Cho \(I_n=\int\limits^1_0\frac{e^{-2nx}}{1+e^{2x}}dx\). Tổng \(I_5+I_6\) bằng : \(\frac{e^6-1}{6.e^6}\) \(\frac{e^8-1}{8.e^8}\) \(\frac{e^{10}-1}{10.e^{10}}\) \(\frac{e^{12}-1}{12.e^{12}}\) Hướng dẫn giải: \(I_n=\int\limits^1_0\dfrac{e^{-2nx}}{1+e^{2x}}dx\Rightarrow I_5=\int_0^1\dfrac{e^{-10x}}{1+e^{2x}}dx,I_6=\int_0^1\dfrac{e^{-12x}}{1+e^{2x}}dx\) \(I_5+I_6=\int_0^1\dfrac{e^{-10x}+e^{-12x}}{1+e^{2x}}dx=\int_0^1\dfrac{e^{-12x}\left(e^{2x}+1\right)}{1+e^{2x}}dx=\int_0^1e^{-12x}dx\) \(=-\dfrac{e^{-12x}}{12}|_0^1=\dfrac{1-e^{-12}}{12}=\dfrac{e^{12}-1}{12e^{12}}\).
Tính tích phân \(I=\)\(\int\limits^2_1\left(x^2+\frac{1}{x^4}\right)dx\). \(I=\frac{19}{8}\) \(I=\frac{21}{8}\) \(I=\frac{23}{8}\) \(I=\frac{25}{8}\) Hướng dẫn giải: Ta có \(\int\limits^2_1\left(x^2+\frac{1}{x^4}\right)dx=\left(\frac{x^3}{3}-\frac{1}{3x^3}\right)_1^2=\left(\frac{8}{3}-\frac{1}{24}\right)=\frac{63}{21}=\frac{21}{8}\) Vậy \(I=\frac{21}{8}\)
Tính tích phân \(I=\)\(\int\limits^1_0\frac{dx}{\sqrt{4-x^2}}\) \(I=\frac{\pi}{3}\) \(I=\frac{\pi}{4}\) \(I=\frac{\pi}{6}\) \(I=\frac{\pi}{8}\) Hướng dẫn giải: Đặt \(x=2\sin t\Rightarrow dx=2\cos tdt;\sqrt{4-x^2}=\sqrt{4\cos^2t}=2\left|\cos t\right|\). Đổi cận: \(x=0\Rightarrow t=0;x=1\Rightarrow t=\dfrac{\pi}{6}\). Ta có \(\int\limits^1_0\dfrac{dx}{\sqrt{4-x^2}}=\int\limits^{\dfrac{\pi}{6}}_0\dfrac{2\cos t}{2\left|\cos t\right|}dt=\int\limits^{\dfrac{\pi}{6}}_0dt=\dfrac{\pi}{6}\)
Tính tích phân \(I=\)\(\int\limits^{a\sqrt{3}}_a\frac{dx}{a^2+x^2}\) \(I=\frac{\pi}{4a}\) \(I=\frac{\pi}{8a}\) \(I=\frac{\pi}{9a}\) \(I=\frac{\pi}{12a}\) Hướng dẫn giải: Đặt \(x=a\tan t\Rightarrow a^2+x^2=a^2\left(1+\tan^2t\right),dx=a\left(1+\tan^2t\right)dt\). Đổi cận: \(x=a\Rightarrow a\tan t=a\Rightarrow t=\dfrac{\pi}{4};x=a\sqrt{3}\Rightarrow t=\dfrac{\pi}{3}\). \(\int\limits^{a\sqrt{3}}_a\dfrac{dx}{a^2+x^2}=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\dfrac{dt}{a}=\dfrac{\pi}{12a}\)
Tính tích phân \(I=\)\(\int\limits^1_0\frac{dx}{e^x+1}\) \(I=\ln\frac{e}{2\left(e+1\right)}\) \(I=\ln\frac{2e}{e+1}\) \(I=\ln\frac{e}{2\left(e-1\right)}\) \(I=\ln\frac{2e}{e-1}\) Hướng dẫn giải: Đặt \(t=e^x+1\) thì \(dt=e^xdx=\left(t-1\right)dx\Rightarrow dx=\dfrac{dt}{t-1}\). Đổi cận \(x|_0^1\Rightarrow t|_2^{e+1}\). \(\int\limits^1_0\frac{dx}{e^x+1}\)=\(\int\limits^{1+e}_2\dfrac{1}{t\left(t-1\right)}dt=\int\limits^{1+e}_2\left(\dfrac{1}{t-1}-\dfrac{1}{t}\right)dt=\ln\dfrac{t-1}{t}|_2^{1+e}=\ln\dfrac{2e}{e+1}\)
Tính tích phân \(I=\)\(\int\limits^{\frac{a}{2}}_0\sqrt{\frac{x}{a-x}}dx\) \(I=a\left(\pi-\frac{1}{2}\right)\) \(I=a\left(\pi+\frac{1}{2}\right)\) \(I=a\left(\frac{\pi-2}{4}\right)\) \(I=a\left(\frac{\pi+2}{4}\right)\) Hướng dẫn giải: Đặt \(x=a\sin^2t\Rightarrow dx=2a\sin t\cos tdt\). Đổi cận \(x|_0^{\dfrac{a}{2}}\Rightarrow t|_0^{\dfrac{\pi}{4}}\) \(\int\limits^{\dfrac{a}{2}}_0\sqrt{\dfrac{x}{a-x}}dx=\int\limits^{\dfrac{\pi}{4}}_0\sqrt{\dfrac{a\sin^2t}{a\cos^2t}}2a\sin t\cos tdt\) \(=\int\limits^{\dfrac{\pi}{4}}_02a\sin^2tdt=a\int\limits^{\dfrac{\pi}{4}}_0\left(1-\cos2t\right)dt\) \(=a\left(t-\dfrac{1}{2}\sin2t\right)|^{\dfrac{\pi}{4}}_0=a\left(\dfrac{\pi}{4}-\dfrac{1}{2}\right)=\dfrac{a\left(\pi-2\right)}{4}\) .
