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) S, c) S, d) Đ

a) \(\int\limits_{ - 2}^2 {f\left( x \right)} dx = \int\limits_{ - 2}^2 {\left( {2x - 4} \right)dx} = - 16\).

b) \(\int\limits_1^4 {{f^2}\left( x \right)} dx = \int\limits_1^4 {{{\left( {2x - 4} \right)}^2}} dx = \left. {\frac{1}{6}{{\left( {2x - 4} \right)}^3}} \right|_1^4 = 12\).

c) \(\int\limits_{ - 2}^5 {\left| {f\left( x \right)} \right|dx} = \int\limits_{ - 2}^5 {\left| {2x - 4} \right|dx} \)\( = \int\limits_{ - 2}^2 {\left( {4 - 2x} \right)dx} + \int\limits_2^5 {\left( {2x - 4} \right)dx} = 16 + 9 = 25 = {3^0}{.5^2}\).

Do đó \({m^2} + {n^2} = 0 + 4 = 4\).

d) \(F\left( a \right) = \int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {\left( {2x - 4} \right)dx} = \left. {\left( {{x^2} - 4x} \right)} \right|_0^a = {a^2} - 4a\)\( = {\left( {a - 2} \right)^2} - 4 \ge - 4\).

Vậy giá trị nhỏ nhất của \(F\left( a \right)\) là \( - 4\) khi \(a = 2\).