìm là \(\dfrac{5}{3}.\)

Theo đề bài ta có

 \(\begin{array}{l}\left\{ \begin{array}{l}f\left( 1 \right) = 10\\f\left( 2 \right) = 20\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}1 + a + b + c = 10\\{2^3} + {2^2}.a + 2b + c = 20\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a + b + c = 9\\4a + 2b + c = 12\end{array} \right. \Rightarrow 3a + b = 3\end{array}\)

\(\begin{array}{l} \Rightarrow \int\limits_0^3 {f'\left( x \right)dx}  = \left. {f\left( x \right)} \right|_0^3 = f\left( 3 \right) - f\left( 0 \right)\\ = {3^3} + {3^2}.a + 3b + c - c = 27 + 9a + 3b\\ = 27 + 3\left( {3a + b} \right) = 27 + 3.3 = 36.\end{array}\)

Ta có \(f\left( x \right) = \int {f'\left( x \right)dx}  = \int {{{\sin }^4}xdx} \)

\(\begin{array}{l} = \int {{{\left( {\dfrac{{1 - \cos 2x}}{2}} \right)}^2}dx} \\ = \dfrac{1}{4}\int {\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)dx} \\ = \dfrac{1}{4}\int {\left( {1 - 2\cos 2x + \dfrac{{1 + \cos 4x}}{2}} \right)dx} \\ = \dfrac{1}{4}\left( {x - \sin 2x + \dfrac{1}{2}x + \dfrac{1}{2}.\dfrac{{\sin 4x}}{4}} \right) + C\\ = \dfrac{{3x}}{8} - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 4x}}{{32}} + C\end{array}\) 

Theo bài ra ta có \(f\left( 0 \right) = 0 \Leftrightarrow C = 0\) \( \Rightarrow f\left( x \right) = \dfrac{{3x}}{8} - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 4x}}{{32}}\).

Vậy \(\int\limits_0^{\frac{\pi }{2}} {f\left( x \right)dx}  = \int\limits_0^{\frac{\pi }{2}} {\left( {\dfrac{{3x}}{8} - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 4x}}{{32}}} \right)dx}  = \dfrac{{3{\pi ^2} - 16}}{{64}}\) (sử dụng MTCT).

Ta có \(v\left( t \right) = \int {a\left( t \right)dt}  = \int {\left( {6 - 3t} \right)dt}  = 6t - \dfrac{{3{t^2}}}{2} + C\).

Theo bài ra ta có: Ô tô đang đứng yên và bắt đầu chuyển động, do đó \(v\left( 0 \right) = 0\) \( \Rightarrow C = 0\).

Khi đó ta có \(v\left( t \right) = 6t - \dfrac{3}{2}{t^2}\), đây là một parabol có bề lõm hướng xuống, đạt giá trị lớn nhất tại \(t = \dfrac{{ - b}}{{2a}} = \dfrac{{ - 6}}{{2.\left( { - \dfrac{3}{2}} \right)}} = 2\).

Vậy quãng đường ô tô đi được từ khi chuyển động đến khi vận tốc của ô tô đạt giá trị lớn nhất là:

\(S = \int\limits_0^2 {v\left( t \right)dt}  = \int\limits_0^2 {\left( {6t - \dfrac{3}{2}{t^2}} \right)dt}  = 8\,\,\left( m \right).\)

Ta có: \(v\left( t \right) = 2t + 3\,\,\left( {m/s} \right)\)

Khi đó quãng đường của vật được tính từ giây thứ nhất đến giây thứ năm là:

\(s\left( t \right) = \int\limits_1^5 {\left( {2t + 3} \right)dt} \) \( = \left. {\left( {{t^2} + 3t} \right)} \right|_1^5\) \( = {5^2} + 3.5 - 1 - 3 = 36\,\,(m).\)

Bước 1:

Theo bài ra ta có:

\(2xf'\left( x \right) = f\left( x \right) + {x^2}\)

\( \Leftrightarrow \sqrt {2x} f'\left( x \right) - \dfrac{1}{{\sqrt {2x} }}f\left( x \right) = \dfrac{{{x^2}}}{{\sqrt {2x} }}\)

\( \Leftrightarrow \dfrac{{f'\left( x \right)\sqrt {2x} {\rm{\;}} - \dfrac{1}{{\sqrt {2x} }}f\left( x \right)}}{{2x}} = \dfrac{{{x^2}}}{{2x\sqrt {2x} }}\)

\( \Leftrightarrow \dfrac{{f'\left( x \right)\sqrt {2x} {\rm{\;}} - f\left( x \right){{\left( {\sqrt {2x} } \right)}^\prime }}}{{2x}} = \dfrac{x}{{2\sqrt {2x} }}\)

\( \Leftrightarrow {\left( {\dfrac{{f\left( x \right)}}{{\sqrt {2x} }}} \right)^\prime } = \dfrac{1}{{2\sqrt 2 }}\sqrt x \)

Bước 2:

Lấy nguyên hàm hai vế ta được:

\(\begin{array}{l}\int {\left( {\dfrac{{f\left( x \right)}}{{\sqrt {2x} }}} \right)'dx}  = \int {\dfrac{1}{{2\sqrt 2 }}\sqrt x dx} \\ \Leftrightarrow \dfrac{{f\left( x \right)}}{{\sqrt {2x} }} = \dfrac{1}{{2\sqrt 2 }}.\dfrac{2}{3}x\sqrt x  + C\\ \Leftrightarrow f\left( x \right) = \dfrac{1}{{3\sqrt 2 }}\sqrt {2x} .x\sqrt x  + C\sqrt {2x} \\ \Leftrightarrow f\left( x \right) = \dfrac{1}{3}{x^2} + C\sqrt {2x} \end{array}\)

Bước 3:

Ta lại có \(f\left( 1 \right) = \dfrac{1}{3} + C\sqrt 2  = 2 \Leftrightarrow C = \dfrac{{5\sqrt 2 }}{6}\) \( \Rightarrow f\left( x \right) = \dfrac{1}{3}{x^2} + \dfrac{{5\sqrt 2 }}{6}\sqrt {2x}  = \dfrac{1}{3}{x^2} + \dfrac{5}{3}\sqrt x \)

Bước 4:

Vậy \(\int\limits_1^4 {f\left( x \right)dx}  = \int\limits_1^4 {\left( {\dfrac{1}{3}{x^2} + \dfrac{5}{3}\sqrt x } \right)dx}  \)\(= \dfrac{{133}}{9}\).

Bước 1: Tìm a

Tại thời điểm $t=20(s), v(20)=0$ nên $v_{0}+20 a=0 \Rightarrow a=-\dfrac{v_{0}}{20}$

Do đó, $v(t)=v_{0}-\dfrac{v_{0}}{20} t$

Bước 2: Tính \(\int\limits_0^{20} {v\left( t \right)dt} \)

Mặt khác, $v(t)=s^{\prime}(t) \Rightarrow \int_{0}^{20} v(t) \mathrm{d} t=\int_{0}^{20} s^{\prime}(t) \mathrm{d} t=\left.s(t)\right|_{0} ^{20}=s(20)-s(0)=120$

Bước 3: Tính \({v_0}\)

Suy ra, $\int_{0}^{20}\left(v_{0}-\dfrac{v_{0}}{20} t\right) \mathrm{d} t=\left.120 \Rightarrow\left(v_{0} t-\dfrac{v_{0}}{40} t^{2}\right)\right|_{0} ^{20}=120$

Từ đó ta có phương trình $20 v_{0}-10 v_{0}=120 \Rightarrow v_{0}=12(\mathrm{~m} / \mathrm{s})$