数学代写代考|杜兰大学微积分MATH1210 Calculus1

这是一次杜兰大学微积分1MATH1210 Calculus1课程的代写成功案例.

微积分（Calculus），数学概念，是高等数学中研究函数的微分（Differentiation）、积分（Integration）以及有关概念和应用的数学分支。它是数学的一个基础学科，内容主要包括极限、微分学、积分学及其应用。

微分学包括求导数的运算，是一套关于变化率的理论。它使得函数、速度、加速度和曲线的斜率等均可用一套通用的符号进行讨论。积分学，包括求积分的运算，为定义和计算面积、体积等提供一套通用的方法。

To start this problem, we will need a partition of the interval given. The problem asks for 4 equally spaced points. So, the partition we need is
$$
P=\left{x_0, x_1, x_2, x_3, x_4\right}={1,2,3,4,5}
$$
The distance between points in this case will be
$$
\Delta x=2-1=3-2=4-3=5-4=1
$$
For a complete answer you did not need to compute the distance for all of the subintervals. So, the approximation of the definite integral is given by
$$
\begin{aligned}
A & \approx f\left(c_1\right) \cdot \Delta x_1+f\left(c_2\right) \cdot \Delta x_2+f\left(c_3\right) \cdot \Delta x_3+f\left(c_4\right) \cdot \Delta x_4 \
& =\Delta x \cdot\left(f\left(c_1\right)+f\left(c_2\right)+f\left(c_3\right)+f\left(c_4\right)\right) \
& =(1) \cdot\left(f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)\right) \
& =\left(f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)\right)
\end{aligned}
$$
Note that the right endpoint of each of the subintervals has been used in the evaluation of the approximation. That is, $c_1=x_1=2, c_2=x_2=3, c_3=x_4=4$, and $c_4=x_4=5$. So,
$$
\begin{aligned}
A & \approx(f(2)+f(3)+f(4)+f(5)) \
& =\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)
\end{aligned}
$$
For the exam, this is far enough. The final answer should be
$$
A \approx \frac{77}{60}
$$

Solution: a. For the first part, the solution is computed as follows.
$$
\begin{aligned}
\sum_{k=1}^{15}(2 \cdot k) & =2 \cdot \sum_{k=1}^{15} k \
& =2 \cdot \frac{(15)(15+1)}{2} \
& =(15)(16) \
& =240
\end{aligned}
$$
b. For the second part, the solution is computed as follows.
$$
\begin{aligned}
\sum_{k=1}^8\left(k^2-3 \cdot k+2\right) & =\sum_{k=1}^8 k^2-3 \cdot \sum_{k=1}^8 k+2 \cdot \sum_{k=1}^8 1 \
& =\frac{(8)(8+1)(2(8)+1)}{6}-3 \cdot \frac{(8)(8+1)}{2}+2 \cdot(8) \
& =112
\end{aligned}
$$

a. The definite integral can be computed as follows:
$$
\begin{aligned}
\int_0^1\left(\sqrt{x}-x^2\right) d x & =\left.\left(\frac{2}{3} \cdot x^{3 / 2}-\frac{1}{3} \cdot x^3\right)\right|0 ^1 \ & =\left(\frac{2}{3} \cdot(1)^{3 / 2}-\frac{1}{3} \cdot(1)^3\right)-\left(\frac{2}{3} \cdot(0)^{3 / 2}-\frac{1}{3} \cdot(0)^3\right) \ & =\left(\frac{2}{3}-\frac{1}{3}\right)-(0) \ & =\frac{1}{3} \end{aligned} $$ b. The definite integral can be computed as follows: $$ \begin{aligned} \int{\pi / 4}^{3 \pi / 4}(\sin (t)+\cos (t)) d t & =\left.(-\cos (t)+\sin (t))\right|_{\pi / 4} ^{3 \pi / 4} \
& =(-\cos (3 \pi / 4)+\sin (3 \pi / 4))-(-\cos (\pi / 4)+\sin (\pi / 4)) \
& =\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)-\left(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right) \
& =\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} \
& =2 \cdot \frac{\sqrt{2}}{2} \
& =\sqrt{2}
\end{aligned}
$$
c. The definite integral can be computed as follows:
$$
\begin{aligned}
\int_1^2\left(\frac{7}{x}+e^{-x}\right) d x & =7 \cdot \int_1^2 \frac{1}{x} d x+\int_1^2 e^{-x} d x \
& =\left.7 \cdot \ln |x|\right|_1 ^2-\left.e^{-x}\right|_1 ^2 \
& =7 \cdot \ln |2|-7 \cdot \ln |1|-e^{-2}+e^{-1} \
& =7 \cdot \ln |2|-\frac{1}{e^2}+\frac{1}{e^1}
\end{aligned}
$$

a. The substitution given results in the following relationships
$$
\begin{aligned}
u & =x+1 \
d u & =d x \
x & =u-1
\end{aligned}
$$
So,
$$
\begin{aligned}
\int(x \cdot \sqrt{x+1}) d x, & =\int((u-1) \cdot \sqrt{u}) d u \
& =\int\left(u^{3 / 2}-u^{1 / 2}\right) d u \
& =\frac{2}{5} \cdot u^{5 / 2}-\frac{2}{3} \cdot u^{3 / 2}+C=\frac{2}{5} \cdot(x+1)^{5 / 2}-\frac{2}{3} \cdot(x+1)^{3 / 2}+C
\end{aligned}
$$
b. The substitution given results in the following relationships
$$
\begin{aligned}
u & =x^3 \
d u & =3 x^2 d x \rightarrow \frac{1}{3} d u=x^2 d x
\end{aligned}
$$
So,
$$
\begin{aligned}
\int\left(x^2 \cdot \sin \left(x^3\right)\right) d x & =\int\left(\frac{1}{3} \cdot \sin (u)\right) d u \
& =\frac{1}{3} \cdot \int \sin (u) d u \
& =-\frac{1}{3} \cdot \cos (u)+C=-\frac{1}{3} \cdot \cos \left(x^3\right)+C
\end{aligned}
$$
b. The substitution given results in the following relationships
$$
\begin{aligned}
u & =-2 \cdot x^2 \
d u & =-4 \cdot x d x \quad \rightarrow \quad-\frac{1}{4} d u=x d x
\end{aligned}
$$
So,
$$
\begin{aligned}
\int\left(x \cdot e^{-2 x^2}\right) d x, & =\int\left(-\frac{1}{4} \cdot e^u\right) d u \
& =-\frac{1}{4} \cdot e^u+C=-\frac{1}{4} \cdot e^{-2 x^2}+C
\end{aligned}
$$

微积分MATH1210 Calculus1代写请认准Belchy

Belchy助力你的留学生涯