Tính:
a) \(\int\limits^2_1\dfrac{1}{x^3}dx\);
b) \(\int\limits^3_1x^{\dfrac{2}{3}}dx\);
c) \(\int\limits^8_1\sqrt[3]{x}dx\).
Những câu hỏi liên quan
Tính cách tích phân sau :
a) \(\int\limits^1_0\left(1+3x\right)^{\dfrac{3}{2}}dx\)
b) \(\int\limits^{\dfrac{1}{2}}_0\dfrac{x^3-1}{x^2-1}dx\)
c) \(\int\limits^2_1\dfrac{ln\left(1+x\right)}{x^2}dx\)
1, I intlimits^1_0dfrac{2x+1}{x^2+x+1}dx
2,intlimits^{dfrac{1}{2}}_0dfrac{5xdx}{left(1-x^2right)^3}
3, intlimits^1_0dfrac{2x}{left(x+1right)^3}dx
4, intlimits^1_0dfrac{4x-2}{left(x^2+1right)left(x+2right)}dx
5, intlimits^1_0dfrac{x^2dx}{x^6-9}
6, intlimits^2_1dfrac{2x-1}{x^2left(x+1right)}dx
Đọc tiếp
1, I = \(\int\limits^1_0\dfrac{2x+1}{x^2+x+1}dx\)
2,\(\int\limits^{\dfrac{1}{2}}_0\dfrac{5xdx}{\left(1-x^2\right)^3}\)
3, \(\int\limits^1_0\dfrac{2x}{\left(x+1\right)^3}dx\)
4, \(\int\limits^1_0\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}dx\)
5, \(\int\limits^1_0\dfrac{x^2dx}{x^6-9}\)
6, \(\int\limits^2_1\dfrac{2x-1}{x^2\left(x+1\right)}dx\)
1/ \(I=\int\limits^1_0\dfrac{2x+1}{x^2+x+1}dx=\int\limits^1_0\dfrac{d\left(x^2+x+1\right)}{x^2+x+1}=ln\left|x^2+x+1\right||^1_0=ln3\)
2/ \(\int\limits^{\dfrac{1}{2}}_0\dfrac{5x}{\left(1-x^2\right)^3}dx=-\dfrac{5}{2}\int\limits^{\dfrac{1}{2}}_0\dfrac{d\left(1-x^2\right)}{\left(1-x^2\right)^3}=\dfrac{5}{4}\dfrac{1}{\left(1-x^2\right)^2}|^{\dfrac{1}{2}}_0=\dfrac{35}{36}\)
3/ \(\int\limits^1_0\dfrac{2x}{\left(x+1\right)^3}dx\Rightarrow\) đặt \(x+1=t\Rightarrow x=t-1\Rightarrow dx=dt;\left\{{}\begin{matrix}x=0\Rightarrow t=1\\x=1\Rightarrow t=2\end{matrix}\right.\)
\(I=\int\limits^2_1\dfrac{2\left(t-1\right)dt}{t^3}=\int\limits^2_1\left(\dfrac{2}{t^2}-\dfrac{2}{t^3}\right)dt=\left(\dfrac{-2}{t}+\dfrac{1}{t^2}\right)|^2_1=\dfrac{1}{4}\)
4/ \(\int\limits^1_0\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}dx\)
Kĩ thuật chung là tách và sử dụng hệ số bất định như sau:
\(\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}=\dfrac{ax+b}{x^2+1}+\dfrac{c}{x+2}=\dfrac{\left(a+c\right)x^2+\left(2a+b\right)x+2b+c}{\left(x^2+1\right)\left(x+2\right)}\)
