Cho hàm số \(f\left( x \right)\) xác định trên \(\mathbb{R}\backslash \left\{ { – 2;1} \right\}\) thỏa mãn \(f’\left( x \right) = \frac{1}{{{x^2} + x – 2}}; f\left( 0 \right) = \frac{1}{3}\) và \(f\left( { – 3} \right) – f\left( 3 \right) = 0\). Tính giá trị biểu thức \(T = f\left( { – 4} \right) + f\left( { – 1} \right) – f\left( 4 \right)\).
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Lời giải:
Báo saiTa có: \(f’\left( x \right) = \frac{1}{{\left( {x – 1} \right)\left( {x + 2} \right)}} = \frac{1}{3}\left( {\frac{1}{{x – 1}} – \frac{1}{{x + 2}}} \right)\).
\(I = f\left( { – 3} \right) – f\left( { – 4} \right) = \int\limits_{ – 4}^{ – 3} {f’\left( x \right){\rm{d}}x} = \left. {\frac{1}{3}\ln \left| {\frac{{x – 1}}{{x + 2}}} \right|} \right|_{ – 4}^{ – 3} = \frac{1}{3}\ln \frac{8}{5}\).
\(J = f\left( 0 \right) – f\left( { – 1} \right) = \int\limits_{ – 1}^0 {f’\left( x \right){\rm{d}}x} = \left. {\frac{1}{3}\ln \left| {\frac{{x – 1}}{{x + 2}}} \right|} \right|_{ – 1}^0 = – \frac{2}{3}\ln 2\)
\(K = f\left( 4 \right) – f\left( 3 \right) = \int\limits_3^4 {f’\left( x \right)} {\rm{d}}x = \left. {\frac{1}{3}\ln \left| {\frac{{x – 1}}{{x + 2}}} \right|} \right|_3^4 = \frac{1}{3}\ln \frac{5}{4}\)
\( – I – J – K = f\left( { – 4} \right) – f\left( { – 3} \right) + f\left( { – 1} \right) – f\left( 0 \right) + f\left( 3 \right) – f\left( 4 \right)\)
\( = \left[ {f\left( { – 4} \right) + f\left( { – 1} \right) – f\left( 4 \right)} \right] – f\left( 0 \right) – \left[ {f\left( { – 3} \right) – f\left( 3 \right)} \right]\)
\(f\left( { – 4} \right) + f\left( { – 1} \right) – f\left( 4 \right) = – I – J – K + f\left( 0 \right) + \left[ {f\left( { – 3} \right) – f\left( 3 \right)} \right]\)
\(T = f\left( { – 4} \right) + f\left( { – 1} \right) – f\left( 4 \right) = – \frac{1}{3}\ln \frac{8}{5} + \frac{2}{3}\ln 2 – \frac{1}{3}\ln \frac{5}{4} + \frac{1}{3} = \frac{1}{3}\ln 2 + \frac{1}{3}\)
