Tính và so sánh:
a) $\int_0^1 2 x d x$ và $2 \int_0^1 x d x$;
b) $\int_0^1\left(x^2+x\right) d x$ và $\int_0^1 x^2 d x+\int_0^1 x d x$;
c) $\int_0^3 x d x$ và $\int_0^1 x d x+\int_1^3 x d x$.
( Mu4-42. Cho hàm so $f(x)$ có đạo hàm trên đoạn $[0 ; 1]$ thỏa mãn $f(1)=0$ và $\int_0^1\left[f^{\prime}(x)\right]^2 d x=\int_0^1(x+1) e^x f(x) d x=\frac{e^2-1}{4}$. Tinh tich phân $I=\int_{0}^1 f(x) d x$.
A. $I=2-e$.
B. $I=\frac{e}{2}$.
C. $l=e-2$.
D. $1=\frac{e-1}{2}$
Tính các tích phân sau:
a) \(\int_0^1x^3\sqrt{1-x^2}dx\)
b) \(\int_1^2\dfrac{dx}{x^2-2x+2}\)
c) \(\int_1^2\dfrac{dx}{\sqrt{4-x^2}}\)
d) \(\int_0^1x\sqrt{x^2+1}dx\)
a.
Đặt \(\sqrt{1-x^2}=u\Rightarrow x^2=1-u^2\Rightarrow xdx=-udu\)
\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=0\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^0_1\left(1-u^2\right).u.\left(-udu\right)=\int\limits^1_0\left(u^2-u^4\right)du=\left(\dfrac{1}{3}u^3-\dfrac{1}{5}u^5\right)|^1_0\)
\(=\dfrac{2}{15}\)
b.
\(\int\limits^2_1\dfrac{dx}{x^2-2x+2}=\int\limits^2_1\dfrac{dx}{\left(x-1\right)^2+1}\)
Đặt \(x-1=tanu\Rightarrow dx=\dfrac{1}{cos^2u}du\)
\(\left\{{}\begin{matrix}x=1\Rightarrow u=0\\x=2\Rightarrow u=\dfrac{\pi}{4}\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{1}{tan^2u+1}.\dfrac{1}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{cos^2u}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0du\)
\(=u|^{\dfrac{\pi}{4}}_0=\dfrac{\pi}{4}\)
c.
\(\int\limits^2_1\dfrac{dx}{\sqrt{4-x^2}}\)
Đặt \(x=2sinu\Rightarrow dx=2cosu.du\)
\(\left\{{}\begin{matrix}x=1\Rightarrow u=\dfrac{\pi}{6}\\x=2\Rightarrow u=\dfrac{\pi}{2}\end{matrix}\right.\)
\(I=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{2cosu.du}{\sqrt{4-4sin^2u}}=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{2cosu.du}{2cosu}=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}du\)
\(=u|^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}=\dfrac{\pi}{3}\)
a\(\int_0^1\dfrac{dx}{x^4+4x^2+3}\)
b \(\int\dfrac{x^2-1}{x^4+1}\)
c\(\int\dfrac{dx}{x\left(x^3+1\right)}\)
d \(\int_0^1\dfrac{xdx}{x^4+x^2+1}\)
a/ \(I=\int\limits^1_0\dfrac{1}{\left(x^2+3\right)\left(x^2+1\right)}dx=\dfrac{1}{2}\int\limits^1_0\left(\dfrac{1}{x^2+1}-\dfrac{1}{x^2+3}\right)dx\)
\(=\dfrac{1}{2}\left(arctanx-\dfrac{1}{\sqrt{3}}arctan\dfrac{x}{\sqrt{3}}\right)|^1_0=\dfrac{\pi}{8}-\dfrac{\pi\sqrt{3}}{36}\)
b/ \(I=\int\dfrac{x^2-1}{x^4+1}dx=\int\dfrac{1-\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}dx\)
Đặt \(x+\dfrac{1}{x}=t\Rightarrow\left(1-\dfrac{1}{x^2}\right)dx=dt\) ; \(x^2+\dfrac{1}{x^2}=t^2-2\)
\(\Rightarrow I=\int\dfrac{dt}{t^2-2}=\int\dfrac{dt}{\left(t-\sqrt{2}\right)\left(t+\sqrt{2}\right)}=\dfrac{1}{2\sqrt{2}}\int\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)dt\)
\(\Rightarrow I=\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{t-\sqrt{2}}{t+\sqrt{2}}\right|+C=\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|+C\)
