\(\int\limits^1_{-1}\frac{x^4}{1+2^x}dx\)
Những câu hỏi liên quan
\(\int\limits^1_{\frac{\sqrt{3}}{3}}\frac{\sqrt{\left(1+x^2\right)^5}}{x^8}dx\)
\(\int\limits^1_{\sqrt{ }3}\)\(\sqrt{\left(1+x^2\right)}\)\(dx\)
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Tính các tích phân sau :
a) \(\int\limits^{\dfrac{\pi}{4}}_0\cos2x.\cos^2xdx\)
b) \(\int\limits^1_{\dfrac{1}{2}}\dfrac{e^x}{e^{2x}-1}dx\)
c) \(\int\limits^1_0\dfrac{x+2}{x^2+2x+1}\ln\left(x+1\right)dx\)
d) \(\int\limits^{\dfrac{\pi}{4}}_0\dfrac{x\sin x+\left(x+1\right)\cos x}{x\sin x+\cos x}dx\)
a)
Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)
\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)
\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)
b)
\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)
\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)
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c)
Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).
Đặt \(x+1=t\)
\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
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d)
\(D=\int ^{\frac{\pi}{4}}_{0}\frac{x\sin x+(x+1)\cos x}{x\sin x+\cos x}dx=\int ^{\frac{\pi}{4}}_{0}dx+\int ^{\frac{\pi}{4}}_{0}\frac{x\cos x}{x\sin x+\cos x}dx\)
Ta có:
\(\int ^{\frac{\pi}{4}}_{0}dx=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|x=\frac{\pi}{4}\)
\(\int ^{\frac{\pi}{4}}_{0}\frac{x\cos xdx}{x\sin x+\cos x}=\int ^{\frac{\pi}{4}}_{0}\frac{d(x\sin x+\cos x)}{x\sin x+\cos x}=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\ln |x\sin x+\cos x|\)
\(=\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)
Suy ra \(D=\frac{\pi}{4}+\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)
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Tính các tích phân sau
1.Iintlimits^{frac{Pi}{4}}_0 (x+1)sin2xdx
2.Iintlimits^2_1frac{x^2+3x+1}{x^2+x}dx
3.Iintlimits^2_1frac{x^2-1}{x^2}lnxdx
4. Iintlimits^1_0xsqrt{2-x^2}dx
5.Iintlimits^1_0frac{left(x+1right)^2}{x^2+1}dx
6. Iintlimits^5_1frac{dx}{1+sqrt{2x-1}}
7. Iintlimits^3_1frac{1+lnleft(x+1right)}{x^2}dx
8.Iintlimits^1_0frac{x^3}{x^4+3x^2+2}dx
9. Iintlimits^{frac{Pi}{4}}_0xleft(1+sin2xright)dx
10. Iintlimits^3_0frac{x}{sqrt{x+1}}dx
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Tính các tích phân sau
1.I=\(\int\limits^{\frac{\Pi}{4}}_0\) (x+1)sin2xdx
2.I=\(\int\limits^2_1\frac{x^2+3x+1}{x^2+x}dx\)
3.I=\(\int\limits^2_1\frac{x^2-1}{x^2}lnxdx\)
4. I=\(\int\limits^1_0x\sqrt{2-x^2}dx\)
5.I=\(\int\limits^1_0\frac{\left(x+1\right)^2}{x^2+1}dx\)
6. I=\(\int\limits^5_1\frac{dx}{1+\sqrt{2x-1}}\)
7. I=\(\int\limits^3_1\frac{1+ln\left(x+1\right)}{x^2}dx\)
8.I=\(\int\limits^1_0\frac{x^3}{x^4+3x^2+2}dx\)
9. I=\(\int\limits^{\frac{\Pi}{4}}_0x\left(1+sin2x\right)dx\)
10. I=\(\int\limits^3_0\frac{x}{\sqrt{x+1}}dx\)
https://i.imgur.com/Pe6vPSJ.jpg
Tính các tích phân sau đây :
a) intlimits^{dfrac{pi}{2}}_0left(x+1right)cosleft(x+dfrac{pi}{2}right)dx
b) intlimits^1_0dfrac{x^2+x+1}{x+1}log_2left(x+1right)dx
c) intlimits^1_{dfrac{1}{2}}dfrac{x^2-1}{x^4+1}dx (đặt tx+dfrac{1}{x} )
d) intlimits^{dfrac{pi}{2}}_0dfrac{sin3xdx}{3+4sin x-cos2x}dx
Đọc tiếp
Tính các tích phân sau đây :
a) \(\int\limits^{\dfrac{\pi}{2}}_0\left(x+1\right)\cos\left(x+\dfrac{\pi}{2}\right)dx\)
b) \(\int\limits^1_0\dfrac{x^2+x+1}{x+1}\log_2\left(x+1\right)dx\)
c) \(\int\limits^1_{\dfrac{1}{2}}\dfrac{x^2-1}{x^4+1}dx\) (đặt \(t=x+\dfrac{1}{x}\) )
d) \(\int\limits^{\dfrac{\pi}{2}}_0\dfrac{\sin3xdx}{3+4\sin x-\cos2x}dx\)
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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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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1)\(\int\limits^1_0\frac{\left(3x^2+2\right)}{x^3+x^2+1}dx\)
2)\(\int\limits^1_0\frac{x}{x^{2+4}}dx\)
1)\(\int\limits^1_0\frac{\left(3x^2+2\right)}{x^3+x^2+1}dx\)
2)\(\int\limits^1_0\frac{x}{x^2+4}dx\)
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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