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Đặt \(t=-x\Rightarrow dx=-dt\)
\(I=\int\limits^{-2}_2\frac{t^{2018}}{e^{-t}+1}\left(-dt\right)=\int\limits^2_{-2}\frac{e^t.t^{2018}}{e^t+1}dt=\int\limits^2_{-2}\frac{e^x.x^{2018}}{e^x+1}dx\)
\(\Rightarrow I+I=\int\limits^2_{-2}\frac{x^{2018}+e^x.x^{2018}}{e^x+1}dx=\int\limits^2_{-2}x^{2018}dx=\frac{2.2^{2019}}{2019}\)
\(\Rightarrow I=\frac{2^{2019}}{2019}\)
Câu 1:
Lấy logarit cơ số tự nhiên 2 vế:
\(x.lny+e^y.x\ge y.lnx+y.e^x\)
\(\Leftrightarrow\frac{lny+e^y}{y}\ge\frac{lnx+e^x}{x}\)
Xét hàm \(f\left(t\right)=\frac{lnt+e^t}{t}\) với \(t>1\)
\(f'\left(t\right)=\frac{\left(e^t+\frac{1}{t}\right).t-lnt-e^t}{t^2}=\frac{t.e^t+1-e^t-lnt}{t^2}\)
Xét \(g\left(t\right)=t.e^t+1-e^t-lnt\Rightarrow g'\left(t\right)=e^t+t.e^t-e^t-\frac{1}{t}\)
\(g'\left(t\right)=t.e^t-\frac{1}{t}=\frac{t^2.e^t-1}{t}>0\) \(\forall t>1\)
\(\Rightarrow g\left(t\right)\) đồng biến \(\Rightarrow g\left(t\right)>g\left(1\right)=1>0\) \(\forall t>1\)
\(\Rightarrow f'\left(t\right)=\frac{g\left(t\right)}{t^2}>0\Rightarrow f\left(t\right)\) đồng biến
\(\Rightarrow f\left(t_1\right)\ge f\left(t_2\right)\Leftrightarrow t_1\ge t_2\)
\(\Rightarrow f\left(y\right)\ge f\left(x\right)\Leftrightarrow y\ge x\) \(\Rightarrow log_xy\ge1>0\)
\(P=log_x\left(xy\right)^{\frac{1}{2}}+log_yx=\frac{1}{2}\left(1+log_xy\right)+\frac{1}{log_xy}\)
\(P=\frac{1}{2}+\frac{1}{2}log_xy+\frac{1}{log_xy}\ge\frac{1}{2}+2\sqrt{\frac{log_xy}{2log_xy}}=\frac{1}{2}+\sqrt{2}\)
\(f'\left(x\right)=\frac{1}{x-1}\Rightarrow\int f'\left(x\right)dx=\int\frac{1}{x-1}dx\)
\(\Rightarrow f\left(x\right)=ln\left|x-1\right|+C\)
\(\Rightarrow f\left(x\right)=\left\{{}\begin{matrix}ln\left|x-1\right|+C_1\left(x>1\right)\\ln\left|x-1\right|+C_2\left(x< 1\right)\end{matrix}\right.\)
\(f\left(0\right)=2018\Leftrightarrow2018=ln\left|0-1\right|+C_2\Rightarrow C_2=2018\)
\(f\left(2\right)=2019\Rightarrow2019=ln\left|2-1\right|+C_1\Rightarrow C_1=2019\)
\(\Rightarrow f\left(x\right)=\left\{{}\begin{matrix}ln\left|x-1\right|+2019\left(x>1\right)\\ln\left|x-1\right|+2018\left(x< 1\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(3\right)=2019+ln2\\f\left(-1\right)=2018+ln2\end{matrix}\right.\) \(\Rightarrow S=1\)
Cho hàm số \(f\left(x\right)=ln2017-ln\dfrac{x+1}{x}\) . Tính tổng S= f'(1)+f'(2)+f'(3)+...+f'(2018)
Câu 1:
\(\int\frac{sinx}{sinx+cosx}dx=\frac{1}{2}\int\frac{sinx+cosx+sinx-cosx}{sinx+cosx}dx=\frac{1}{2}\int dx-\frac{1}{2}\int\frac{cosx-sinx}{sinx+cosx}dx\)
\(=\frac{1}{2}x-\frac{1}{2}\int\frac{d\left(sinx+cosx\right)}{sinx+cosx}=\frac{1}{2}x-\frac{1}{2}ln\left|sinx+cosx\right|+C\)
\(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{2}\\b=-\frac{1}{2}\end{matrix}\right.\)
\(\int cos^2xdx=\int\left(\frac{1}{2}+\frac{1}{2}cos2x\right)dx=\frac{1}{2}x+\frac{1}{4}sin2x+C\)
\(\Rightarrow\left\{{}\begin{matrix}c=\frac{1}{2}\\d=2\end{matrix}\right.\) \(\Rightarrow I=5\)
Câu 2:
\(I=\int\left(sin\left(lnx\right)-cos\left(lnx\right)\right)dx=\int sin\left(lnx\right)dx-\int cos\left(lnx\right)dx=I_1-I_2\)
Xét \(I_2=\int cos\left(lnx\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=cos\left(lnx\right)\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-\frac{1}{x}sin\left(lnx\right)dx\\v=x\end{matrix}\right.\)
