Biết \(\int\limits_0^{\frac{\pi }{3}} {\frac{{1 - \cos 2x}}{{1 + \cos 2x}}dx} = a\sqrt 3 + \frac{\pi }{b}\) \(\left( {a,{\mkern 1mu} {\mkern 1mu} b \in \mathbb{Z}} \right)\). Tính \(a + b\).

Trả lời: −1

Ta có $\left\{\begin{array}{l}\underset{2}{\overset{7}{\int }}\left[2f\left(x\right)+3g\left(x\right)\right]dx=1\\ \underset{2}{\overset{7}{\int }}\left[f\left(x\right)-2g\left(x\right)\right]dx=4\end{array}\right.$ $\iff \left\{\begin{array}{l}2\underset{2}{\overset{7}{\int }}f\left(x\right)dx+3\underset{2}{\overset{7}{\int }}g\left(x\right)dx=1\\ \underset{2}{\overset{7}{\int }}f\left(x\right)dx-2\underset{2}{\overset{7}{\int }}g\left(x\right)dx=4\end{array}\right.$ $\iff \left\{\begin{array}{l}\underset{2}{\overset{7}{\int }}f\left(x\right)dx=2\\ \underset{2}{\overset{7}{\int }}g\left(x\right)dx=-1\end{array}\right.$

Do đó \(\int\limits_2^7 {f\left( x \right)dx - 3\int\limits_7^2 {g\left( x \right)dx} } \)\( = \int\limits_2^7 {f\left( x \right)dx + 3\int\limits_2^7 {g\left( x \right)dx} } = 2 - 3 = - 1\).

a) S, b) Đ, c) S, d) Đ

a) \(\int {{2^x}{\rm{d}}x} = \frac{{{2^x}}}{{\ln 2}} + C\).

b) \(\int {{{\rm{e}}^{2x}}{\rm{d}}x} = \frac{{{{\rm{e}}^{2x}}}}{2} + C\).

c) \[\int {{e^x}\left( {{e^x}--{\rm{ }}1} \right)} dx = \int {\left( {{e^{2x}} - {e^x}} \right)} dx = \frac{1}{2}{e^{2x}} - {e^x} + C\].

d) \(\int {{e^{3x}}{{.3}^x}dx} = \int {{{\left( {3{e^3}} \right)}^x}dx} = \frac{{{{\left( {3{e^3}} \right)}^x}}}{{\ln \left( {3{e^3}} \right)}} + C = \frac{{{{\left( {3{e^3}} \right)}^x}}}{{3 + \ln 3}} + C\).