45 bài tập trắc nghiệm tích phân mức độ vận dụng, vận dụng cao

Ta có: \(g'\left( x \right) = 2f'\left( x \right) - 2\left( {1 - x} \right) = 2\left[ {f'\left( x \right) - 1 + x} \right]\).

\(g'\left( x \right) = 0 \Leftrightarrow f'\left( x \right) = 1 - x\).

Từ hình vẽ ta thấy, trên đoạn \(\left[ { - 4;3} \right]\) thì đường thẳng \(y = 1 - x\) cắt đồ thị hàm số \(y = f'\left( x \right)\) tại \(3\) điểm lần lượt có hoành độ \( - 4; - 1;3\) nên ta so sánh các giá trị \(g\left( { - 4} \right),g\left( { - 1} \right),g\left( 3 \right)\).

Ta thấy: \({S_1} = \int\limits_{ - 4}^{ - 1} {\left[ {1 - x - f'\left( x \right)} \right]dx}  = \int\limits_{ - 4}^{ - 1} {\left[ { - g'\left( x \right)} \right]dx}  = g\left( { - 4} \right) - g\left( { - 1} \right) > 0 \Rightarrow g\left( { - 4} \right) > g\left( { - 1} \right)\).

\({S_2} = \int\limits_{ - 1}^3 {\left[ {f'\left( x \right) - 1 + x} \right]dx}  = \int\limits_{ - 1}^3 {\left[ {g'\left( x \right)} \right]dx}  = g\left( 3 \right) - g\left( { - 1} \right) > 0 \Rightarrow g\left( 3 \right) > g\left( { - 1} \right)\).

Do đó \(g\left( { - 1} \right)\) là GTNN của hàm số \(y = g\left( x \right)\) hay hàm số đạt GTNN tại \({x_0} =  - 1\).

Chọn B

Đặt \(\left\{ \begin{array}{l}u = \ln x\\dv = {x^3}dx\end{array} \right.\)\( \Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{x}dx\\v = \dfrac{{{x^4}}}{4}\end{array} \right.\)

Khi đó \(\int\limits_1^e {{x^3}} \ln xdx = \left. {\ln x.\dfrac{{{x^4}}}{4}} \right|_1^e - \dfrac{1}{4}\int\limits_1^e {{x^3}dx}  = \dfrac{{{e^4}}}{4} - \dfrac{1}{4}.\dfrac{1}{4}\left. {{x^4}} \right|_1^e = \dfrac{{{e^4}}}{4} - \dfrac{1}{{16}}\left( {{e^4} - 1} \right) = \dfrac{{3{e^4} + 1}}{{16}}\).

Do đó \(a = 4,b = 16 \Rightarrow ab = 64\).

Chọn A

Đặt \(\left\{ \begin{array}{l}u = {x^4} - 8x\\dv = f\left( x \right)dx\end{array} \right.\)\( \Rightarrow \left\{ \begin{array}{l}du = 4{x^3} - 8\\v = F\left( x \right)\end{array} \right.\).

Khi đó \(12 = \int\limits_1^3 {\left( {{x^4} - 8x} \right)f\left( x \right)dx} \)\( = \left. {\left[ {\left( {{x^4} - 8x} \right).F\left( x \right)} \right]} \right|_1^3 - \int\limits_1^3 {\left( {4{x^3} - 8} \right)F\left( x \right)dx} \)

                \(\begin{array}{l} = 57.F\left( 3 \right) - \left( { - 7} \right).F\left( 1 \right) - 4\int\limits_1^3 {\left( {{x^3} - 2} \right)F\left( x \right)dx}  = 57.5 + 7.3 - 4I = 306 - 4I\\ \Rightarrow 12 = 306 - 4I \Leftrightarrow I = \dfrac{{147}}{2}\end{array}\)

Chọn A

Ta có: \(I = \int\limits_0^1 {\frac{{{x^2} + 2x}}{{{{\left( {x + 3} \right)}^2}}}dx} \) \( = \int\limits_0^1 {\frac{{{x^2} + 2x}}{{{x^2} + 6x + 9}}dx} \) \( = \int\limits_0^1 {\left[ {1 - \frac{{4x + 9}}{{{{\left( {x + 3} \right)}^2}}}} \right]dx} \) \( = \left. x \right|_0^1 - \int\limits_0^1 {\frac{{4x + 9}}{{{{\left( {x + 3} \right)}^2}}}dx} \) \( = 1 - J\)

