Bài 5 trang 127 SGK Giải tích 12. Tính

a) $A={\int }_0^3\frac{x}{\sqrt{1+x}}dx$

• Đặt $t=\sqrt{1+x}\iff t^2=1+x\Rightarrow \left\{\begin{array}{l}2tdt=dx\\ x=t^2-1\end{array}\right..$
• Đổi cận: $x=0\Rightarrow t=1; x=3\Rightarrow t=2.$
• Khi đó $A={\int }_1^2\frac{t^2-1}{t}2tdt=2{\int }_1^2\left(t^2-1\right)dt$$=2{\overline{)\left(\frac{t^3}{3}-t\right)}}_1^2=\frac{8}{3}.$

b) $B={\int }_1^{64}\frac{1+\sqrt{x}}{\sqrt[3]{x}}dx$

• Đặt $\sqrt[3]{x}=t\iff x=t^3\iff dx=3t^{2}dt.$
• Đổi cận : $x=1\Rightarrow t=1; x=64\Rightarrow t=\sqrt[3]{64}=4.$
• Khi đó $B={\int }_1^4\frac{1+\sqrt{t^3}}{t}3t^{2}dt=3{\int }_1^4\left(1+t^{\frac{3}{2}}\right)tdt$$=3{\int }_1^4\left(t+t^{\frac{5}{2}}\right)dt$$\mathrm{}=3{\overline{)\left(\frac{t^2}{2}+\frac{2t^{{\displaystyle \frac{7}{2}}}}{7}\right)}}_1^4=\frac{1839}{14}.$

c) $C={\int }_0^2{x}^2{e}^{3x}dx$

• Đặt $\left\{\begin{array}{l}u=x^2\\ dv=e^{3x}dx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=2xdx\\ v=\frac{e^{3x}}{3}\end{array}\right..$
• Khi đó : $C={\overline{)\frac{x^2{e}^{3x}}{3}}}_0^2-{\int }_0^2\frac{e^{3x}}{3}2xdx$$=\frac{4e^6}{3}-\frac{2}{3}{\int }_0^{2}x{e}^{3x}dx=\frac{4e^6}{3}-\frac{2}{3}I.$
• Tính $I={\int }_0^{2}x{e}^{3x}dx :$ Đặt $\left\{\begin{array}{l}{u}^{\text{'}}=x\\ dv=e^{3x}dx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du\text{'}=dx\\ v=\frac{e^{3x}}{3}\end{array}\right.$$\Rightarrow I={\overline{)\frac{x{e}^{3x}}{3}}}_0^2-{\int }_0^2\frac{e^{3x}}{3}dx=\frac{2e^6}{3}-{\overline{)\frac{e^{3x}}{9}}}_0^2$$=\frac{2e^6}{3}-\frac{e^6}{9}+\frac{1}{9}.$
• Vậy $C=\frac{4e^6}{3}-\frac{2}{3}\left(\frac{2e^6}{3}-\frac{e^6}{9}+\frac{1}{9}\right)$$=\frac{26e^6}{27}-\frac{2}{27}=\frac{2}{27}\left(13e^6-1\right).$

d) $D={\int }_0^{\mathrm{\pi }}\sqrt{1+\sin 2x}dx$$={\int }_0^{\mathrm{\pi }}\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}dx$$={\int }_0^{\mathrm{\pi }}\sqrt{{\left(\sin x+\cos x\right)}^2}dx$$={\int }_0^{\mathrm{\pi }}\sqrt{{\left[\sqrt{2}\sin \left(x+\frac{\mathrm{\pi }}{4}\right)\right]}^2}dx$$=\sqrt{2}{\int }_0^{\mathrm{\pi }}\left|\sin \left(x+\frac{\mathrm{\pi }}{4}\right)\right|dx$

+) Mặt khác xét trên đoạn $\left[0;\mathrm{\pi }\right]$ thì  $\sin \left(x+\frac{\mathrm{\pi }}{4}\right)\ge 0 khi \forall x\in \left[0;\frac{3\mathrm{\pi }}{4}\right] ;$$\sin \left(x+\frac{\mathrm{\pi }}{4}\right)<0 khi x\in \left(\frac{3\mathrm{\pi }}{4};\mathrm{\pi }\right) .$

+) Do đó : $D=\sqrt{2}{\int }_0^{\frac{3\mathrm{\pi }}{4}}\left|\sin \left(x+\frac{\mathrm{\pi }}{4}\right)\right|dx+\sqrt{2}{\int }_{\frac{3\mathrm{\pi }}{4}}^{\mathrm{\pi }}\left|\sin \left(x+\frac{\mathrm{\pi }}{4}\right)\right|dx$$=\sqrt{2}{\int }_0^{\frac{3\mathrm{\pi }}{4}}\sin \left(x+\frac{\mathrm{\pi }}{4}\right)dx+\sqrt{2}{\int }_{\frac{3\mathrm{\pi }}{4}}^{\mathrm{\pi }}-\sin \left(x+\frac{\mathrm{\pi }}{4}\right)dx$$=-{\overline{)\sqrt{2}\cos \left(x+\frac{\mathrm{\pi }}{4}\right)}}_0^{\frac{3\mathrm{\pi }}{4}}+{\overline{)\sqrt{2}\cos \left(x+\frac{\mathrm{\pi }}{4}\right)}}_{\frac{3\mathrm{\pi }}{4}}^{\mathrm{\pi }}$$=-\sqrt{2}\left(-1-\frac{\sqrt{2}}{2}\right)+\sqrt{2}\left(-\frac{\sqrt{2}}{2}-\left(-1\right)\right)=2\sqrt{2.}$