高等数学（第七版）同济大学 习题4-1 个人解答

高等数学（第七版）同济大学 习题4-1

 

$\begin{array}{rlr}& 1.利用求导运算验证下列等式：& \end{array}$

$\begin{array}{rlr}& (1)\int \frac{1}{\sqrt{x^2+1}}dx=ln(x+\sqrt{x^2+1})+C；& \\ & & \\ & (2)\int \frac{1}{x^2\sqrt{x^2-1}}dx=\frac{\sqrt{x^2-1}}{x}+C；& \\ & & \\ & (3)\int \frac{2x}{(x^2+1)(x+1{)}^2}dx=arctanx+\frac{1}{x+1}+C；& \\ & & \\ & (4)\int secxdx=ln|tanx+secx|+C；& \\ & & \\ & (5)\int xcosxdx=xsinx+cosx+C；& \\ & & \\ & (6)\int e^{x}sinxdx=\frac{1}{2}{e}^x(sinx-cosx)+C.& \end{array}$

解：

$\begin{array}{rlr}& (1)[ln(x+\sqrt{x^2+1})+C{]}^{\prime }=\frac{1}{x+\sqrt{x^2+1}}\cdot \left(1+\frac{x}{\sqrt{x^2+1}}\right)=\frac{1}{\sqrt{x^2+1}}& \\ & & \\ & (2){\left(\frac{\sqrt{x^2-1}}{x}+C\right)}^{\prime }=\frac{\frac{x}{\sqrt{x^2-1}}\cdot x-\sqrt{x^2-1}}{x^2}=\frac{1}{x^2\sqrt{x^2-1}}& \\ & & \\ & (3){\left(arctanx+\frac{1}{x+1}+C\right)}^{\prime }=\frac{1}{x^2+1}-\frac{1}{(x+1{)}^2}=\frac{2x}{(x^2+1)(x+1{)}^2}& \\ & & \\ & (4)(ln|tanx+secx|+C{)}^{\prime }=\frac{1}{tanx+secx}\cdot (se{c}^{2}x+secxtanx)=secx& \\ & & \\ & (5)(xsinx+cosx+C{)}^{\prime }=sinx+xcosx-sinx=xcosx& \\ & & \\ & (6){\left[\frac{1}{2}{e}^x(sinx-cosx)+C\right]}^{\prime }=\frac{1}{2}{e}^x(sinx-cosx)+\frac{1}{2}{e}^x(cosx+sinx)=e^{x}sinx& \end{array}$

---

$\begin{array}{rlr}& 2.求下列不定积分：& \end{array}$

$\begin{array}{rlr}& (1)\int \frac{dx}{x^2}；(2)\int x\sqrt{x}dx；& \\ & & \\ & (3)\int \frac{dx}{\sqrt{x}}；(4)\int x^2\sqrt[3]{x}dx；& \\ & & \\ & (5)\int \frac{dx}{x^2\sqrt{x}}；(6)\int \sqrt[m]{x^n}dx；& \\ & & \\ & (7)\int 5x^{3}dx；(8)\int (x^2-3x+2)dx；& \\ & & \\ & (9)\int \frac{dh}{\sqrt{2gh}}(g是常数)；(10)\int (x^2+1{)}^{2}dx；& \\ & & \\ & (11)\int (\sqrt{x}+1)(\sqrt{x^3}-1)dx；(12)\int \frac{(1-x{)}^2}{\sqrt{x}}dx；& \\ & & \\ & (13)\int \left(2e^x+\frac{3}{x}\right)dx；(14)\int \left(\frac{3}{1+x^2}-\frac{2}{\sqrt{1-x^2}}\right)dx；& \\ & & \\ & (15)\int e^x\left(1-\frac{e^{-x}}{\sqrt{x}}\right)dx；(16)\int 3^x{e}^{x}dx；& \\ & & \\ & (17)\int \frac{2\cdot 3^x-5\cdot 2^x}{3^x}dx；(18)\int secx(secx-tanx)dx；& \\ & & \\ & (19)\int co{s}^2\frac{x}{2}dx；(20)\int \frac{dx}{1+cos2x}；& \\ & & \\ & (21)\int \frac{cos2x}{cosx-sinx}dx；(22)\int \frac{cos2x}{co{s}^{2}xsi{n}^{2}x}dx；& \\ & & \\ & (23)\int co{t}^{2}xdx；(24)\int cos\theta (tan\theta +sec\theta )d\theta ；& \\ & & \\ & (25)\int \frac{x^2}{x^2+1}dx；(26)\int \frac{3x^4+2x^2}{x^2+1}dx.& \end{array}$