c/ \(I=\int\dfrac{dx}{x\left(x^3+1\right)}=\int\dfrac{x^2dx}{x^3\left(x^3+1\right)}\)
Đặt \(x^3+1=t\Rightarrow3x^2dx=dt\)
\(\Rightarrow I=\dfrac{1}{3}\int\dfrac{dt}{\left(t-1\right)t}=\dfrac{1}{3}\int\left(\dfrac{1}{t-1}-\dfrac{1}{t}\right)dt=\dfrac{1}{3}ln\left|\dfrac{t-1}{t}\right|+C\)
\(\Rightarrow I=\dfrac{1}{3}ln\left|\dfrac{x^3}{x^3+1}\right|+C\)
d/ \(I=\int\limits^1_0\dfrac{xdx}{x^4+x^2+1}\)
Đặt \(x^2=t\Rightarrow2xdx=dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=1\Rightarrow t=1\end{matrix}\right.\)
\(I=\dfrac{1}{2}\int\limits^1_0\dfrac{dt}{t^2+t+1}=\dfrac{1}{2}\int\limits^1_0\dfrac{dt}{\left(t+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}=\dfrac{2}{3}\int\limits^1_0\dfrac{dt}{\dfrac{4}{3}\left(t+\dfrac{1}{2}\right)^2+1}\)
Đặt \(t+\dfrac{1}{2}=\dfrac{\sqrt{3}}{2}tanu\Rightarrow dt=\dfrac{\sqrt{3}}{2}.\dfrac{du}{cos^2u}\); \(\left\{{}\begin{matrix}t=0\Rightarrow u=\dfrac{\pi}{6}\\t=1\Rightarrow u=\dfrac{\pi}{3}\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{2}{3}.\dfrac{\sqrt{3}}{2}\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{du}{cos^2u\left(tan^2u+1\right)}=\dfrac{\sqrt{3}}{3}\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}du=\dfrac{\pi\sqrt{3}}{18}\)
Cho hàm số y = f(x) có đạo hàm liên tục trên đoạn [0;1] thỏa mãn f(1) = 1,\(\int_0^1xf\left(x\right)dx=\dfrac{1}{5}\), \(\int_0^1\left[f'\left(x\right)\right]^2dx=\dfrac{9}{5}\) Tính tích phân \(I=\int_0^1f\left(x\right)dx\)
Đang học Lý mà thấy bài nguyên hàm hay hay nên nhảy vô luôn :b
\(I_1=\int\limits^1_0xf\left(x\right)dx\)
\(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow\int xf\left(x\right)dx=\dfrac{1}{2}x^2f\left(x\right)-\dfrac{1}{2}\int x^2f'\left(x\right)dx\)
\(\Rightarrow\int\limits^1_0xf\left(x\right)dx=\dfrac{1}{2}x^2|^1_0-\dfrac{1}{2}\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{2}\int\limits^1_0\left[f'\left(x\right)\right]^2dx=\dfrac{3}{10}\Rightarrow\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{3}{5}\)
Đoạn này hơi rối xíu, ông để ý kỹ nhé, nhận thấy ta có 2 dữ kiện đã biết, là: \(\int\limits^1_0\left[f'\left(x\right)\right]^2dx=\dfrac{9}{5}and\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{3}{5}\) có gì đó liên quan đến hằng đẳng thức, nên ta sẽ sử dụng luôn
\(\int\limits^1_0\left[f'\left(x\right)+tx^2\right]^2dx=0\)
\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx+2t\int\limits^1_0x^2f'\left(x\right)dx+t^2\int\limits^1_0x^4dx=0\)
\(\Leftrightarrow\dfrac{9}{5}+\dfrac{6}{5}t+\dfrac{1}{5}t^2=0\) \(\left(\int\limits^1_0x^4dx=\dfrac{1}{5}x^5|^1_0=\dfrac{1}{5}\right)\)\(\)\(\Leftrightarrow t=-3\Rightarrow\int\limits^1_0\left[f'\left(x\right)-3x^2\right]^2dx=0\)
\(\Leftrightarrow f'\left(x\right)=3x^2\Leftrightarrow f\left(x\right)=x^3+C\)
\(\Rightarrow\int\limits^1_0f\left(x\right)dx=\int\limits^1_0x^3dx=\dfrac{1}{4}x^4|^1_0=\dfrac{1}{4}\)
P/s: Có gì ko hiểu hỏi mình nhé !
