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高等数学（第七版）同济大学习题4-1 个人解答

2022-08-03 14:37:00 【Navigator_Z】

高等数学（第七版）同济大学习题4-1

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

$\begin{aligned} &\ \ (1)\ \ \int \frac{1}{\sqrt{x^2+1}}dx=ln(x+\sqrt{x^2+1})+C；\\\\ &\ \ (2)\ \ \int \frac{1}{x^2\sqrt{x^2-1}}dx=\frac{\sqrt{x^2-1}}{x}+C；\\\\ &\ \ (3)\ \ \int \frac{2x}{(x^2+1)(x+1)^2}dx=arctan\ x+\frac{1}{x+1}+C；\\\\ &\ \ (4)\ \ \int sec\ xdx=ln\ |tan\ x+sec\ x|+C；\\\\ &\ \ (5)\ \ \int xcos\ xdx=xsin\ x+cos\ x+C；\\\\ &\ \ (6)\ \ \int e^xsin\ xdx=\frac{1}{2}e^x(sin\ x-cos\ x)+C. & \end{aligned}$

解：

$\begin{aligned} &\ \ (1)\ [ln(x+\sqrt{x^2+1})+C]'=\frac{1}{x+\sqrt{x^2+1}}\cdot \left(1+\frac{x}{\sqrt{x^2+1}}\right)=\frac{1}{\sqrt{x^2+1}}\\\\ &\ \ (2)\ \left(\frac{\sqrt{x^2-1}}{x}+C\right)'=\frac{\frac{x}{\sqrt{x^2-1}}\cdot x-\sqrt{x^2-1}}{x^2}=\frac{1}{x^2\sqrt{x^2-1}}\\\\ &\ \ (3)\ \left(arctan\ x+\frac{1}{x+1}+C\right)'=\frac{1}{x^2+1}-\frac{1}{(x+1)^2}=\frac{2x}{(x^2+1)(x+1)^2}\\\\ &\ \ (4)\ (ln\ |tan\ x+sec\ x|+C)'=\frac{1}{tan\ x+sec\ x}\cdot (sec^2\ x+sec\ xtan\ x)=sec\ x\\\\ &\ \ (5)\ (xsin\ x+cos\ x+C)'=sin\ x+xcos\ x-sin\ x=xcos\ x\\\\ &\ \ (6)\ \left[\frac{1}{2}e^x(sin\ x-cos\ x)+C\right]'=\frac{1}{2}e^x(sin\ x-cos\ x)+\frac{1}{2}e^x(cos\ x+sin\ x)=e^xsin\ x & \end{aligned}$

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

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

解：

$\begin{aligned} &\ \ (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^3dx=5\frac{x^{3+1}}{3+1}+C=\frac{5}{4}x^4+C\\\\ &\ \ (8)\ \int (x^2-3x+2)dx=\int x^2dx-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)^2dx=\int (x^4+2x^2+1)dx=\int x^4dx+2\int x^2dx+\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^2dx+\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^xdx+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}}=3arctan\ x-2arcsin\ x+C\\\\ &\ \ (15)\ \int e^x\left(1-\frac{e^{-x}}{\sqrt{x}}\right)dx=\int e^xdx-\int x^{-\frac{1}{2}}dx=e^x-2x^{\frac{1}{2}}+C\\\\ &\ \ (16)\ \int 3^xe^xdx=\int (3e)^xdx=\frac{(3e)^x}{ln(3e)}+C=\frac{3^xe^x}{ln\ 3+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)^xdx=2x-\frac{5}{ln\ \frac{2}{3}}\left(\frac{2}{3}\right)^x+C=2x-\frac{5}{ln\ 2-ln\ 3}\left(\frac{2}{3}\right)^x+C\\\\ &\ \ (18)\ \int sec\ x(sec\ x-tan\ x)dx=\int sec^2\ xdx-\int sec\ xtan\ xdx=tan\ x-sec\ x+C\\\\ &\ \ (19)\ \int cos^2 \frac{x}{2}dx=\int \frac{1+cos\ x}{2}dx=\frac{x+sin\ x}{2}+C\\\\ &\ \ (20)\ \int \frac{dx}{1+cos\ 2x}=\int \frac{sec^2\ x}{2}dx=\frac{1}{2}tan\ x+C\\\\ &\ \ (21)\ \int \frac{cos\ 2x}{cos\ x-sin\ x}dx=\int \frac{cos^2\ x-sin^2\ x}{cos\ x-sin\ x}dx=sin\ x-cos\ x+C\\\\ &\ \ (22)\ \int \frac{cos\ 2x}{cos^2\ xsin^2\ x}dx=\int \frac{cos^2\ x-sin^2\ x}{cos^2\ xsin^2\ x}dx=\int (csc^2\ x-sec^2\ x)dx=\\\\ &\ \ \ \ \ \ \ \ \ \int csc^2\ xdx-\int sec^2\ xdx=-cot\ x-tan\ x+C\\\\ &\ \ (23)\ \int cot^2\ xdx=\int csc^2\ xdx-\int dx=-cot\ x-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-arctan\ x+C\\\\ &\ \ (26)\ \int \frac{3x^4+2x^2}{x^2+1}dx=\int 3x^2dx-\int dx+\int \frac{1}{x^2+1}dx=x^3-x+arctan\ x+C & \end{aligned}$

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

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

解：

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

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

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

解：

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

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

解：

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

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

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

解：

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

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

解：

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