解：

$\begin{array}{rlr}& (1)\int \frac{dx}{x^2}=\int x^{-2}dx=\frac{x^{-2+1}}{-2+1}+C=-\frac{1}{x}+C& \\ & & \\ & (2)\int x\sqrt{x}dx=\int x^{\frac{3}{2}}dx=\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+C=\frac{2}{5}{x}^{\frac{5}{2}}+C& \\ & & \\ & (3)\int \frac{dx}{\sqrt{x}}=\int x^{-\frac{1}{2}}dx=\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C=2\sqrt{x}+C& \\ & & \\ & (4)\int x^2\sqrt[3]{x}dx=\int x^{\frac{7}{3}}dx=\frac{x^{\frac{7}{3}+1}}{\frac{7}{3}+1}+C=\frac{3}{10}{x}^{\frac{10}{3}}+C& \\ & & \\ & (5)\int \frac{dx}{x^2\sqrt{x}}=\int x^{-\frac{5}{2}}dx=\frac{x^{-\frac{5}{2}+1}}{-\frac{5}{2}+1}+C=-\frac{2}{3}{x}^{-\frac{3}{2}}+C& \\ & & \\ & (6)\int \sqrt[m]{x^n}dx=\int x^{\frac{n}{m}}dx=\frac{x^{\frac{n}{m}+1}}{\frac{n}{m}+1}+C=\frac{m}{m+n}{x}^{\frac{m+n}{m}}+C& \\ & & \\ & (7)\int 5x^{3}dx=5\frac{x^{3+1}}{3+1}+C=\frac{5}{4}{x}^4+C& \\ & & \\ & (8)\int (x^2-3x+2)dx=\int x^{2}dx-3\int xdx+2\int dx=\frac{1}{3}{x}^3-\frac{3}{2}{x}^2+2x+C& \\ & & \\ & (9)\int \frac{dh}{\sqrt{2gh}}=\frac{1}{\sqrt{2g}}\int h^{-\frac{1}{2}}dh=\sqrt{\frac{2h}{g}}+C& \\ & & \\ & (10)\int (x^2+1{)}^{2}dx=\int (x^4+2x^2+1)dx=\int x^{4}dx+2\int x^{2}dx+\int dx=\frac{1}{5}{x}^5+\frac{2}{3}{x}^3+x+C& \\ & & \\ & (11)\int (\sqrt{x}+1)(\sqrt{x^3}-1)dx=\int (x^2+x^{\frac{3}{2}}-x^{\frac{1}{2}}-1)dx=& \\ & & \\ & \int x^{2}dx+\int x^{\frac{3}{2}}dx-\int x^{\frac{1}{2}}dx-\int dx=\frac{1}{3}{x}^3+\frac{2}{5}{x}^{\frac{5}{2}}-\frac{2}{3}{x}^{\frac{3}{2}}-x+C& \\ & & \\ & (12)\int \frac{(1-x{)}^2}{\sqrt{x}}dx=\int (x^{\frac{3}{2}}-2x^{\frac{1}{2}}+x^{-\frac{1}{2}})dx=\int x^{\frac{3}{2}}dx-2\int x^{\frac{1}{2}}dx+\int x^{-\frac{1}{2}}dx=\frac{2}{5}{x}^{\frac{5}{2}}-\frac{4}{3}{x}^{\frac{3}{2}}+2x^{\frac{1}{2}}+C& \\ & & \\ & (13)\int \left(2e^x+\frac{3}{x}\right)dx=2\int e^{x}dx+3\int \frac{dx}{x}=2e^x+3ln|x|+C& \\ & & \\ & (14)\int \left(\frac{3}{1+x^2}-\frac{2}{\sqrt{1-x^2}}\right)dx=3\int \frac{dx}{1+x^2}-2\int \frac{dx}{\sqrt{1-x^2}}=3arctanx-2arcsinx+C& \\ & & \\ & (15)\int e^x\left(1-\frac{e^{-x}}{\sqrt{x}}\right)dx=\int e^{x}dx-\int x^{-\frac{1}{2}}dx=e^x-2x^{\frac{1}{2}}+C& \\ & & \\ & (16)\int 3^x{e}^{x}dx=\int (3e{)}^{x}dx=\frac{(3e{)}^x}{ln(3e)}+C=\frac{3^x{e}^x}{ln3+1}+C& \\ & & \\ & (17)\int \frac{2\cdot 3^x-5\cdot 2^x}{3^x}dx=2\int dx-5\int {\left(\frac{2}{3}\right)}^{x}dx=2x-\frac{5}{ln\frac{2}{3}}{\left(\frac{2}{3}\right)}^x+C=2x-\frac{5}{ln2-ln3}{\left(\frac{2}{3}\right)}^x+C& \\ & & \\ & (18)\int secx(secx-tanx)dx=\int se{c}^{2}xdx-\int secxtanxdx=tanx-secx+C& \\ & & \\ & (19)\int co{s}^2\frac{x}{2}dx=\int \frac{1+cosx}{2}dx=\frac{x+sinx}{2}+C& \\ & & \\ & (20)\int \frac{dx}{1+cos2x}=\int \frac{se{c}^{2}x}{2}dx=\frac{1}{2}tanx+C& \\ & & \\ & (21)\int \frac{cos2x}{cosx-sinx}dx=\int \frac{co{s}^{2}x-si{n}^{2}x}{cosx-sinx}dx=sinx-cosx+C& \\ & & \\ & (22)\int \frac{cos2x}{co{s}^{2}xsi{n}^{2}x}dx=\int \frac{co{s}^{2}x-si{n}^{2}x}{co{s}^{2}xsi{n}^{2}x}dx=\int (cs{c}^{2}x-se{c}^{2}x)dx=& \\ & & \\ & \int cs{c}^{2}xdx-\int se{c}^{2}xdx=-cotx-tanx+C& \\ & & \\ & (23)\int co{t}^{2}xdx=\int cs{c}^{2}xdx-\int dx=-cotx-x+C& \\ & & \\ & (24)\int cos\theta (tan\theta +sec\theta )d\theta =\int sin\theta d\theta +\int d\theta =-cos\theta +\theta +C& \\ & & \\ & (25)\int \frac{x^2}{x^2+1}dx=\int dx-\int \frac{1}{x^2+1}dx=x-arctanx+C& \\ & & \\ & (26)\int \frac{3x^4+2x^2}{x^2+1}dx=\int 3x^{2}dx-\int dx+\int \frac{1}{x^2+1}dx=x^3-x+arctanx+C& \end{array}$