\(\Rightarrow\left\{{}\begin{matrix}a+c=0\\2a+b=4\\2b+c=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=0\\a=-c=2\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^1_0\left(\dfrac{2x}{x^2+1}-\dfrac{2}{x+2}\right)dx=\int\limits^1_0\dfrac{d\left(x^2+1\right)}{x^2+1}-2\int\limits^1_0\dfrac{d\left(x+2\right)}{x+2}=ln\dfrac{8}{9}\)
5/ \(\int\limits^1_0\dfrac{x^2dx}{x^6-9}\Rightarrow\) đặt \(x^3=t\Rightarrow3x^2dx=dt\Rightarrow x^2dx=\dfrac{1}{3}dt;\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=1\Rightarrow t=1\end{matrix}\right.\)
\(I=\dfrac{1}{3}\int\limits^1_0\dfrac{dt}{t^2-9}=\dfrac{1}{18}\int\limits^1_0\left(\dfrac{1}{t-3}-\dfrac{1}{t+3}\right)dt=\dfrac{1}{18}ln\left|\dfrac{t-3}{t+3}\right||^1_0=-\dfrac{1}{18}ln2\)
6/ Tương tự câu 4, sử dụng hệ số bất định ta tách được:
\(\int\limits^2_1\dfrac{2x-1}{x^2\left(x+1\right)}dx=\int\limits^2_1\left(\dfrac{3x-1}{x^2}-\dfrac{3}{x+1}\right)dx=\int\limits^2_1\left(\dfrac{3}{x}-\dfrac{1}{x^2}-\dfrac{3}{x+1}\right)dx\)
\(=\left(3ln\left|\dfrac{x}{x+1}\right|+\dfrac{1}{x}\right)|^2_1=3ln\dfrac{4}{3}-\dfrac{1}{2}\)
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Nếu \(\int\limits^2_1\) f(x) dx = -2 và \(\int\limits^3_2\) f(x) dx =1 thì \(\int\limits^3_1\) f(x) dx bằng
A. -3
B. -1
C. 1
D. 3
\(\int\limits^3_1f\left(x\right)dx=-2+1=-1\)
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Tính :
a) intlimits^2_{-1}left(5x^2-x+e^{0,5x}right)dx
b) intlimits^2_{0,5}left(2sqrt{x}+dfrac{3}{x^2}+cos xright)dx
c) intlimits^2_1dfrac{dx}{sqrt{2x+3}} (đặt tsqrt{2x+3} )
d) intlimits^2_1sqrt[3]{3x^3+4}x^2dx (đặt tsqrt[3]{3x^3+4} )
e) intlimits^2_{-2}left(x-2right)left|xright|dx
g) intlimits^0_1xcos xdx
h) intlimits^{dfrac{pi}{2}}_{dfrac{pi}{6}}dfrac{1+sin2x+cos2x}{sin x+cos x}dx
i) intlimits^{dfrac{pi}{2}}_0e^xsin xdx
k) intlimits^e_1x^2ln^2xdx
Đọc tiếp
Tính :
a) \(\int\limits^2_{-1}\left(5x^2-x+e^{0,5x}\right)dx\)
b) \(\int\limits^2_{0,5}\left(2\sqrt{x}+\dfrac{3}{x^2}+\cos x\right)dx\)
c) \(\int\limits^2_1\dfrac{dx}{\sqrt{2x+3}}\) (đặt \(t=\sqrt{2x+3}\) )
d) \(\int\limits^2_1\sqrt[3]{3x^3+4}x^2dx\) (đặt \(t=\sqrt[3]{3x^3+4}\) )
e) \(\int\limits^2_{-2}\left(x-2\right)\left|x\right|dx\)
g) \(\int\limits^0_1x\cos xdx\)
h) \(\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{1+\sin2x+\cos2x}{\sin x+\cos x}dx\)
i) \(\int\limits^{\dfrac{\pi}{2}}_0e^x\sin xdx\)
k) \(\int\limits^e_1x^2\ln^2xdx\)
Bài tập: Tính.
