Tính tích phân hàm phân nhánh (VDC)

Vấn đề 5. Tính tích phân hàm phân nhánh.
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Lời Giải:
Ta có $I=\displaystyle\int\limits_{-1}^0 f(x)\mathrm{\,d}x+\displaystyle\int\limits_0^2 f(x)\mathrm{\,d}x=\displaystyle\int\limits_{-1}^0 \mathrm{e}^{2x}\mathrm{\,d}x+\displaystyle\int\limits_0^2 (x+1)\mathrm{\,d}x=\dfrac{9\mathrm{e}^2-1}{2\mathrm{e}^2}$.
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Lời Giải:
Ta có $f'(x)=\dfrac{2}{2x-1}$.
$ \to f(x)=\displaystyle\int\limits \dfrac{2}{2x-1}\mathrm{\,d}x=\ln |2x-1|+C=\begin{cases}& \ln (1-2x)+C_1 &; x<\dfrac{1}{2}\\& \ln (2x-1)+C_2 &; x>\dfrac{1}{2}\end{cases}.$.
$f(0)=1 \to \ln (1-2 \cdot 0)+C_1=1 \to C_1=1$.
$f(1)=2 \to \ln (2 \cdot 1-1)+C_2=2 \to C_2=2$.
Do đó $f(x)=\begin{cases}&\ln (1-2x)+1khix<\dfrac{1}{2}\\&\ln (2x-1)+2khix>\dfrac{1}{2}\end{cases} \to \begin{cases}&f(-1)=\ln 3+1\\&f(3)=\ln 5+2\end{cases}$.
$ \to f(-1)+f(3)=3+\ln 5+\ln 3=3+\ln 15$.
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Lời Giải:
Ta có $f'(x)=\dfrac{1}{x^2+x-2}=\dfrac{1}{3}\left(\dfrac{1}{x-1}-\dfrac{1}{x+2}\right)$.

$ \to f(x)=\displaystyle\int \dfrac{1}{x^2+x-2}\mathrm{\,d}x=\begin{cases}\dfrac{1}{3}\left[\ln (1-x)-\ln (-x-2)\right]+C_1 &; x<-2\\\dfrac{1}{3}\left[\ln (1-x)-\ln (x+2)\right]+C_2 &; -2< x< 1\\\dfrac{1}{3}\left[\ln (x-1)-\ln (x+2)\right]+C_3 &; x>1\end{cases}$.

$f(0)=\dfrac{1}{3} \to \dfrac{1}{3}\left[\ln (1-0)-\ln (0+2)\right]+C_2=\dfrac{1}{3} \to C_2=\dfrac{1}{3}\ln 2+\dfrac{1}{3}$.

$f(-3)-f(3)=0 \to C_1-C_3=\dfrac{1}{3}\ln \dfrac{1}{10}$.

Ta có $f(-4)+f(-1)-f(4)=\dfrac{1}{3}\ln \dfrac{5}{2}+\dfrac{1}{3}\ln 2-\dfrac{1}{3}\ln \dfrac{1}{2}+C_2+C_1-C_3=\dfrac{1}{3}\ln 2+\dfrac{1}{3}$.
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Lời Giải:
Ta có $f'(x)=\dfrac{1}{x\left(\ln x-1\right)}$.
$ \to f(x)=\displaystyle\int \dfrac{1}{x\left(\ln x-1\right)}\mathrm{\,d}x=\displaystyle\int \dfrac{\mathrm{d}\left(\ln x-1\right)}{\left(\ln x-1\right)}=\ln \left|\ln x-1\right|+C$
$=\begin{cases}&\ln \left(1-\ln x\right)+C_1khix \in (0; e)\\&\ln \left(\ln x-1\right)+C_2khix \in (\mathrm{e}; +\infty)\end{cases}$.
$f\left(\dfrac{1}{\mathrm{e}^2}\right)=\ln 6 \to \ln \left(1-\ln \dfrac{1}{\mathrm{e}^2}\right)+C_1=\ln 6 \to C_1=\ln 2$.
$f(\mathrm{e}^2)=3 \to \ln \left(\ln \mathrm{e}^2-1\right)+C_2=3 \to C_2=3$.
Do đó $f(x)=\begin{cases}&\ln \left(1-\ln x\right)+\ln 2khix \in (0; e)\\&\ln \left(\ln x-1\right)+3khix \in (\mathrm{e}; +\infty)\end{cases} \to \begin{cases}&f\left(\dfrac{1}{\mathrm{e}}\right)=\ln 2+\ln 2\\&f(\mathrm{e}^3)=\ln 2+3\end{cases}$.
$ \to f\left(\dfrac{1}{\mathrm{e}}\right)+f(\mathrm{e}^3)=3\left(\ln 2+1\right)$.
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Lời Giải:
Với $x$ thuộc vào mỗi khoảng $\left(-\dfrac{\pi}{4}+k\pi; -\dfrac{\pi}{4}+k\pi \right), k \in \mathbb{Z}$ ta có\\
$F(x)=\displaystyle\int \dfrac{\mathrm{\,d}x}{1+\sin 2x}=\displaystyle\int \dfrac{\mathrm{\,d}x}{\left(\sin x+\cos x\right)^2}=\displaystyle\int \dfrac{\mathrm{\,d}x}{2\cos^2\left(x+\dfrac{\pi}{4}\right)}=\dfrac{1}{2}\tan \left(x+\dfrac{\pi}{4}\right)+C$.
$0; -\dfrac{\pi}{12} \in \left(-\dfrac{\pi}{4}; \dfrac{\pi}{4}\right)$ nên $F(0)-F\left(-\dfrac{\pi}{12}\right)=\dfrac{1}{2}\tan \left(x-\dfrac{\pi}{4}\right)\bigg|_{-\tfrac{\pi}{12}}^0 =-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\xrightarrow{F(0)=1}F\left(-\dfrac{\pi}{12}\right)=\dfrac{3}{2}-\dfrac{\sqrt{3}}{2}$.
$\pi; \dfrac{11\pi}{12} \in \left(\dfrac{\pi}{4}; \dfrac{5\pi}{4}\right)$ nên $F(\pi)-F\left(\dfrac{11\pi}{12}\right)=\dfrac{1}{2}\tan \left(x-\dfrac{\pi}{4}\right)\bigg|_{\tfrac{11\pi}{12}}^{\pi} =-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\xrightarrow{F(\pi)=0}F\left(\dfrac{11\pi}{12}\right)=\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}$.
Vậy $P=F\left(-\dfrac{\pi}{12}\right)-F\left(\dfrac{11\pi}{12}\right)=1$.