tính các tích phân
1.\(\int_0^1\dfrac{4x+2}{x^2+x+1}dx\)
2.\(\int_0^1\dfrac{4x+1}{\left(2-x\right)^4}dx\)
3.\(\int_0^1\dfrac{x^2+1}{\left(x^3+3x\right)^3}dx\)
Câu 1:
Ta có \(I_1=\int ^{1}_{0}\frac{4x+2}{x^2+x+1}dx=2\int ^{1}_{0}\frac{2x+1}{x^2+x+1}dx\)
\(=2\int ^{1}_{0}\frac{d(x^2+x+1)}{x^2+x+1}=2.\left.\begin{matrix} 1\\ 0\end{matrix}\right|\ln |x^2+x+1|=2\ln 3\)
Câu 2:
\(I_2=\int ^{1}_{0}\frac{4x+1}{(2-x)^4}dx=\int ^{1}_{0}\frac{4(x-2)+9}{(2-x)^4}dx\)
\(=4\int ^{1}_{0}\frac{dx}{(x-2)^3}+9\int \frac{dx}{(2-x)^4}=4\int ^{1}_{0}\frac{d(x-2)}{(x-2)^3}-9\int ^{1}_{0}\frac{d(2-x)}{(2-x)^4}\)
\(=4\int ^{-1}_{-2}\frac{dt}{t^3}-9\int ^{1}_{2}\frac{dk}{k^4}\) với \(x-2=t; 2-x=k\)
\(=4.\left.\begin{matrix} -1\\ -2\end{matrix}\right|\frac{t^{-3+1}}{-3+1}-9.\left.\begin{matrix} 1\\ 2\end{matrix}\right|\frac{k^{-4+1}}{-4+1}=\frac{9}{8}\)
Câu 3:
Phân số \(\frac{x^2+1}{(x^3+3x)^3}\) không xác định trên \([0;1]\); hàm không liên tục nên không có tích phân.
