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【例】过坐标原点作曲线<span class="mathquill-embedded-latex">y=lnx</span>的切线，该切线与曲线<span class="mathquill-embedded-latex">y=lnx</span>及<span class="mathquill-embedded-latex">x</span>轴围成平面图形<span class="mathquill-embedded-latex">D</span>。
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<p>
（1）求<span class="mathquill-embedded-latex">D</span>的面积<span class="mathquill-embedded-latex">A；</span>
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<p>
（2）求<span class="mathquill-embedded-latex">D</span>绕直线<span class="mathquill-embedded-latex">x=e</span>旋转一周所得旋转体的体积<span class="mathquill-embedded-latex">V</span>。
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<p>
【分析】先求出切点坐标及切线方程，再用定积分求面积面积<span class="mathquill-embedded-latex">A；</span>旋转体体积可用一大立体（圆锥）体积减去一小立体体积进行计算，为了帮助理解，可先画草图。
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<p>
解：（1）设切点的横坐标为<span class="mathquill-embedded-latex">{x}_{0}</span>，由曲线<span class="mathquill-embedded-latex">y=lnx</span>在点<span class="mathquill-embedded-latex">({x}_{0},ln{x}_{0})</span>处的切线方程是
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<p style="text-align: center;">
<span class="mathquill-embedded-latex">y=ln{x}_{0}+\frac {1} {{x}_{0}}(x-{x}_{0})</span>
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由该切线过原点知<span class="mathquill-embedded-latex">ln{x}_{0}-1=0</span>，从而<span class="mathquill-embedded-latex">{x}_{0}=e</span>。所以该切线的方程为
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">y=\frac {1} {e}x</span>.
</p>
<p>
平面图形<span class="mathquill-embedded-latex">D</span>的面积是
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<p style="text-align: center;">
<span class="mathquill-embedded-latex">A=\int ^{1}_{0} {({e}^{y}-ey)}dy=({e}^{y}-\frac {e} {2}{y}^{2}){|}^{1}_{0}=\frac {1} {2}e-1.</span>
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<p>
<br/>
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<p>
（2）切线<span class="mathquill-embedded-latex">y=\frac {1} {e}x</span>与<span class="mathquill-embedded-latex">x</span>轴及直线<span class="mathquill-embedded-latex">x=e</span>所围成的三角形绕直线<span class="mathquill-embedded-latex">x=e旋转所得的圆锥体体积为</span>
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">{V}_{1}=\frac {1} {3}\pi {e}^{2}</span>
</p>
<p>
曲线<span class="mathquill-embedded-latex">y=lnx</span>与<span class="mathquill-embedded-latex">x</span>轴及直线<span class="mathquill-embedded-latex">x=e</span>所围成的图形绕直线<span class="mathquill-embedded-latex">x=e</span>旋转所得的旋转体体积为
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">{V}_{2}=\int ^{1}_{0} {\pi {(e-{e}^{y})}^{2}}dy=\int ^{1}_{0} {\pi ({e}^{2}-2{e}^{y+1}+{e}^{2y})}dy</span>
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">=\pi {e}^{2}+\pi (-2{e}^{y+1}+\frac {1} {2}{e}^{2y}){|}^{1}_{0}=-\frac {\pi } {2}({e}^{2}+1)+2\pi e.</span>
</p>
<p>
因此所求旋转体的体积为
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">V={V}_{1}-{V}_{2}=\frac {1} {3}\pi {e}^{2}+\frac {\pi } {2}({e}^{2}+1)-2\pi e</span>
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">=\frac {\pi } {6}(5{e}^{2}-12e+3).</span>
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<p>
<span class="mathquill-embedded-latex"><br/></span>
