Tính tích phân hàm ẩn dựa vào tính chất (VDC)

6. Tính tích phân hàm ẩn dựa vào tính chất (VDC)

Vấn đề 6. Tính tích phân dựa vào tính chất.
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Lời Giải:
Do $f(x)$ là hàm lẻ nên $f(-x)=-f(x)$.
Xét $A=\displaystyle\int\limits_{-2}^0 f(-x)\mathrm{\,d}x=2$. Đặt $t=-x \to \mathrm{\,d}t=-\mathrm{\,d}x$.
Đổi cận: $\begin{cases}&x=-2 \to t=2\\&x=0 \to t=0\end{cases}$.
Khi đó $A=-\displaystyle\int\limits_2^0 f(t)\mathrm{\,d}t=\displaystyle\int\limits_0^2 f(t)\mathrm{\,d}t=\displaystyle\int\limits_0^2 f(x)\mathrm{\,d}x$.
Xét $B=\displaystyle\int\limits_1^2 f(-2x)\mathrm{\,d}x=-\displaystyle\int\limits_1^2 f(2x)\mathrm{\,d}x$. Đặt $u=2x \to \mathrm{\,d}u=2\mathrm{\,d}x$.
Đổi cận: $\begin{cases}&x=1 \to u=2\\&x=2 \to u=4\end{cases}$.
Khi đó $B=-\dfrac{1}{2}\displaystyle\int\limits_2^4 f(u)\mathrm{\,d}u=-\dfrac{1}{2}\displaystyle\int\limits_2^4 f(x)\mathrm{\,d}x \to \displaystyle\int\limits_2^4 f(x)\mathrm{\,d}x=-2B=-2 \cdot 4=-8$.
Vậy $I=\displaystyle\int\limits_0^4 f(x)\mathrm{\,d}x=\displaystyle\int\limits_0^2 f(x)\mathrm{\,d}x+\displaystyle\int\limits_2^4 f(x)\mathrm{\,d}x=2-8=-6$.
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Lời Giải:
Vì $f(x)$ là hàm số chẵn nên $\displaystyle\int\limits_1^3 f(-2x)\mathrm{\,d}x=\displaystyle\int\limits_1^3 f(2x)\mathrm{\,d}x=3$.
Xét $K=\displaystyle\int\limits_1^3 f(2x)\mathrm{\,d}x=3$. Đặt $t=2x \to \mathrm{\,d}t=2\mathrm{\,d}x$.
Đổi cận: $\begin{cases}&x=1 \to t=2\\&x=3 \to t=6\end{cases}$.
Khi đó $K=\dfrac{1}{2}\displaystyle\int\limits_2^6 f(t)\mathrm{\,d}t=\dfrac{1}{2}\displaystyle\int\limits_2^6 f(x)\mathrm{\,d}x \to \displaystyle\int\limits_2^6 f(x)\mathrm{\,d}x=2K=6$.
Vậy $I=\displaystyle\int\limits_{-1}^6 f(x)\mathrm{\,d}x=\displaystyle\int\limits_{-1}^2 f(x)\mathrm{\,d}x+\displaystyle\int\limits_2^6 f(x)\mathrm{\,d}x=8+6=14$.
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Lời Giải:
Đặt $t=(3+7)-x \to \mathrm{\,d}t=-\mathrm{\,d}x$.
Đổi cận $\begin{cases}&x=7 \to t=3\\&x=3 \to t=7\end{cases}$.
Khi đó $I=-\displaystyle\int\limits_7^3 (10-t)f(10-t)\mathrm{\,d}t=\displaystyle\int\limits_3^7 (10-t)f(10-t)\mathrm{\,d}t=\displaystyle\int\limits_3^7 (10-x)f(10-x)\mathrm{\,d}x$.
$\overset{f(x)=f(10-x)}=\displaystyle\int\limits_3^7 (10-x)f(x)\mathrm{\,d}x=10\displaystyle\int\limits_3^7 f(x)\mathrm{\,d}x-\displaystyle\int\limits_3^7 xf(x)\mathrm{\,d}x=10\displaystyle\int\limits_3^7 f(x)\mathrm{\,d}x-I$.
Suy ra $2I=10\displaystyle\int\limits_3^7 f(x)\mathrm{\,d}x=10 \cdot 4=40 \to I=20$.
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Lời Giải:
Đặt $x=-t \to \mathrm{\,d}x=-\mathrm{\,d}t$.
Đổi cận $\begin{cases}&x=-\pi \to t=\pi \\&x=\pi \to t=-\pi\end{cases}$.
