Câu hỏi
16/12/2024 1Tính các tích phân sau:
a)
b)
c)
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Lời giải của Vua Trắc Nghiệm
a) Ta có \(\int\limits_0^{2\pi } {\left( {2x + \cos x} \right)dx} \)\( = \int\limits_0^{2\pi } {2xdx} + \int\limits_0^{2\pi } {\cos xdx} \)\( = \left. {{x^2}} \right|_0^{2\pi } + \left. {\sin x} \right|_0^{2\pi }\)\( = 4{\pi ^2}\).
b) Ta có \(\int\limits_1^2 {\left( {{3^x} – \frac{3}{x}} \right)dx} \)\( = \int\limits_1^2 {{3^x}dx} – \int\limits_1^2 {\frac{3}{x}dx} \)
\( = \left. {\frac{{{3^x}}}{{\ln 3}}} \right|_1^2 – \left. {3\ln \left| x \right|} \right|_1^2 = \frac{{{3^2}}}{{\ln 3}} – \frac{3}{{\ln 3}} – 3\ln 2 + 3\ln 1 = \frac{6}{{\ln 3}} – 3\ln 2\).
c) Ta có \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\left( {\frac{1}{{{{\cos }^2}x}} – \frac{1}{{{{\sin }^2}x}}} \right)dx} \)\( = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{1}{{{{\cos }^2}x}}dx} – \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{1}{{{{\sin }^2}x}}dx} \)
\( = \left. {\tan x} \right|_{\frac{\pi }{6}}^{\frac{\pi }{3}} + \left. {\cot x} \right|_{\frac{\pi }{6}}^{\frac{\pi }{3}}\)\( = \tan \frac{\pi }{3} – \tan \frac{\pi }{6} + \cot \frac{\pi }{3} – \cot \frac{\pi }{6}\)\( = \sqrt 3 – \frac{{\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3} – \sqrt 3 = 0\).