Nội dung bài giảng
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) \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}sin3xcos5xdx\)
Giải:
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.