Khi đó $I=-\displaystyle\int\limits_{\pi}^{-\pi} \dfrac{f(-t)}{2018^{-t}+1}\mathrm{\,d}t=\displaystyle\int\limits_{-\pi}^{\pi} \dfrac{f(-t)}{2018^{-t}+1}\mathrm{\,d}t$
$=\displaystyle\int\limits_{-\pi}^{\pi} \dfrac{2018^tf(-t)}{1+2018^t}\mathrm{\,d}t=\displaystyle\int\limits_{-\pi}^{\pi} \dfrac{2018^xf(-x)}{1+2018^x}\mathrm{\,d}x$.
Vì $y=f(x)$ là hàm số chẵn trên đoạn $\left[-\pi; \pi \right]$
nên $f(-x)=f(x) \to I=\displaystyle\int\limits_{-\pi}^{\pi} \dfrac{2018^xf(x)}{2018^x+1}\mathrm{\,d}x$.
Vậy $2I=\displaystyle\int\limits_{-\pi}^{\pi} \dfrac{f(x)}{2018^x+1}\mathrm{\,d}x+\displaystyle\int\limits_{-\pi}^{\pi} \dfrac{2018^xf(x)}{2018^x+1}\mathrm{\,d}x$
$=\displaystyle\int\limits_{-\pi}^{\pi} f(x)\mathrm{\,d}x=2\displaystyle\int\limits_0^{\pi} f(x)\mathrm{\,d}x=2.2018 \to I=2018$.
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Lời Giải:
Gọi $I=\displaystyle\int\limits_0^{\pi} \dfrac{x\sin^{2018}x}{\sin^{2018}x+\cos^{2018}x}\mathrm{\,d}x$.
Đặt $t=\pi -x \to \mathrm{\,d}t=-\mathrm{\,d}x$. Đổi cận $\begin{cases}&x=0 \to t=\pi \\&x=\pi \to t=0\end{cases}$.
Khi đó $I=-\displaystyle\int\limits_{\pi}^0 \dfrac{\left(\pi -t\right)\sin^{2018}\left(\pi -t\right)}{\sin^{2018}\left(\pi -t\right)+\cos^{2018}\left(\pi -t\right)}\mathrm{\,d}t$
$=\displaystyle\int\limits_0^{\pi} \dfrac{\left(\pi -t\right)\sin^{2018}t}{\sin^{2018}t+\cos^{2018}t}\mathrm{\,d}t=\displaystyle\int\limits_0^{\pi} \dfrac{\left(\pi -x\right)\sin^{2018}x}{\sin^{2018}x+\cos^{2018}x}\mathrm{\,d}x$.
Suy ra $2I=\displaystyle\int\limits_0^{\pi} \dfrac{x\sin^{2018}x}{\sin^{2018}x+\cos^{2018}x}\mathrm{\,d}x+\displaystyle\int\limits_0^{\pi} \dfrac{\left(\pi -x\right)\sin^{2018}x}{\sin^{2018}x+\cos^{2018}x}\mathrm{\,d}x$
$=\displaystyle\int\limits_0^{\pi} \dfrac{\pi \sin^{2018}x}{\sin^{2018}x+\cos^{2018}x}\mathrm{\,d}x$.
$ \to I=\dfrac{\pi}{2}\displaystyle\int\limits_0^{\pi} \dfrac{\sin^{2018}x}{\sin^{2018}x+\cos^{2018}x}\mathrm{\,d}x$
$=\dfrac{\pi}{2}\left[\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \dfrac{\sin^{2018}x}{\sin^{2018}x+\cos^{2018}x}\mathrm{\,d}x+\displaystyle\int\limits_{\tfrac{\pi}{2}}^{\pi} \dfrac{\sin^{2018}x}{\sin^{2018}x+\cos^{2018}x}\mathrm{\,d}x\right]$.
Đặt $x=u+\dfrac{\pi}{2}$ ta suy ra $\displaystyle\int\limits_{\tfrac{\pi}{2}}^{\pi} \dfrac{\sin^{2018}x}{\sin^{2018}x+\cos^{2018}x}\mathrm{\,d}x$
$=\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \dfrac{\cos^{2018}u}{\sin^{2018}u+\cos^{2018}u}\mathrm{\,d}u=\displaystyle\int\limits_{\tfrac{\pi}{2}}^{\pi} \dfrac{\cos^{2018}x}{\sin^{2018}x+\cos^{2018}x}\mathrm{\,d}x$.
Vậy $I=\dfrac{\pi}{2}\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \mathrm{\,d}x=\dfrac{\pi^2}{4} \to \begin{cases}&a=2\\&b=4\end{cases} \to P=8$.