Tính tích phân \(I=\)\(\int\limits^1_0x^3\sqrt{1-x^2}dx\) \(I=\frac{2}{15}\) \(I=\frac{4}{15}\) \(I=\frac{7}{15}\) \(I=\frac{8}{15}\) Hướng dẫn giải: Đặt \(t=\sqrt{1-x^2}\Rightarrow t^2=1-x^2\Rightarrow x^2=1-t^2,2xdx=-2tdt\). Đổi cận \(x|_0^1\Rightarrow t|_1^0\). Do đó \(\int\limits^1_0x^3\sqrt{1-x^2}dx=-\int\limits^1_0x^2\sqrt{1-x^2}\left(-xdx\right)=\int\limits^1_0\left(1-t^2\right)t^2dt\) \(=\int\limits^1_0\left(t^2-t^4\right)dt=\left(\dfrac{t^3}{3}-\dfrac{t^5}{5}\right)|_0^1=\dfrac{2}{15}\)
\(\int\limits^2_0\left|x^2-x\right|dx\) bằng : \(\frac{1}{2}\) \(1\) \(\frac{1}{4}\) \(\frac{1}{3}\) Hướng dẫn giải:
Tính \(I=\)\(\int\limits^4_2x.\ln\left(x-1\right)dx\). \(I=7,5\ln3+4\) \(I=\dfrac{15\ln3}{2}-4\) \(I=\dfrac{15\ln3}{2}+2\) \(I=7,5\ln3-2\) Hướng dẫn giải: \(\int\limits^4_2x.\ln\left(x-1\right)dx=\int\limits^4_2\ln\left(x-1\right)\left(\dfrac{x^2-1}{2}\right)'dx=\dfrac{\left(x^2-1\right)\ln\left(x-1\right)}{2}|_2^4-\int\limits^4_2\dfrac{x^2-1}{2}\left(\ln\left(x-1\right)\right)'dx\) \(=\dfrac{15\ln3}{2}-\int\limits^4_2\dfrac{x^2-1}{2}.\dfrac{1}{x-1}dx=\dfrac{15\ln3}{2}-\int\limits^4_2\dfrac{x+1}{2}dx\) \(=\dfrac{15\ln3}{2}-\dfrac{\left(x+1\right)^2}{4}|_2^4=\dfrac{15\ln3}{2}-4\)
Tính \(I=\) \(\int\limits^1_0x.e^{2x}dx\) \(I=\)\(\frac{e^2+1}{2}\) \(I=\)\(\frac{e^2-1}{2}\) \(I=\)\(\frac{e^2+1}{4}\) \(I=\)\(\frac{e^2-1}{4}\) Hướng dẫn giải: \(\int\limits^1_0x.e^{2x}dx=\int\limits^1_0x\left(\dfrac{e^{2x}}{2}\right)'dx=\dfrac{xe^{2x}}{2}|_0^1-\int\limits^1_0\dfrac{e^{2x}}{2}\left(x\right)'dx=\dfrac{e^2}{2}-\int\limits^1_0\dfrac{e^{2x}}{2}dx\) \(=\dfrac{e^2}{2}-\dfrac{e^{2x}}{4}|_0^1=\dfrac{e^2}{2}-\left(\dfrac{e^2}{4}-\dfrac{1}{4}\right)=\dfrac{e^2+1}{4}\)
