Tính các tích phân sau : a) b) c) d) e) g)

Bài 1 (SGK trang 112)

Gửi bởi: Sách Giáo Khoa Vào 26 tháng 12 2018 lúc 11:13:41

Lý thuyết

Câu hỏi

Tính các tích phân sau :

a) $\int\limits^{\dfrac{1}{2}}_{-\dfrac{1}{2}}\sqrt[3]{\left(1-x\right)^2dx}$

b) $\int\limits^{\dfrac{\pi}{2}}_0\sin\left(\dfrac{\pi}{4}-x\right)dx$

c) $\int\limits^2_{\dfrac{1}{2}}\dfrac{1}{x\left(x+1\right)}dx$

d) $\int\limits^2_0x\left(x+1\right)^2dx$

e) $\int\limits^2_{\dfrac{1}{2}}\dfrac{1-3x}{\left(x+1\right)^2}dx$

g) $\int\limits^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}\sin3x\cos5xdx$

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