Giải Bài 4.32 trang 29 SGK Toán 12

Lời giải

Áp dụng công thức tính tích phân $\displaystyle\int_a^b f(x)\,dx = F(b)-F(a)$, với $F$ là một nguyên hàm của $f$.

a) $\displaystyle\int_1^4 (x^3-2\sqrt{x})\,dx$

Nguyên hàm: $\displaystyle\int (x^3-2x^{\frac12})\,dx = \dfrac{x^4}{4} - 2\cdot\dfrac{x^{\frac32}}{\frac32} = \dfrac{x^4}{4} - \dfrac{4}{3}x\sqrt{x}$.

$$\int_1^4 (x^3-2\sqrt{x})\,dx = \left[\dfrac{x^4}{4} - \dfrac{4}{3}x\sqrt{x}\right]_1^4 = \left(\dfrac{256}{4} - \dfrac{4}{3}\cdot 4\cdot 2\right) - \left(\dfrac14 - \dfrac43\right)$$
$$= \left(64 - \dfrac{32}{3}\right) - \left(\dfrac14 - \dfrac43\right) = 64 - \dfrac{32}{3} - \dfrac14 + \dfrac43 = \dfrac{653}{12}.$$

b) $\displaystyle\int_0^{\frac{\pi}{2}} (\cos x-\sin x)\,dx$

Nguyên hàm: $\displaystyle\int(\cos x-\sin x)\,dx = \sin x + \cos x$.

$$\int_0^{\frac{\pi}{2}} (\cos x-\sin x)\,dx = \big[\sin x + \cos x\big]_0^{\frac{\pi}{2}} = (1+0) - (0+1) = 0.$$

c) $\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \dfrac{dx}{\sin^2 x}$

Nguyên hàm: $\displaystyle\int \dfrac{dx}{\sin^2 x} = -\cot x$.

$$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \dfrac{dx}{\sin^2 x} = \big[-\cot x\big]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -\cot\dfrac{\pi}{4} + \cot\dfrac{\pi}{6} = -1 + \sqrt{3} = \sqrt{3}-1.$$

d) $\displaystyle\int_1^{16} \dfrac{x-1}{\sqrt{x}}\,dx$

Biến đổi: $\dfrac{x-1}{\sqrt{x}} = \dfrac{x}{\sqrt{x}} - \dfrac{1}{\sqrt{x}} = \sqrt{x} - x^{-\frac12}$.

Nguyên hàm: $\displaystyle\int\left(x^{\frac12} - x^{-\frac12}\right)dx = \dfrac{2}{3}x\sqrt{x} - 2\sqrt{x}$.

$$\int_1^{16} \dfrac{x-1}{\sqrt{x}}\,dx = \left[\dfrac{2}{3}x\sqrt{x} - 2\sqrt{x}\right]_1^{16} = \left(\dfrac{2}{3}\cdot 16\cdot 4 - 2\cdot 4\right) - \left(\dfrac23 - 2\right)$$
$$= \left(\dfrac{128}{3} - 8\right) - \left(\dfrac23 - 2\right) = \dfrac{128}{3} - 8 - \dfrac23 + 2 = 42 - 6 = 36.$$

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