Question

Tính các tích phân sau:

a) \(\int\limits^4_1\left(x^3-2\sqrt{x}\right)dx\);             b) \(\int\limits^{\dfrac{\pi}{2}}_0\left(\cos x-\sin x\right)dx\);              c) \(\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{6}}\dfrac{dx}{\sin^2x}\);                  d) \(\int\limits^{16}_1\dfrac{x-1}{\sqrt{x}}dx\).

Answer 1

a) \(\int\limits_1^4 {\left( {{x^3} - 2\sqrt x } \right)dx}  = \left( {\frac{{{x^4}}}{4} - \frac{{4x\sqrt x }}{3}} \right)\left| \begin{array}{l}4\\1\end{array} \right. = \frac{{{4^4}}}{4} - \frac{{4.4\sqrt 4 }}{3} - \frac{1}{4} + \frac{{4.1\sqrt 1 }}{3} = \frac{{653}}{{12}}\)

b) \(\int\limits_0^{\frac{\pi }{2}} {\left( {\cos x - \sin x} \right)dx}  = \left( {\sin x + \cos x} \right)\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. = \sin \frac{\pi }{2} + \cos \frac{\pi }{2} - \sin 0 - \cos 0 = 1 - 1 = 0\)

c) \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{dx}}{{{{\sin }^2}x}}}  =  - \cot x\left| \begin{array}{l}\frac{\pi }{4}\\\frac{\pi }{6}\end{array} \right. =  - \cot \frac{\pi }{4} + \cot \frac{\pi }{6} =  - 1 + \sqrt 3 \)

d) \(\int\limits_1^{16} {\frac{{x - 1}}{{\sqrt x }}dx}  = \int\limits_1^{16} {\left( {{x^{\frac{1}{2}}} - {x^{\frac{{ - 1}}{2}}}} \right)dx}  = \left( {\frac{{2x\sqrt x }}{3} - 2\sqrt x } \right)\left| \begin{array}{l}16\\1\end{array} \right. = \frac{{2.16\sqrt {16} }}{3} - 2\sqrt {16}  - \frac{2}{3} + 2 = 36\)