Bài tập 1 trang 112 - SGK Giải tích 12: Bài 2. Tích phân...

Bài 1. Tính các tích phân sau:

a)$\int_{\frac{-1}{2}}^{\frac{1}{2}}\sqrt[3]{ (1-x)^{2}}dx$                    b) $\int_{0}^{\frac{\pi}{2}}sin(\frac{\pi}{4}-x)dx$

c)$\int_{\frac{1}{2}}^{2}\frac{1}{x(x+1)}dx$                        d) $\int_{0}^{2}x(x+1)^{2}dx$

e)$\int_{\frac{1}{2}}^{2}\frac{1-3x}{(x+1)^{2}}dx$                        g) $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}sin3xcos5xdx$

:

a) $\int_{\frac{-1}{2}}^{\frac{1}{2}}\sqrt[3]{ (1-x)^{2}}dx$ = $-\int_{\frac{-1}{2}}^{\frac{1}{2}}(1-x)^{\frac{2}{3}}d(1-x)=-\frac{3}{5}(1-x)^{\frac{5}{3}}|_{\frac{-1}{2}}^{\frac{1}{2}}$

 = $-\frac{3}{5}\left [ \frac{1}{2\sqrt[3]{4}}-\frac{3\sqrt[3]{9}}{2\sqrt[3]{4}} \right ]=\frac{3}{10\sqrt[3]{4}}(3\sqrt[3]{9}-1)$

b) $\int_{0}^{\frac{\pi}{2}}sin(\frac{\pi}{4}-x)dx$=$-\int_{0}^{\frac{\pi}{2}}sin(\frac{\pi}{4}-x)d(\frac{\pi}{4}-x)$ = $cos(\frac{\pi}{4}-x)|_{0}^{\frac{\pi}{2}}$

= $cos(\frac{\pi}{4}-\frac{\pi}{2})-cos\frac{\pi}{4}=0$

c)$\int_{\frac{1}{2}}^{2}\frac{1}{x(x+1)}dx$=$\int_{\frac{1}{2}}^{2}(\frac{1}{x}-\frac{1}{x+1})dx =ln\left | \frac{x}{x+1} \right ||_{\frac{1}{2}}^{2}=ln2$

d)$\int_{0}^{2}x(x+1)^{2}dx$= $\int_{0}^{2}(x^{3}+2x^{2}+x)dx=(\frac{x^{4}}{4}+\frac{2}{3}x^{3}+\frac{x^{2}}{2})|_{0}^{2}$

  = $\frac{16}{4}+\frac{16}{3}+2= 11\tfrac{1}{3}$

e)$\int_{\frac{1}{2}}^{2}\frac{1-3x}{(x+1)^{2}}dx$= $\int_{\frac{1}{2}}^{2}\frac{-3(x+1)+4}{(x+1)^{2}}dx=\int_{\frac{1}{2}}^{2}\left [ \frac{-3}{x+1}+\frac{4}{(x+1)^{2}} \right ]dx$

= $\left ( -3.ln\left | x+1 \right |-\frac{4}{x+1} \right )|_{\frac{1}{2}}^{2}= \frac{4}{3}-3ln2$

g)Ta có $f(x) = sin3xcos5x$ là hàm số lẻ.

Vì $f(-x) = sin(-3x)cos(-5x)$

                 $= -sin3xcos5x = -f(x)$

nên:

         $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}sin3xcos5x =0$

Chú ý: Có thể tính trực tiếp bằng cách đặt $x= -t$ hoặc biến đổi thành tổng.