1/ I=\(\int_{-2}^2\left|x^2-1\right|dx\)
2/ I= \(\int_1^e\sqrt{x}.lnxdx\)
3/ I= \(\int_0^{\dfrac{\pi}{2}}\left(e^{sinx}+cosx\right)cosxdx\)
4/ I= \(\int_0^{\dfrac{pi}{2}}\dfrac{sin2x}{\sqrt{cos^2x+4sin^2x}}dx\)
5/ I= \(\int_0^{\dfrac{\pi}{4}}\sqrt{2}cos\sqrt{x}dx\)
6/ I= \(\int_1^{\sqrt{e}}\dfrac{1}{x\sqrt{1-ln^2x}}dx\)
7/ I= \(\int_{-\dfrac{\pi}{4}}^{\dfrac{\pi}{4}}\dfrac{sin^6x+cos^6x}{6^x+1}dx\)
Nhìn đề dữ dội y hệt cr của tui z :( Để làm từ từ
Lập bảng xét dấu cho \(\left|x^2-1\right|\) trên đoạn \(\left[-2;2\right]\)
x | -2 | -1 | 1 | 2 |
\(x^2-1\) | 0 | 0 |
\(\left(-2;-1\right):+\)
\(\left(-1;1\right):-\)
\(\left(1;2\right):+\)
\(\Rightarrow I=\int\limits^{-1}_{-2}\left|x^2-1\right|dx+\int\limits^1_{-1}\left|x^2-1\right|dx+\int\limits^2_1\left|x^2-1\right|dx\)
\(=\int\limits^{-1}_{-2}\left(x^2-1\right)dx-\int\limits^1_{-1}\left(x^2-1\right)dx+\int\limits^2_1\left(x^2-1\right)dx\)
\(=\left(\dfrac{x^3}{3}-x\right)|^{-1}_{-2}-\left(\dfrac{x^3}{3}-x\right)|^1_{-1}+\left(\dfrac{x^3}{3}-x\right)|^2_1\)
Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính
2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)
\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)
\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)
3/ \(I=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx+\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)
Xét \(A=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx\)
\(t=\sin x\Rightarrow dt=\cos x.dx\Rightarrow A=\int\limits^{\dfrac{\pi}{2}}_0e^t.dt=e^{\sin x}|^{\dfrac{\pi}{2}}_0\)
Xét \(B=\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)
\(=\int\limits^{\dfrac{\pi}{2}}_0\dfrac{1+\cos2x}{2}.dx=\dfrac{1}{2}.\int\limits^{\dfrac{\pi}{2}}_0dx+\dfrac{1}{2}\int\limits^{\dfrac{\pi}{2}}_0\cos2x.dx\)
\(=\dfrac{1}{2}x|^{\dfrac{\pi}{2}}_0+\dfrac{1}{2}.\dfrac{1}{2}\sin2x|^{\dfrac{\pi}{2}}_0\)
I=A+B=...
I=\(\int_0^1\)\(\frac{dx}{\sqrt{3+2x-x^2}}\)
J=\(\int_0^1\)xln(2x+1)dx
K=\(\int_0^1\)\(ln\left(x^3-3x+2\right)dx\)
cho f(x) dương liên tục trên [0;1] f(0)=1. Biết \(3\int_0^1\left[f'\left(x\right)\left[f\left(x\right)\right]^2+\frac{1}{9}\right]dx\le2\int_0^1\sqrt{f'\left(x\right)}f\left(x\right)dx\) . Tính \(\int_0^1\left[f\left(x\right)\right]^3dx\)
\(3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]dx\le2\int\limits^1_0\sqrt{f'\left(x\right)}f\left(x\right)dx\) (1)
Ta lại có:
\(3f'\left(x\right).f^2\left(x\right)+\frac{1}{3}\ge2\sqrt{f'\left(x\right)}.f\left(x\right)\)
\(\Rightarrow3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]\ge2\int\limits^1_0\sqrt{f'\left(x\right)}.f\left(x\right)dx\) (2)
Từ (1); (2) \(\Rightarrow3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]dx=2\int\limits^1_0\sqrt{f'\left(x\right)}.f\left(x\right)dx\)
Dấu "=" xảy ra khi và chỉ khi:
\(3f'\left(x\right).f^2\left(x\right)=\frac{1}{3}\Rightarrow3\int f'\left(x\right).f^2\left(x\right)dx=\int\frac{1}{3}dx\)
\(\Rightarrow f^3\left(x\right)=\frac{x}{3}+C\)
Thay \(x=0\Rightarrow f^3\left(0\right)=C\Rightarrow C=1\)
\(\Rightarrow f^3\left(x\right)=\frac{x}{3}+1\Rightarrow\int\limits^1_0f^3\left(x\right)dx=\int\limits^1_0\left(\frac{x}{3}+1\right)dx=\frac{7}{6}\)
\(\int_0^1\)\(\dfrac{x-2}{\left(x+1\right)^2}dx\)