Với \(J = \int\limits_0^1 {\frac{{4x + 9}}{{{{\left( {x + 3} \right)}^2}}}dx} \) \( = \int\limits_0^1 {\frac{{4x + 12 - 3}}{{{{\left( {x + 3} \right)}^2}}}dx} \) \( = \int\limits_0^1 {\left[ {\frac{4}{{x + 3}} - \frac{3}{{{{\left( {x + 3} \right)}^2}}}} \right]dx} \) \( = 4.\int\limits_0^1 {\frac{{dx}}{{x + 3}}}  - 3.\int\limits_0^1 {\frac{{dx}}{{{{\left( {x + 3} \right)}^2}}}} \)

\( = 4\int\limits_0^1 {\frac{{d\left( {x + 3} \right)}}{{x + 3}}}  - 3.\int\limits_0^1 {\frac{{d\left( {x + 3} \right)}}{{{{\left( {x + 3} \right)}^2}}}} \) \( = \left. {\left( {4\ln \left| {x + 3} \right| + \frac{3}{{x + 3}}} \right)} \right|_0^1\) \( = 4\ln 4 + \frac{3}{4} - 4\ln 3 - 1 = 4\ln \frac{4}{3} - \frac{1}{4}\)

Do đó \(I = 1 - J = 1 - \left( {4\ln \frac{4}{3} - \frac{1}{4}} \right) = \frac{5}{4} - 4\ln \frac{4}{3}\)\( \Rightarrow a = 5,b = 3\) \( \Rightarrow {a^2} + {b^2} = {5^2} + {3^2} = 34\).

Chọn D.

Giả sử phương trình đường parabol (P) là \(y = a{x^2} + bx + c,\left( {a \ne 0} \right)\)

(P) có đỉnh là \(I\left( {2;7} \right)\), đồng thời đi qua điểm (3;6) nên ta có hệ phương trình:

\(\left\{ \begin{array}{l} - \frac{b}{{2a}} = 2\\4a + 2b + c = 7\\9a + 3b + c = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}4a + b = 0\\4a + 2b + c = 7\\9a + 3b + c = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a =  - 1\\b = 4\\c = 3\end{array} \right. \Rightarrow \left( P \right):y =  - {x^2} + 4x + 3\)

Ta có hàm số sau: \(v\left( t \right) = \left\{ \begin{array}{l} - {t^2} + 4t + 3,\,\,\,\,\,0 \le t \le 3\\\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,t > 3\,\end{array} \right.\)

Quãng đường cần tìm là :

\(\begin{array}{l}S = \int\limits_0^7 {v\left( t \right)dt}  = \int\limits_0^3 {v\left( t \right)dt}  + \int\limits_3^7 {v\left( t \right)dt}  = \int\limits_0^3 {\left( { - {t^2} + 4t + 3} \right)dt}  + \int\limits_3^7 {6dt} \\ = \left. {\left( { - \frac{1}{3}{t^3} + 2{t^2} + 3t} \right)} \right|_0^3 + 6.\left( {7 - 3} \right) = 18 + 24 = 42\,\,\left( {km} \right)\end{array}\)

Chọn: B

Ta có \(\int\limits_0^1 {\dfrac{{dx}}{{\sqrt {2x + m} }} = \int\limits_0^1 {{{\left( {2x + m} \right)}^{ - \frac{1}{2}}}dx}  = \dfrac{1}{2}\left. {\dfrac{{{{\left( {2x + m} \right)}^{\frac{1}{2}}}}}{{\dfrac{1}{2}}}} \right|} _0^1 = \sqrt {2 + m}  - \sqrt m \)

Từ đề bài ta có \(I \ge 1 \Leftrightarrow \sqrt {2 + m}  - \sqrt m  \ge 1\)  \(\left( {m > 0} \right)\)

\(\sqrt {2 + m}  \ge \sqrt m  + 1 \Leftrightarrow 2 + m \ge m + 2\sqrt m  + 1 \Leftrightarrow 2\sqrt m  \le 1 \Leftrightarrow m \le \dfrac{1}{4} \Rightarrow 0 < m \le \dfrac{1}{4}.\)

Chọn A

Đặt \(t = \ln x \Rightarrow x = {e^t} \Rightarrow dx = {e^t}dt\). Đổi cận \(\left\{ \begin{array}{l}x = 1 \Rightarrow t = 0\\x = 3 \Rightarrow t = \ln 3\end{array} \right.\).