b, \(\int\limits^{\dfrac{\pi}{6}}_0cos2xdx\) d, \(\int\limits^2_1\dfrac{dx}{\left(2x-1\right)^2}\)
c, \(\int\limits^1_{-1}\left(2x+1\right)^3dx\)
\(b=\dfrac{1}{2}\int\limits^{\dfrac{\pi}{6}}_0cos2xd\left(2x\right)=\dfrac{1}{2}sin2x|^{\dfrac{\pi}{6}}_0=\dfrac{\sqrt{3}}{4}\)
\(c=\dfrac{1}{2}\int\limits^1_{-1}\left(2x+1\right)^3d\left(2x+1\right)=\dfrac{1}{2}.\dfrac{1}{4}\left(2x+1\right)^4|^1_{-1}=10\)
\(d=\dfrac{1}{2}\int\limits^2_1\dfrac{d\left(2x-1\right)}{\left(2x-1\right)^2}=-\dfrac{1}{2}.\dfrac{1}{\left(2x-1\right)}|^2_1=\dfrac{1}{3}\)
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1) \(\int\limits^2_1\dfrac{\sqrt{x^2-1}}{x}dx\)
2) \(\int x.\sqrt{1-x^4}dx\)
3)\(\int\dfrac{1}{x^2.\sqrt{25-x^2}}dx\)
4) \(\int\dfrac{\sqrt{x^2-9}}{x^3}dx\)
Tính :
a) intlimits^{dfrac{pi}{2}}_0cos2x.sin^2dx
b) intlimits^1_{-1}left|2^x-2^{-x}right|dx
c) intlimits^2_1dfrac{left(x+1right)left(x+2right)left(x+3right)}{x^2}dx
d) intlimits^2_0dfrac{1}{x^2-2x-3}dx
e) intlimits^{dfrac{pi}{2}}_0left(sin x+cos xright)^2dx
g) intlimits^{pi}_0left(x+sin xright)^2dx
Đọc tiếp
Tính :
a) \(\int\limits^{\dfrac{\pi}{2}}_0\cos2x.\sin^2dx\)
b) \(\int\limits^1_{-1}\left|2^x-2^{-x}\right|dx\)
c) \(\int\limits^2_1\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x^2}dx\)
d) \(\int\limits^2_0\dfrac{1}{x^2-2x-3}dx\)
e) \(\int\limits^{\dfrac{\pi}{2}}_0\left(\sin x+\cos x\right)^2dx\)
g) \(\int\limits^{\pi}_0\left(x+\sin x\right)^2dx\)
a)
Ta có:
∫π20cos2xsin2xdx=12∫π20cos2x(1−cos2x)dx=12∫π20[cos2x−1+cos4x2]dx=14∫π20(2cos2x−cos4x−1)dx=14[sin2x−sin4x4−x]π20=−14.π2=−π8∫0π2cos2xsin2xdx=12∫0π2cos2x(1−cos2x)dx=12∫0π2[cos2x−1+cos4x2]dx=14∫0π2(2cos2x−cos4x−1)dx=14[sin2x−sin4x4−x]0π2=−14.π2=−π8
b)
Ta có: Xét 2x – 2-x ≥ 0 ⇔ x ≥ 0.
Ta tách thành tổng của hai tích phân:
∫1−1|2x−2−x|dx=−∫0−1(2x−2−x)dx+∫10(2x−2−x)dx=−(2xln2+2−xln2)∣∣0−1+(2xln2+2−xln2)∣∣10=1ln2∫−11|2x−2−x|dx=−∫−10(2x−2−x)dx+∫01(2x−2−x)dx=−(2xln2+2−xln2)|−10+(2xln2+2−xln2)|01=1ln2
c)
∫21(x+1)(x+2)(x+3)x2dx=∫21x3+6x2+11x+6x2dx=∫21(x+6+11x+6x2)dx=[x22+6x+11ln|x|−6x]∣∣21=(2+12+11ln2−3)−(12+6−6)=212+11ln2∫12(x+1)(x+2)(x+3)x2dx=∫12x3+6x2+11x+6x2dx=∫12(x+6+11x+6x2)dx=[x22+6x+11ln|x|−6x]|12=(2+12+11ln2−3)−(12+6−6)=212+11ln2
d)
∫201x2−2x−3dx=∫201(x+1)(x−3)dx=14∫20(1x−3−1x+1)dx=14[ln|x−3|−ln|x+1|]∣∣20=14[1−ln2−ln3]=14(1−ln6)∫021x2−2x−3dx=∫021(x+1)(x−3)dx=14∫02(1x−3−1x+1)dx=14[ln|x−3|−ln|x+1|]|02=14[1−ln2−ln3]=14(1−ln6)
e)
∫π20(sinx+cosx)2dx=∫π20(1+sin2x)dx=[x−cos2x2]∣∣π20=π2+1∫0π2(sinx+cosx)2dx=∫0π2(1+sin2x)dx=[x−cos2x2]|0π2=π2+1
g)
I=∫π0(x+sinx)2dx∫π0(x2+2xsinx+sin2x)dx=[x33]∣∣π0+2∫π0xsinxdx+12∫π0(1−cos2x)dxI=∫0π(x+sinx)2dx∫0π(x2+2xsinx+sin2x)dx=[x33]|0π+2∫0πxsinxdx+12∫0π(1−cos2x)dx