\(\Rightarrow I_2=x.cos\left(lnx\right)+\int sin\left(lnx\right)dx=x.cos\left(lnx\right)+I_1\)
\(\Rightarrow I=I_1-\left(x.cos\left(lnx\right)+I_1\right)=-x.cos\left(lnx\right)+C\)
b/ \(I=\int\limits sin\left(lnx\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=sin\left(lnx\right)\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{1}{x}cos\left(lnx\right)dx\\v=x\end{matrix}\right.\)
\(\Rightarrow I=x.sin\left(lnx\right)-\int cos\left(lnx\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=cos\left(lnx\right)\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-\frac{1}{x}sin\left(lnx\right)dx\\v=x\end{matrix}\right.\)
\(\Rightarrow I=x\left[sin\left(lnx\right)-cos\left(lnx\right)\right]-I\)
\(\Rightarrow I=\frac{1}{2}x\left[sin\left(lnx\right)-cos\left(lnx\right)\right]|^{e^{\pi}}_1=\frac{1}{2}\left(e^{\pi}+1\right)\)
\(\Rightarrow a=2;b=\pi;c=1\)
Đặt \(3-2x=t\Rightarrow dx=-\frac{1}{2}dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=3\\x=2\Rightarrow t=-1\end{matrix}\right.\)
\(\Rightarrow P=\int\limits^{-1}_3\left[f\left(t\right)+2019\right].\left(-\frac{1}{2}\right)dt=\frac{1}{2}\int\limits^3_{-1}f\left(t\right)dt+\int\limits^3_{-1}\frac{2019}{2}dt\)
\(=\frac{15}{2}+\frac{2019}{2}.4=\frac{8091}{2}\)
Câu 1:
Đặt \(\sqrt{lnx+1}=t\Rightarrow lnx=t^2-1\Rightarrow\frac{dx}{x}=2tdt\)
\(\Rightarrow I=\int3t.2t.dt=6\int t^2dt=2t^3+C\)
\(=2\sqrt{\left(lnx+1\right)^3}+C=2\left(lnx+1\right)\sqrt{lnx+1}+C\)
\(=ln\left(x.e\right)^2\sqrt{ln\left(x.e\right)+0}\Rightarrow a=2;b=0\)
Câu 2:
\(\int\limits^b_ax^{-\frac{1}{2}}dx=2x^{\frac{1}{2}}|^b_a=2\left(\sqrt{b}-\sqrt{a}\right)=2\Rightarrow\sqrt{b}-\sqrt{a}=1\)
Ta có hệ: \(\left\{{}\begin{matrix}\sqrt{b}-\sqrt{a}=1\\a^2+b^2=17\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=4\\a=1\end{matrix}\right.\) (lưu ý loại cặp nghiệm âm do \(\frac{1}{\sqrt{x}}\) chỉ xác định trên miền (a;b) dương)
Câu 4:
\(\int\frac{3x+a}{x^2+4}dx=\frac{3}{2}\int\frac{2x}{x^2+4}dx+a\int\frac{1}{x^2+4}dx\)
\(=\frac{3}{2}ln\left(x^2+4\right)+\frac{a}{2}arctan\left(\frac{x}{2}\right)+C\)
\(\Rightarrow a=2\)
\(\Rightarrow I=\int\limits^{\frac{e}{4}}_1ln\left(x\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{1}{x}dx\\v=x\end{matrix}\right.\)
\(\Rightarrow I=x.lnx|^{\frac{e}{4}}_1-\int\limits^{\frac{e}{4}}_1dx=\frac{e}{4}.ln\left(\frac{e}{4}\right)-\frac{e}{4}+1=-\frac{ln\left(2^e\right)}{2}+1\)
Câu 5:
\(f'\left(x\right)=\int f''\left(x\right)dx=-\frac{1}{4}\int x^{-\frac{3}{2}}dx=\frac{1}{2\sqrt{x}}+C\)
\(f'\left(2\right)=\frac{1}{2\sqrt{2}}+C=2+\frac{1}{2\sqrt{2}}\Rightarrow C=2\)
\(\Rightarrow f'\left(x\right)=\frac{1}{2\sqrt{x}}+2\)
\(\Rightarrow f\left(x\right)=\int f'\left(x\right)dx=\int\left(\frac{1}{2\sqrt{x}}+2\right)dx=\sqrt{x}+2x+C_1\)
\(f\left(4\right)=\sqrt{4}+2.4+C_1=10\Rightarrow C_1=0\)
\(\Rightarrow f\left(x\right)=2x+\sqrt{x}\)
\(\Rightarrow F\left(x\right)=\int f\left(x\right)dx=\int\left(2x+\sqrt{x}\right)dx=x^2+\frac{2}{3}\sqrt{x^3}+C_2\)
\(F\left(1\right)=1+\frac{2}{3}+C_2=1+\frac{2}{3}\Rightarrow C_2=0\)
\(\Rightarrow F\left(x\right)=x^2+\frac{2}{3}\sqrt{x^3}\Rightarrow\int\limits^1_0\left(x^2+\frac{2}{3}\sqrt{x^3}\right)dx=\frac{3}{5}\)