\(\begin{array}{l} \Rightarrow \int\limits_1^3 {\dfrac{{\left( {3x + 1} \right)dx}}{{3{x^2} + x\ln x}}}  = \int\limits_0^{\ln 3} {\dfrac{{\left( {3{e^t} + 1} \right){e^t}dt}}{{3{e^{2t}} + {e^t}t}}}  = \int\limits_0^{\ln 3} {\dfrac{{3{e^t} + 1dt}}{{3{e^t} + t}}}  = \int\limits_0^{\ln 3} {\dfrac{{d\left( {3{e^t} + t} \right)}}{{3{e^t} + t}}} \\ = \left. {\ln \left| {3{e^t} + t} \right|} \right|_0^{\ln 3} = \ln \left| {9 + \ln 3} \right| - \ln 3 = \ln \dfrac{{9 + \ln 3}}{3} = \ln \left( {3 + \dfrac{{\ln 3}}{3}} \right) \Rightarrow \left\{ \begin{array}{l}a = 3\\b = 3\\c = 3\,\,\left( {tm} \right)\end{array} \right. \Rightarrow a + b + c = 9\end{array}\)

Chọn D.

Đặt \(t =  - x \Rightarrow dt =  - dx \Leftrightarrow dx =  - dt\)

Đổi cận \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x =  - 1 \Rightarrow t = 1\end{array} \right.\)

Khi đó \(I = \int\limits_0^{ - 1} {x{{(x - 1)}^2}dx}  = \int\limits_0^1 {\left( { - t} \right){{\left( { - t - 1} \right)}^2}\left( { - dt} \right)}  = \int\limits_0^1 {t{{\left( {t + 1} \right)}^2}dt} \)

Chọn D

Ta có \(\left[ {2f\left( x \right) + 1 - {x^2}} \right]f'\left( x \right) = 2x\left[ {1 + f\left( x \right)} \right]\)

\(\begin{array}{l} \Leftrightarrow 2f\left( x \right).f'\left( x \right) + f'\left( x \right)\left( {1 - {x^2}} \right) = 2x.\left( {1 + f\left( x \right)} \right)\\ \Leftrightarrow 2f\left( x \right).f'\left( x \right) = \left( {{x^2} - 1} \right)f'\left( x \right) + 2x\left( {1 + f\left( x \right)} \right)\\ \Leftrightarrow {\left[ {{f^2}\left( x \right)} \right]^\prime } = {\left[ {\left( {{x^2} - 1} \right)\left( {f\left( x \right) + 1} \right)} \right]^\prime }\end{array}\)

Lấy nguyên hàm hai vế ta được \({f^2}\left( x \right) = \left( {{x^2} - 1} \right)\left( {f\left( x \right) + 1} \right) + C\)

Lại có \(f\left( 1 \right) = 1 \Rightarrow 1 = \left( {1 - 1} \right).2 + C \Rightarrow C = 1\)

Nên \({f^2}\left( x \right) = \left( {{x^2} - 1} \right)\left( {f\left( x \right) + 1} \right) + 1\)

\(\begin{array}{l} \Leftrightarrow {f^2}\left( x \right) = {x^2}f\left( x \right) + {x^2} - f\left( x \right)\\ \Leftrightarrow f\left( x \right)\left( {{x^2} - f\left( x \right)} \right) + {x^2} - f\left( x \right) = 0\\ \Leftrightarrow \left( {{x^2} - f\left( x \right)} \right)\left( {f\left( x \right) + 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}f\left( x \right) =  - 1\left( {ktm} \right)\\f\left( x \right) = {x^2}\left( {tm} \right)\end{array} \right.\end{array}\)

Suy ra \(\int\limits_0^1 {f\left( x \right)dx}  = \int\limits_0^1 {{x^2}dx}  = \dfrac{1}{3}.\)

Chọn C

Sử dụng MTCT ta tính được:

\(\int\limits_{ - 2}^0 {f'\left( x \right)dx}  = \int\limits_{ - 2}^0 {\left( {x - 1} \right)\left( {{x^2} - 3} \right)\left( {{x^4} - 1} \right)}  < 0\)

\( \Rightarrow f\left( 0 \right) - f\left( { - 2} \right) < 0 \Leftrightarrow f\left( 0 \right) < f\left( { - 2} \right)\,\,\left( 1 \right)\)

\(\eqalign{
& \int\limits_0^2 {f'\left( x \right)dx} = \int\limits_{ - 2}^0 {\left( {x - 1} \right)\left( {{x^2} - 3} \right)\left( {{x^4} - 1} \right)} < 0 \cr
& \Rightarrow f\left( 2 \right) - f\left( 0 \right) < 0 \Leftrightarrow f\left( 2 \right) < f\left( 0 \right)\,\,\left( 2 \right) \cr} \)

Từ (1) và (2) ta có: \(f\left( 2 \right) < f\left( 0 \right) < f\left( { - 2} \right).\)

Chọn C.