</p>
<p>
【例】某建筑工地打地基时，需用汽锤将桩打进土层。汽锤每次击打，都将克服土层对桩的阻力而做功。设土层对桩的阻力的大小与桩被打进地下的深度成正比(比例系数为<span class="mathquill-embedded-latex">k</span>，<span class="mathquill-embedded-latex">k<0)</span>，汽锤第一次击打将桩打进地下<span class="mathquill-embedded-latex">a</span>（米）。根据设计方案，要求汽锤每次击打桩时所做的功与前一次击打时所做的功之比为常数<span class="mathquill-embedded-latex">k(0<k<1)。问：</span>
</p>
<p>
（1）汽锤击打桩3次后，可将桩打进地下多深？
</p>
<p>
（2）基击打次数不限，汽锤至多能将桩打进地下多深？
</p>
<p>
【分析】本题属于变力做功问题，可用定积分进行计算，而击打次数不限，相当于求数列的极限。
</p>
<p>
解：设第<span class="mathquill-embedded-latex">n</span>次击打后，桩被打进地下<span class="mathquill-embedded-latex">{x}_{n}</span>，第<span class="mathquill-embedded-latex">n</span>次击打时，汽锤所做的功为<span class="mathquill-embedded-latex">{W}_{n}(n=1,2,3,\cdots )</span>，通过求<span class="mathquill-embedded-latex">\sum ^{n}_{i=1} {{W}_{i}}</span>直接求出<span class="mathquill-embedded-latex">{x}_{n}</span>。因为当桩被打进地下的深度为<span class="mathquill-embedded-latex">x</span>时，土层对桩的阻力的大小为<span class="mathquill-embedded-latex">kx</span>，所以
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">{W}_{1}=\int ^{{x}_{1}}_{0} {kxdx=\frac {k} {2}}{x}^{2}_{1}=\frac {k} {2}{a}^{2}</span>,
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<p style="text-align: center;">
<span class="mathquill-embedded-latex">{W}_{2}=\int ^{{x}_{2}}_{{x}_{1}} {kxdx}</span>,
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">{W}_{3}=\int ^{{x}_{3}}_{{x}_{2}} {kxdx}</span>,
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<p style="text-align: center;">
…
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">{W}_{n}=\int ^{{x}_{n}}_{{x}_{n-1}} {kxdx}</span>,,
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<p>
相加得
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">\sum ^{n}_{i=1} {{W}_{i}}={W}_{1}+{W}_{2}+{W}_{3}+\cdots +{W}_{n}</span>
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">=\int ^{{x}_{1}}_{0} {kxdx}+\int ^{{x}_{2}}_{{x}_{1}} {kxdx}+\int ^{{x}_{3}}_{{x}_{2}} {kxdx}+\cdots +\int ^{{x}_{n}}_{{x}_{n-1}} {kxdx}</span>
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">=\int ^{{x}_{n}}_{0} {kxdx}=\frac {k} {2}{x}^{2}_{n},</span>
</p>
<p>
又由题设知，<span class="mathquill-embedded-latex">{W}_{i+1}=r{W}_{i},i=1,2,\cdots ,n.</span> 因此
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">\sum ^{n}_{i=1} {{W}_{i}}={W}_{1}+r{W}_{1}+{r}^{2}{W}_{1}+\cdots +{r}^{n-1}{W}_{1}</span>
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">=(1+r+{r}^{2}+\cdot +{r}^{n-1})\cdot \frac {1} {2}k{a}^{2}</span>
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">=\frac {1-{r}^{n}} {1-r}\cdot \frac {1} {2}k{a}^{2}.</span>
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<p>
从而
</p>
<p style="text-align: center;">
<span class="mathquill-embedded-latex">{x}_{n}=a\sqrt {\frac {1-{r}^{n}} {1-r}}(m)</span>
</p>
<p>
<br/>
</p>
<p>
因此
</p>
<p>
（1）<span class="mathquill-embedded-latex">{x}_{3}=\sqrt {1+r+{r}^{2}}a(m)</span>，即汽锤击打桩3次后，可将桩打进地下<span class="mathquill-embedded-latex">\sqrt {1+r+{r}^{2}}a(m)</span>。
</p>
<p>
（2）<span class="mathquill-embedded-latex">{\lim}_{n\to \infty }{x}_{n}=\frac {a} {\sqrt {1-r}}(m)</span>.
</p>
<p>
（3）$$\sum_{i=1}^{100}i$$.
</p>
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