Câu hỏi của Võ KEN

Question

Giải dùm e mấy câu tích phân vs ạ

Nguyễn Việt Lâm · Accepted Answer

11. $I=\int\limits^2_1x\sqrt{x^2+1}dx$Đặt $\sqrt{x^2+1}=t\Leftrightarrow x^2=t^2-1\Rightarrow xdx=tdt$ ; $\left\{{}\begin{matrix}x=1\Rightarrow t=\sqrt{2}\x=2\Rightarrow t=\sqrt{5}\end{matrix}\right.$$I=\int\limits^{\sqrt{5}}_{\sqrt{2}}t.tdt=\int\limits^{\sqrt{5}}_{\sqrt{2}}t^2dt=\dfrac{1}{3}t^3|^{\sqrt{5}}_{\sqrt{2}}=\dfrac{1}{3}\left(5\sqrt{5}-2\sqrt{2}\right)$12. Đặt $\sqrt[3]{8-4x}=t\Rightarrow x=\dfrac{8-t^3}{4}\Rightarrow dx=-\dfrac{3}{4}t^2dt$ ; $\left\{{}\begin{matrix}x=0\Rightarrow t=2\x=2\Rightarrow t=0\end{matrix}\right.$$I=\int\limits^0_2t.\left(-\dfrac{3}{4}t^2dt\right)=\dfrac{3}{4}\int\limits^2_0t^3dt=\dfrac{3}{16}t^4|^2_0=3$13. Đặt $\sqrt{3-2x}=t\Rightarrow x=\dfrac{3-t^2}{2}\Rightarrow dx=-tdt$ ; $\left\{{}\begin{matrix}x=0\Rightarrow t=\sqrt{3}\x=1\Rightarrow t=1\end{matrix}\right.$$I=\int\limits^1_{\sqrt{3}}\dfrac{-tdt}{t}=\int\limits^{\sqrt{3}}_1dt=t|^{\sqrt{3}}_1=\sqrt{3}-1$Câu trả lời: 11. $I=\int\limits^2_1x\sqrt{x^2+1}dx$Đặt $\sqrt{x^2+1}=t\Leftrightarrow x^2=t^2-1\Rightarrow xdx=tdt$ ; $\left\{{}\begin{matrix}x=1\Rightarrow t=\sqrt{2}\x=2\Rightarrow t=\sqrt{5}\end{matrix}\right.$$I=\int\limits^{\sqrt{5}}_{\sqrt{2}}t.tdt=\int\limits^{\sqrt{5}}_{\sqrt{2}}t^2dt=\dfrac{1}{3}t^3|^{\sqrt{5}}_{\sqrt{2}}=\dfrac{1}{3}\left(5\sqrt{5}-2\sqrt{2}\right)$12. Đặt $\sqrt[3]{8-4x}=t\Rightarrow x=\dfrac{8-t^3}{4}\Rightarrow dx=-\dfrac{3}{4}t^2dt$ ; $\left\{{}\begin{matrix}x=0\Rightarrow t=2\x=2\Rightarrow t=0\end{matrix}\right.$$I=\int\limits^0_2t.\left(-\dfrac{3}{4}t^2dt\right)=\dfrac{3}{4}\int\limits^2_0t^3dt=\dfrac{3}{16}t^4|^2_0=3$13. Đặt $\sqrt{3-2x}=t\Rightarrow x=\dfrac{3-t^2}{2}\Rightarrow dx=-tdt$ ; $\left\{{}\begin{matrix}x=0\Rightarrow t=\sqrt{3}\x=1\Rightarrow t=1\end{matrix}\right.$$I=\int\limits^1_{\sqrt{3}}\dfrac{-tdt}{t}=\int\limits^{\sqrt{3}}_1dt=t|^{\sqrt{3}}_1=\sqrt{3}-1$

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP