Câu hỏi
18/12/2024 3Tính các tích phân sau:
a) \(\int\limits_1^4 {\left( {{x^3} – 2\sqrt x } \right)dx} \);
b) \(\int\limits_0^{\frac{\pi }{2}} {\left( {\cos x – \sin x} \right)dx} \);
c) \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{dx}}{{{{\sin }^2}x}}} \);
d) \(\int\limits_1^{16} {\frac{{x – 1}}{{\sqrt x }}dx} \).
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Lời giải của Vua Trắc Nghiệm
a) Ta có \(\int\limits_1^4 {\left( {{x^3} – 2\sqrt x } \right)dx} \)\( = \int\limits_1^4 {{x^3}dx} – 2\int\limits_1^4 {{x^{\frac{1}{2}}}dx} \)\( = \left. {\left( {\frac{{{x^4}}}{4} – \frac{4}{3}{x^{\frac{3}{2}}}} \right)} \right|_1^4\)\( = \frac{{160}}{3} + \frac{{13}}{{12}} = \frac{{653}}{{12}}\).
b) Ta có \(\int\limits_0^{\frac{\pi }{2}} {\left( {\cos x – \sin x} \right)dx} \)\( = \int\limits_0^{\frac{\pi }{2}} {\cos xdx} – \int\limits_0^{\frac{\pi }{2}} {\sin xdx} \)\( = \left. {\left( {\sin x + \cos x} \right)} \right|_0^{\frac{\pi }{2}} = 1 – 1 = 0\).
c) Ta có \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{dx}}{{{{\sin }^2}x}}} = \left. { – \cot x} \right|_{\frac{\pi }{6}}^{\frac{\pi }{4}} = – 1 + \sqrt 3 \).
d) Ta có \(\int\limits_1^{16} {\frac{{x – 1}}{{\sqrt x }}dx} \)\( = \int\limits_1^{16} {\sqrt x dx} – \int\limits_1^{16} {\frac{1}{{\sqrt x }}dx} \)\( = \int\limits_1^{16} {{x^{\frac{1}{2}}}dx} – \int\limits_1^{16} {{x^{ – \frac{1}{2}}}dx} \)\( = \left. {\left( {\frac{2}{3}{x^{\frac{3}{2}}} – 2{x^{\frac{1}{2}}}} \right)} \right|_1^{16}\)\( = \frac{{104}}{3} + \frac{4}{3} = 36\).