---

$\begin{array}{rlr}& 3.含有未知函数的导数的方程称为微分方程，例如方程\frac{dy}{dx}=f(x).其中\frac{dy}{dx}为未知函数的导数，& \\ & & \\ & f(x)为已知函数。如果将函数y=\phi (x)代入微分方程，使微分方程称为恒等式，那么函数y=\phi (x)& \\ & & \\ & 就称为该微分方程的解。求下列微分方程满足所给条件的解：& \end{array}$

$\begin{array}{rlr}& (1)\frac{dy}{dx}=(x-2{)}^2，y{|}_{x=2}=0；& \\ & & \\ & (2)\frac{d^{2}x}{d{y}^2}=\frac{2}{t^3}，\frac{dx}{dt}{|}_{t=1}=1，x{|}_{t=1}=1.& \end{array}$

解：

$\begin{array}{rlr}& (1)y=\int (x-2{)}^{2}dx=\frac{1}{3}(x-2{)}^3+C，由y{|}_{x=2}=0，得C=0，因此，y=\frac{1}{3}(x-2{)}^3.& \\ & & \\ & (2)\frac{dx}{dt}=\int \frac{2}{t^3}dt=-\frac{1}{t^2}+C_0，由\frac{dx}{dt}{|}_{t=1}=1，得C_0=2，因此，\frac{dx}{dt}=-\frac{1}{t^2}+2，& \\ & & \\ & x=\int \left(-\frac{1}{t^2}+2\right)dt=\frac{1}{t}+2t+C_1，由x{|}_{t=1}=1，得C_1=-2，因此，x=\frac{1}{t}+2t-2.& \end{array}$

