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| （未顯示由 15 位使用者於中間所作的 189 次修訂） |
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| {{SandboxNote}}
| | #REDIRECT [[主命名空间沙盒]] |
| <math>
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| \max \left [ f\left ( \frac{\sqrt{6}}{2}\right ),f(1),f(2) \right ]=\frac{3}{4}\left (2 \sqrt{6}-3\right ),\min \left [ f\left ( \frac{\sqrt{6}}{2}\right ),f(1),f(2)\right ]=
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| 所以 \frac{0}{2-1}< \int_{1}^{2}f(x)\mathrm{d}x< \frac{\frac{3}{4}\left (2 \sqrt{6}-3\right )}{2-1}
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| 即0< \int_{1}^{2}(2x^{3}-x^{4})\mathrm{d}x< \frac{3}{4}\left (2 \sqrt{6}-3\right )
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| 4.(1)F(x)=\int_{0}^{x}\sqrt{1+t}\mathrm{d}t
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| 解：F^{'}(x)=\sqrt{1+x}
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| (2)F(x)=\int_{x}^{-1}te^{-t}\mathrm{d}t
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| 解：原式=-\int_{-1}^{x}te^{-t}\mathrm{d}t
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| =-\int_{-1}^{0}te^{-t}\mathrm{d}t-\int_{0}^{x}te^{-t}\mathrm{d}t
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| 所以F^{'}(x)=\left ( -\int_{-1}^{0}te^{-t}\mathrm{d}t-\int_{0}^{x}te^{-t}\mathrm{d}t\right )^{'}=\left (-\int_{-1}^{0}te^{-t}\mathrm{d}t \right )^{'}-\left ( \int_{0}^{x}te^{-t}\mathrm{d}t\right )^{'}
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| =-xe^{-x}
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| (3)F(x)=\int_{0}^{x^{2}}\frac{1}{\sqrt{1+t^{4}}}\mathrm{d}t
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| 解：F^{'}(x)=\frac{1}{\sqrt{1+ x^{8}}}\cdot 2x
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| =\frac{2x}{\sqrt{1+ x^{8}}}
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| <math>
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| ----
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| <\math>
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| 5.(4)\int_{-2}^{3}\left ( x-1\right )^{3}\mathrm{d}x
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| 解：\int \left ( x-1\right )^{3}\mathrm{d}x=\frac{1}{4}\left ( x-1\right )^{4}+C
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| 故原式=\left [ \frac{1}{4}\left ( x-1\right )^{4}\right ]_{-2}^{3}
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| =\frac{1}{4}\left ( 3-1\right )^{4}-\frac{1}{4}\left ( -2-1\right )^{4}=-\frac{73}{4}
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| (13)\int_{-1}^{2}\left | 2x\right |\mathrm{d}x
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| 解：\int_{-1}^{2}\left | 2x\right |\mathrm{d}x=\int_{0}^{2}2x\mathrm{d}x-\int_{-1}^{0}2x\mathrm{d}x
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| \int 2x\mathrm{d}x=x^{2}+C
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| 故原式=\left [ x^{2}\right ]_{0}^{2}-\left [ x^{2}\right ]_{-1}^{0}=5
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| 6.(7)\int_{0}^{\ln 2}\sqrt{e^{x}-1}\mathrm{d}x
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| 解：设\int\sqrt{e^{x}-1}\mathrm{d}x=I
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| 令t=\sqrt{e^{x}-1}，则x=\ln \left ( t^{2}+1\right )，\mathrm{d}x=\frac{2t}{t^{2}+1}，
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| I=\int \frac{2t^{2}}{t^{2}+1}\mathrm{d}t=2t-2\arctan t+C=2\sqrt{e^{x}-1}-2\arctan \sqrt{e^{x}-1}+C
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| 故原式=\left [ 2\sqrt{e^{x}-1}-2\arctan \sqrt{e^{x}-1}\right ]_{0}^{\ln 2}=2-\frac{\pi}{2}
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| (10)\int_{1}^{2}\frac{\sqrt{x^{2}-1}}{x}\mathrm{d}x
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| 解：设\int\frac{\sqrt{x^{2}-1}}{x}\mathrm{d}x=I
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| 令x=\frac{1}{\cos t}，则t=\arccos \frac{1}{x}，\mathrm{d}x=\frac{\sin t}{\cos^{2}t}\mathrm{d}t，
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| I=\int\frac{\sin t}{\cos t}\cdot \cos t\cdot \frac{\sin t}{\cos^{2}t}\mathrm{d}t
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| =\int\tan^{2}t\mathrm{d}t=\tan t -t+C=\sqrt{x^{2}-1}-\arccos \frac{1}{x}+C
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| 故原式=\left [ \sqrt{x^{2}-1}-\arccos \frac{1}{x}\right ]_{1}^{2}=\sqrt{3}-\frac{\pi}{3}
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| 12.(1)\int_{1}^{e}\ln x\mathrm{d}x
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| 解：\int\ln x\mathrm{d}x=x\ln x-x+C
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| 故原式=\left [x\ln x-x\right ]_{1}^{e}=1
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| 12.(5)\int_{0}^{\frac{\pi}{2}}x\sin x\mathrm{d}x
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| 解：\int x\sin x\mathrm{d}x=\cos x-x\cos x+C
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| 故原式=\left [\cos x-x\cos x\right ]_{0}^{\frac{\pi}{2}}=1
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| <\math>
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