Tính :J=∫π0xsinxdxJ=∫0πxsinxdx
Đặt u = x ⇒ u’ = 1 và v’ = sinx ⇒ v = -cos x
Suy ra:
J=[−xcosx]∣∣π0+∫π0cosxdx=π+[sinx]∣∣π0=πJ=[−xcosx]|0π+∫0πcosxdx=π+[sinx]|0π=π
Do đó:
I=π33+2π+12[x−sin2x2]∣∣π30=π33+2π+π2=2π3+15π6
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Tính các tích phân:
a) intlimits^1_0dfrac{xe^x+1+x}{e^x+1}dx
b)intlimits^{dfrac{pi}{2}}_0dfrac{1-sinleft(xright)}{1+cosleft(xright)}dx
c)intlimits^2_1dfrac{left(x-1right)lnleft(xright)}{x^2}dx
d)intlimits^e_1ln( x + 1)dx
Đọc tiếp
Tính các tích phân:
a) \(\int\limits^1_0\)\(\dfrac{xe^x+1+x}{e^x+1}\)dx
b)\(\int\limits^{\dfrac{\pi}{2}}_0\)\(\dfrac{1-\sin\left(x\right)}{1+\cos\left(x\right)}\)dx
c)\(\int\limits^2_1\)\(\dfrac{\left(x-1\right)ln\left(x\right)}{x^2}\)dx
d)\(\int\limits^e_1\)ln( x + 1)dx
Câu a)
\(I=\int ^{1}_{0}\frac{x(e^x+1)+1}{e^x+1}dx=\int ^{1}_{0}xdx+\int ^{1}_{0}\frac{dx}{e^x+1}\)
\(=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2}{2}+\int ^{1}_{0}\frac{d(e^x)}{e^x(e^x+1)}=\frac{1}{2}+\left.\begin{matrix} 1\\ 0\end{matrix}\right|\ln\left | \frac{e^x}{e^x+1} \right |\)
\(\Leftrightarrow I=\frac{3}{2}+\ln 2-\ln (e+1)\)
Câu d)
\(I=\int ^{e}_{1}\ln(x+1)d(x)=\int ^{e}_{1}\ln (x+1)d(x+1)\)
Đặt \(\left\{\begin{matrix} u=\ln (x+1)\\ dv=d(x+1)\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{d(x+1)}{x+1}\\ v=x+1\end{matrix}\right.\)
\(\Rightarrow I=\left.\begin{matrix} e\\ 1\end{matrix}\right|(x+1)\ln (x+1)-\int ^{e}_{1}d(x+1)\)
\(=(e+1)\ln \left ( \frac{e+1}{e} \right )-2\ln \left (\frac{2}{e}\right )\)
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Câu b)
Đặt \(\tan \frac{x}{2}=t\). Ta có:
\(\left\{\begin{matrix} dt=d\left ( \tan \frac{x}{2} \right )=\frac{1}{2\cos ^2\frac{x}{2}}dx=\frac{t^2+1}{2}dx\rightarrow dx=\frac{2dt}{t^2+1}\\\ \cos x=\frac{1-t^2}{t^2+1}\end{matrix}\right.\)
\( I=\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{1}{1+\cos x}dx}_{A}+\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{d(\cos x)}{\cos x+1}}_{B}\)
Có \(B=\int ^{\frac{\pi}{2}}_{0}\frac{d(\cos x+1)}{\cos x+1}=\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\ln |\cos x+1|=-\ln 2\)
\(A=\int ^{1}_{0}\frac{2dt}{(t^2+1)\frac{2}{t^2+1}}=\int ^{1}_{0}dt=1\)
\(\Rightarrow I=A+B=1-\ln 2\)
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Câu c)
Xét \(I=\underbrace{\int ^{2}_{1}\frac{\ln xdx}{x}}_{A}-\underbrace{\int ^{2}_{1}\frac{\ln xdx}{x^2}}_{B}\)
Có \( A=\int ^{2}_{1}\ln xd(\ln x)=\left.\begin{matrix} 2\\ 1\end{matrix}\right|\frac{\ln ^2 x}{2}=\frac{\ln ^2 2}{2}\)
