a). SAI

\[\int\limits_0^2 {\left( {3{x^2} - 2{{\rm{e}}^x}} \right){\rm{d}}x} = 3\int\limits_0^2 {{x^2}{\rm{d}}x} - 2\int\limits_0^2 {{{\rm{e}}^x}{\rm{d}}x} = \left. {{x^3}} \right|_0^2 - \left. {2{{\rm{e}}^x}} \right|_0^2 = 8 - \left( {2{{\rm{e}}^2} - 2} \right) = 10 - 2{{\rm{e}}^2}\].

b). ĐÚNG

Lượng nước chảy ra khỏi bồn chứa trong \(10\) phút đầu tiên là

\[V = \int\limits_0^{10} {k\left( t \right){\rm{d}}t} = \int\limits_0^{10} {\left( {250 - 6t} \right){\rm{d}}t} = \left. {\left( {250t - 3{t^2}} \right)} \right|_0^{10} = 2200\] (lít).

c). ĐÚNG

Ta có \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\left( {{{\tan }^2}x - {{\cot }^2}x} \right){\rm{d}}x} = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - 1 - \frac{1}{{{{\sin }^2}x}} + 1} \right){\rm{d}}x} = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}} \right){\rm{d}}x} \)

\[ = \left. {\left( {\tan x + \cot x} \right)} \right|_{\frac{\pi }{6}}^{\frac{\pi }{4}} = 2 - \left( {\frac{1}{{\sqrt 3 }} + \sqrt 3 } \right) = 2 - \frac{4}{3}\sqrt 3 \]. Do đó \(a + b + c = 9\).

d). SAI

\(\int\limits_0^{\frac{\pi }{3}} {\left| {{{\cos }^2}x - \sin x\cos x} \right|{\rm{d}}x} = \int\limits_0^{\frac{\pi }{3}} {\left( {\left| {\cos x} \right|.\left| {\cos x - \sin x} \right|} \right){\rm{d}}x} \)

Ta có \(\cos x > 0\) với mọi \(x \in \left[ {0;\frac{\pi }{3}} \right]\) và phương trình \(\cos x - \sin x = 0\) có nghiệm \(x = \frac{\pi }{4}\) trên đoạn \(\left[ {0;\frac{\pi }{3}} \right]\).

Do đó \(\int\limits_0^{\frac{\pi }{3}} {\left( {\left| {\cos x} \right|.\left| {\cos x - \sin x} \right|} \right){\rm{d}}x} = \int\limits_0^{\frac{\pi }{4}} {\left[ {\cos x.\left( {\cos x - \sin x} \right)} \right]{\rm{d}}x} - \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\left[ {\cos x.\left( {\cos x - \sin x} \right)} \right]{\rm{d}}x} \)

\( = \int\limits_0^{\frac{\pi }{4}} {\left( {{{\cos }^2}x - \sin x\cos x} \right){\rm{d}}x} - \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\left( {{{\cos }^2}x - \sin x\cos x} \right){\rm{d}}x} \)

\( = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\left( {1 + \cos 2x - \sin 2x} \right){\rm{d}}x} - \frac{1}{2}\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\left( {\left( {1 + \cos 2x - \sin 2x} \right)} \right){\rm{d}}x} \)

\( = \left. {\frac{1}{2}\left( {x + \frac{1}{2}\sin 2x + \frac{1}{2}\cos 2x} \right)} \right|_0^{\frac{\pi }{4}} - \left. {\frac{1}{2}\left( {x + \frac{1}{2}\sin 2x + \frac{1}{2}\cos 2x} \right)} \right|_{\frac{\pi }{4}}^{\frac{\pi }{3}}\)

\( = \frac{\pi }{8} + \frac{1}{4} - \left[ {\frac{\pi }{6} + \frac{{\sqrt 3 }}{8} - \frac{1}{8} - \left( {\frac{\pi }{8} + \frac{1}{4}} \right)} \right] = \frac{\pi }{8} + \frac{1}{4} - \frac{\pi }{6} - \frac{{\sqrt 3 }}{8} + \frac{3}{8} + \frac{\pi }{8} = \frac{5}{8} + \frac{\pi }{{12}} - \frac{{\sqrt 3 }}{8}\)

Do đó \(a = \frac{5}{8},\;b = \frac{1}{{12}},\;c = \frac{1}{8} \Rightarrow a - c + b = \frac{1}{2} - \frac{1}{{12}} = \frac{5}{{12}}\).

Hàm số \(F(x) = 2{x^3} - 2x + 1\) là nguyên hàm của hàm số nào sau đây?