Ta có: \(\int\limits_0^{\dfrac{\pi }{4}} {\left[ {f(2x) - \sin x} \right]} dx = \int\limits_0^{\dfrac{\pi }{4}} {f\left( {2x} \right)} dx - \int\limits_0^{\dfrac{\pi }{4}} {\sin x} dx = I - J\)

Tính \(I = \int\limits_0^{\dfrac{\pi }{4}} {f\left( {2x} \right)dx} \).

Đặt \(t = 2x \Rightarrow dt = 2dx\)\( \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {f\left( t \right).\dfrac{{dt}}{2}}  = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {f\left( t \right)dt}  = \dfrac{1}{2}.4 = 2\).

Tính \(J = \int\limits_0^{\dfrac{\pi }{4}} {\sin x} dx =  - \left. {\cos x} \right|_0^{\dfrac{\pi }{4}} =  - \left( {\dfrac{{\sqrt 2 }}{2} - 1} \right) = 1 - \dfrac{{\sqrt 2 }}{2}\).

Vậy \(I - J = 2 - \left( {1 - \dfrac{{\sqrt 2 }}{2}} \right) = 1 + \dfrac{{\sqrt 2 }}{2}\) hay \(\int\limits_0^{\dfrac{\pi }{4}} {\left[ {f(2x) - \sin x} \right]} dx = 1 + \dfrac{{\sqrt 2 }}{2}\).

Chọn D

Ta có:  \(I = \int\limits_1^2 {\dfrac{{\ln x}}{{{x^2}}}} dx\)

Đặt \(\ln x = t \Rightarrow x = {e^t} \Rightarrow dt = \dfrac{1}{x}dx\). Đổi cận: \(\left\{ \begin{array}{l}x = 1 \Rightarrow t = 0\\x = 2 \Rightarrow t = \ln 2\end{array} \right..\)

\( \Rightarrow I = \int\limits_0^{\ln 2} {\dfrac{t}{{{e^t}}}dt = } \int\limits_0^{\ln 2} {t{e^{ - t}}dt} \)

Đặt: \(\left\{ \begin{array}{l}u = t\\dv = {e^{ - t}}dt\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dt\\v =  - {e^{ - t}}\end{array} \right.\)

\(\begin{array}{l} \Rightarrow I = \left. { - t{e^{ - t}}} \right|_0^{\ln 2} + \int\limits_0^{\ln 2} {{e^{ - t}}dt}  =  - \ln 2.{e^{ - \ln 2}} - \left. {{e^{ - t}}} \right|_0^{\ln 2} =  - \dfrac{1}{2}\ln 2 - {e^{ - \ln 2}} + 1 =  - \dfrac{1}{2}\ln 2 + \dfrac{1}{2}.\\ \Rightarrow \left\{ \begin{array}{l}a =  - \dfrac{1}{2}\\b = 1\\c = 2\end{array} \right. \Rightarrow S = 2a + 3b + c = 4.\end{array}\)

Chọn  A.

Đặt  \(\left\{ \begin{array}{l}\ln \left( {x - 1} \right) = u\\\dfrac{1}{{{{\left( {x - 1} \right)}^2}}}dx = dv\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{1}{{x - 1}}dx = du\\ - \dfrac{1}{{x - 1}} = v\end{array} \right.\)

 Ta có \(\int\limits_2^{e + 1} {\dfrac{{\ln \left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}dx = \ln \left( {x - 1} \right).\left. {\left( { - \dfrac{1}{{x - 1}}} \right)} \right|} _2^{e + 1} + \int\limits_2^{e + 1} {\dfrac{1}{{{{\left( {x - 1} \right)}^2}}}dx} \)

                               \( =  - \dfrac{1}{e} - \left. {\dfrac{1}{{x - 1}}} \right|_2^{e + 1} =  - \dfrac{1}{e} - \dfrac{1}{e} + 1 = 1 - 2.{e^{ - 1}}\)

Suy ra \(a = 1;\,\,\,b =  - 2 \Rightarrow a + b =  - 1.\)

Chọn B.