Với \(B\) đặt \(\left\{\begin{matrix} u=\ln x\\ dv=\frac{dx}{x^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dx}{x}\\ v=\frac{-1}{x}\end{matrix}\right.\)
\(\Rightarrow B=\left.\begin{matrix} 2\\ 1\end{matrix}\right|\frac{-\ln x}{x}+\int ^{2}_{1}\frac{dx}{x^2}=\left.\begin{matrix} 2\\ 1\end{matrix}\right|\left ( \frac{-\ln x}{x}-\frac{1}{x} \right )=\frac{1}{2}-\frac{\ln 2}{2}\)
\(\Rightarrow I=A-B=\frac{\ln ^2 2+\ln 2-1}{2}\)
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câu 2: a) cho hàm số y=f(x) có đạo hàm liên tục trên[1;2] thỏa mãn \(\int\limits^2_1\) f'(x)dx=10 và \(\int\limits^2_1\) \(\dfrac{f'\left(x\right)}{f\left(x\right)}\) dx=ln2.Biết f(x)>0 ∀x∈[1;2] . tính f(2)b)Tính diện tích hinh phẳng giới hạn bởi...
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2a. Đề sai, nhìn biểu thức \(\dfrac{f'\left(x\right)}{f'\left(x\right)}dx\) là thấy
2b. Đồ thị hàm số không cắt Ox trên \(\left(0;1\right)\) nên diện tích cần tìm:
\(S=\int\limits^1_0\left(x^4-5x^2+4\right)dx=\dfrac{38}{15}\)
3a. Phương trình (P) theo đoạn chắn:
\(\dfrac{x}{4}+\dfrac{y}{-1}+\dfrac{z}{-2}=1\)
3b. Câu này đề sai, đề cho mặt phẳng (Q) rồi thì sao lại còn viết pt mặt phẳng (Q) nữa?
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3b. \(\overrightarrow{n_{\left(Q\right)}}=\left(3;-1;-2\right)\)
Do (P) song song (Q) nên (P) cũng nhận \(\left(3;-1;-2\right)\) là 1 vtpt
Do đó pt (P) có dạng:
\(3\left(x-0\right)-1\left(y-0\right)-2\left(z-1\right)=0\)
\(\Leftrightarrow3x-y-2z+2=0\)
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Tính tích phân bằng định nghĩa và các tính chất:
1. \(\int\limits^e_1\left(x+\frac{1}{x}+\frac{1}{x^2}\right)dx\)
2. \(\int\limits^2_1\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)dx\)
3. \(\int\limits^2_1\frac{2x^3-4x+5}{x}dx\)
4. \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx\)
1/ \(\int\limits^e_1\left(x+\frac{1}{x}+\frac{1}{x^2}\right)dx=\left(\frac{x^2}{2}+lnx-\frac{1}{x}\right)|^e_1=\frac{e^2}{2}-\frac{1}{e}+\frac{3}{2}\)
2/ \(\int\limits^2_1\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)dx=\int\limits^2_1\left(x\sqrt{x}+1\right)dx=\int\limits^2_1\left(x^{\frac{3}{2}}+1\right)dx\)
\(=\left(\frac{2}{5}.x^{\frac{5}{2}}+x\right)|^2_1=\frac{8\sqrt{2}-7}{5}\)
3/
\(\int\limits^2_1\frac{2x^3-4x+5}{x}dx=\int\limits^2_1\left(2x^2-4+\frac{5}{x}\right)dx=\left(\frac{2}{3}x^3-4x+5lnx\right)|^2_1=\frac{2}{3}+5ln2\)
4/ \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx=\int\limits^2_1\left(6x^2-2x\right)dx=\left(2x^3-x^2\right)|^2_1=11\)
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