---

$\begin{array}{rlr}& 4.汽车以20m/s的速度在直道上行驶，刹车后匀减速行驶了50m停住，求刹车加速度。可执行下列步骤：& \end{array}$

$\begin{array}{rlr}& (1)求微分方程\frac{d^{2}s}{d{t}^2}=-k满足条件\frac{ds}{dt}{|}_{t=0}=20及s{|}_{t=0}=0的解；& \\ & & \\ & (2)求使\frac{ds}{dt}=0的t值及相应的s值；& \\ & & \\ & (3)求使s=50的k值。& \end{array}$

解：

$\begin{array}{rlr}& (1)\frac{ds}{dt}=\int -kdt=-kt+C_0，由\frac{ds}{dt}{|}_{t=0}=20，得C_0=20，因此，\frac{ds}{dt}=-kt+20，& \\ & & \\ & s=\int (-kt+20)dt=-\frac{1}{2}k{t}^2+20t+C_1，由s{|}_{t=0}=0，得C_1=0，因此，s=-\frac{1}{2}k{t}^2+20t.& \\ & & \\ & (2)令\frac{ds}{dt}=0，得t=\frac{20}{k}.& \\ & & \\ & (3)当t=\frac{20}{k}时，s=50，即-\frac{1}{2}k{\left(\frac{20}{k}\right)}^2+\frac{400}{k}=50，得k=4，因此，刹车加速度为-4m/s^2.& \end{array}$

---

$\begin{array}{rlr}& 5.一曲线通过点(e^2,3)，且在任一点处的切线的斜率等于该点横坐标的倒数，求该曲线的方程。& \end{array}$

解：

$\begin{array}{rlr}& 设曲线方程为y=f(x)，则点(x,y)处的切线斜率为f^{\prime }(x)，且f^{\prime }(x)=\frac{1}{x}，因此，f(x)=\int \frac{1}{x}dx=ln|x|+C，& \\ & & \\ & 又因曲线过点(e^2,3)，有f(e^2)=ln|e^2|+C=3，得C=1，则曲线方程为y=lnx+1& \end{array}$

---

$\begin{array}{rlr}& 6.一物体由静止开始运动，经ts后的速度是3t^{2}m/s，问& \end{array}$

$\begin{array}{rlr}& (1)在3s后物体离开出发点的距离是多少？& \\ & & \\ & (2)物体走完360m需要多少时间？& \end{array}$

解：

$\begin{array}{rlr}& (1)设物体自原点沿横轴正向由静止开始运动，移动函数为s=s(t)，则s(t)=\int v(t)dt=\int 3t^{2}dt=t^3+C，& \\ & & \\ & 由于s(0)=0，所以s(t)=t^3，则s(3)=27，所以3s后物体离开出发点的距离是27m。& \\ & & \\ & (2)由t^3=360，得t=\sqrt[3]{360}\approx 7.1，则物体走完360m需要大约7.1s& \end{array}$

---

$\begin{array}{rlr}& 7.证明函数arcsin(2x-1)，arccos(1-2x)和2arctan\sqrt{\frac{x}{1-x}}都是\frac{1}{\sqrt{x-x^2}}的原函数。& \end{array}$

解：

$\begin{array}{rlr}& [arcsin(2x-1){]}^{\prime }=\frac{2}{\sqrt{1-(2x-1{)}^2}}=\frac{1}{\sqrt{x-x^2}}& \\ & & \\ & [arccos(1-2x){]}^{\prime }=-\frac{-2}{\sqrt{1-(1-2x{)}^2}}=\frac{1}{\sqrt{x-x^2}}& \\ & & \\ & {\left(2arctan\sqrt{\frac{x}{1-x}}\right)}^{\prime }=2\frac{1}{1+\frac{x}{1-x}}\cdot \frac{1}{2}\sqrt{\frac{1-x}{x}}\cdot \frac{1}{(1-x{)}^2}=\frac{1}{\sqrt{x-x^2}}& \end{array}$