\(\begin{array}{l}\int\limits_0^1 {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]dx}  = \int\limits_0^1 {\left[ {{e^x}f\left( x \right) + {e^x}f'\left( x \right)} \right]dx} \\ = \int\limits_0^1 {\left[ {{e^x}f\left( x \right)} \right]'dx}  = \left. {\left[ {{e^x}f\left( x \right)} \right]} \right|_0^1 = e.f\left( 1 \right) - f\left( 0 \right) = e - 1\\ \Rightarrow a = 1;\,\,\,b =  - 1 \Rightarrow {a^{2019}} + {b^{2019}} = {1^{2019}} + {\left( { - 1} \right)^{2019}} = 1 - 1 = 0\end{array}\)

Chọn: C

Ta có \(f'\left( x \right) = 2{\sin ^2}x + 3 = 1 - \cos 2x + 3 = 4 - \cos 2x\).

\( \Rightarrow f\left( x \right) = \int {f'\left( x \right)dx}  = \int {\left( {4 - \cos 2x} \right)dx}  = 4x - \dfrac{1}{2}\sin 2x + C\)

Theo giả thiết có \(f\left( 0 \right) = 4 \Leftrightarrow 4.0 - \dfrac{1}{2}\sin 0 + C = 4 \Leftrightarrow C = 4\).

\(\begin{array}{l} \Rightarrow f\left( x \right) = 4x - \dfrac{1}{2}\sin 2x + 4\\ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {f\left( x \right)dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\left( {4x - \dfrac{1}{2}\sin 2x + 4} \right)dx} \\ = \left. {\left( {2{x^2} + \dfrac{1}{4}\cos 2x + 4x} \right)} \right|_0^{\dfrac{\pi }{4}} = 2\dfrac{{{\pi ^2}}}{{16}} + \pi  - \dfrac{1}{4} = \dfrac{{{\pi ^2} + 8\pi  - 2}}{8}\end{array}\).

Chọn C

Đặt \(t = 5x \Rightarrow \left\{ \begin{array}{l}dx = \dfrac{{dt}}{5}\\x = \dfrac{t}{5}\end{array} \right.\).

Đổi cận: Với \(x = 0 \Rightarrow t = 0\); với \(x = 1 \Rightarrow t = 5\).

Khi đó: \(\int\limits_0^1 {xf\left( {5x} \right){\rm{d}}x}  = 1 \Leftrightarrow \int\limits_0^5 {\dfrac{t}{5}f\left( t \right)\dfrac{{{\rm{dt}}}}{5}}  = 1 \Leftrightarrow \int\limits_0^5 {t.f\left( t \right){\rm{d}}t}  = 25\)

Do đó \(\int\limits_0^5 {x.f\left( x \right){\rm{d}}x}  = 25\,\) (vì tích phân không phụ thuộc vào biến)

Đặt: \(\left\{ \begin{array}{l}u = f\left( x \right)\\{\rm{d}}v = x{\rm{d}}x\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\rm{d}}u = f'\left( x \right){\rm{d}}x\\v = \dfrac{{{x^2}}}{2}\end{array} \right.\).

Ta có: \(\int\limits_0^5 {x.f\left( x \right){\rm{d}}x}  = 25\, \Leftrightarrow \dfrac{{{x^2}}}{2}.f\left( x \right)\left| \begin{array}{l}5\\0\end{array} \right. - \dfrac{1}{2}\int\limits_0^5 {{x^2}.f'\left( x \right){\rm{d}}x}  = 25\)

\( \Leftrightarrow \dfrac{{25}}{2} - \dfrac{1}{2}\int\limits_0^5 {{x^2}.f'\left( x \right){\rm{d}}x}  = 25 \Leftrightarrow \int\limits_0^5 {{x^2}.f'\left( x \right){\rm{d}}x}  =  - 25\).

Chọn D.

\(f\left( {4x} \right) = f\left( x \right) + 4{x^3} + 2x \Leftrightarrow f\left( {4x} \right) - f\left( x \right) = 4{x^3} + 2x\)

\( \Rightarrow f\left( x \right)\) có dạng \(f\left( x \right) = a{x^3} + b{x^2} + cx + d\)

Vì \(f\left( 0 \right) = 2 \Rightarrow d = 2\).

