Tính tích phân bằng công thức Newton–Leibniz

a) \(\int\limits_1^4 \left( x^3 - 2\sqrt{x} \right)dx = \left( \frac{x^4}{4} - \frac{4x\sqrt{x}}{3} \right)\Bigg|_1^4\) \(= \left(\frac{4^4}{4} - \frac{4 \cdot 4\sqrt{4}}{3}\right) - \left(\frac{1}{4} - \frac{4 \cdot 1 \cdot \sqrt{1}}{3}\right)\) \(= \left(64 - \frac{32}{3}\right) - \left(\frac{1}{4} - \frac{4}{3}\right)\) \(= 64 - \frac{32}{3} - \frac{1}{4} + \frac{4}{3} = 64 - \frac{28}{3} - \frac{1}{4} = \frac{768 - 112 - 3}{12} = \dfrac{653}{12}\) b) \(\int\limits_0^{\frac{\pi}{2}} \left( \cos x - \sin x \right)dx = \left( \sin x + \cos x \right)\Bigg|_0^{\frac{\pi}{2}}\) \(= \left(\sin\frac{\pi}{2} + \cos\frac{\pi}{2}\right) - \left(\sin 0 + \cos 0\right)\) \(= (1 + 0) - (0 + 1) = 0\) c) \(\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{dx}{\sin^2 x} = \left( -\cot x \right)\Bigg|_{\frac{\pi}{6}}^{\frac{\pi}{4}}\) \(= -\cot\frac{\pi}{4} + \cot\frac{\pi}{6} = -1 + \sqrt{3}\) d) Tách hàm số: \(\frac{x-1}{\sqrt{x}} = x^{\frac{1}{2}} - x^{-\frac{1}{2}}\) \(\int\limits_1^{16} \frac{x-1}{\sqrt{x}}\,dx = \int\limits_1^{16} \left( x^{\frac{1}{2}} - x^{-\frac{1}{2}} \right)dx = \left( \frac{2x\sqrt{x}}{3} - 2\sqrt{x} \right)\Bigg|_1^{16}\) \(= \left(\frac{2 \cdot 16\sqrt{16}}{3} - 2\sqrt{16}\right) - \left(\frac{2 \cdot 1 \cdot \sqrt{1}}{3} - 2\sqrt{1}\right)\) \(= \left(\frac{128}{3} - 8\right) - \left(\frac{2}{3} - 2\right)\) \(= \frac{128}{3} - 8 - \frac{2}{3} + 2 = \frac{126}{3} - 6 = 42 - 6 = 36\)