Ta có:

\(\begin{array}{l}f\left( {4x} \right) - f\left( x \right) = 4{x^3} + 2x\\ \Rightarrow \left( {64a{x^3} + 16b{x^2} + 4cx + 2} \right) - \left( {a{x^3} + b{x^2} + cx + 2} \right) = 4{x^3} + 2x\\ \Rightarrow 63a{x^3} + 15b{x^2} + 3cx = 4{x^3} + 2x\\ \Rightarrow \left\{ \begin{array}{l}63a = 4\\15b = 0\\3c = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \dfrac{4}{{63}}\\b = 0\\c = \dfrac{2}{3}\end{array} \right.\\ \Rightarrow f\left( x \right) = \dfrac{4}{{63}}{x^3} + \dfrac{2}{3}x + 2\end{array}\)

Vậy \(\int\limits_0^1 {f\left( x \right)dx}  = \int\limits_0^1 {\left( {\dfrac{4}{{63}}{x^3} + \dfrac{2}{3}x + 2} \right)dx}  = \dfrac{{148}}{{63}}\).

Chọn A.

Ta có :

\(\begin{array}{l}\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{1}{{1 + \tan x}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{1}{{1 + \dfrac{{\sin x}}{{\cos x}}}}dx} \\ = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{1}{{\dfrac{{\cos x + \sin x}}{{\cos x}}}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\cos x}}{{\sin x + \cos x}}dx} \\ = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\sin x + \cos x + \cos x - \sin x}}{{2\left( {\sin x + \cos x} \right)}}dx} \\ = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{4}} {\left( {1 + \dfrac{{\cos x - \sin x}}{{\sin x + \cos x}}} \right)dx} \\ = \dfrac{1}{2}\left[ {x + \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{d\left( {\sin x + \cos x} \right)}}{{\sin x + \cos x}}} } \right]\\ = \dfrac{1}{2}\left. {\left( {x + \ln \left| {\sin x + \cos x} \right|} \right)} \right|_0^{\dfrac{\pi }{4}}\\ = \dfrac{1}{2}\left( {\dfrac{\pi }{4} + \ln \sqrt 2 } \right) = \dfrac{\pi }{8} + \dfrac{1}{2}\ln \sqrt 2 \\ = \dfrac{\pi }{8} + \dfrac{1}{4}\ln 2\\ \Rightarrow a = \dfrac{1}{8},b = \dfrac{1}{4} \Rightarrow \dfrac{a}{b} = \dfrac{1}{2}\end{array}\)

Chọn A.

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).

Chọn C.

Ta có:

\(\begin{array}{l}\,\,\,\,\,\,\,\left( {\sqrt x  + \sqrt {x + 1} } \right)f'\left( x \right) = 1\,\,\,\,\forall x \ge  - 1\\ \Leftrightarrow f'\left( x \right) = \dfrac{1}{{\sqrt x  + \sqrt {x + 1} }}\,\,\,\,\forall x \ge  - 1\\ \Leftrightarrow f'\left( x \right) = \sqrt {x + 1}  - \sqrt x \,\,\,\,\forall x \ge  - 1\end{array}\)

\( \Rightarrow f\left( x \right) = \int {\left( {\sqrt {x + 1}  - \sqrt x } \right)dx}  = \dfrac{2}{3}\left( {\left( {x + 1} \right)\sqrt {x + 1}  - x\sqrt x } \right) + C\)

Mà \(f\left( 0 \right) = \dfrac{2}{3}\)\( \Rightarrow \dfrac{2}{3}\left( {1 - 0} \right) + C = \dfrac{2}{3} \Rightarrow C = 0\).

\( \Rightarrow f\left( x \right) = \dfrac{2}{3}\left( {\left( {x + 1} \right)\sqrt {x + 1}  - x\sqrt x } \right)\).

Khi đó ta có:

\(\begin{array}{l}\int\limits_0^1 {f\left( x \right)dx}  = \dfrac{2}{3}\int\limits_0^1 {\left( {\left( {x + 1} \right)\sqrt {x + 1}  - x\sqrt x } \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{2}{3}.\dfrac{2}{5}\left. {\left( {{{\left( {x + 1} \right)}^2}\sqrt {x + 1}  - {x^2}\sqrt x } \right)} \right|_0^1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{4}{{15}}\left[ {\left( {4\sqrt 2  - 1} \right) - \left( {1 - 0} \right)} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{16\sqrt 2  - 8}}{{15}}\\ \Rightarrow \left\{ \begin{array}{l}a = 16\\b =  - 8\end{array} \right.\end{array}\)

Vậy \(T = a + b = 16 + \left( { - 8} \right) = 8.\)

Chọn D.