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AP微积分AB:梯形和

学习AP微积分AB的概念、例题和解释

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3诊断测试 164实践测试今日问题抽认卡从概念中学习

例子问题

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例子问题1:梯形的金额

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\\&\text{Using }2\text{ intervals.}\end{align*}$

可能的答案:

-63.97

-105.17

-10.84

-3.29

正确答案:

-10.84

解释：

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-4}^{1.4}(11sin(17cos(x)))dx\\&\text{So the interval is }[-4,1.4]\text{ the subintervals have length }\frac{1.4-(-4)}{2}=\frac{27}{10}\\&\text{and the x-values are:}\\&[-4,-\frac{13}{10},\frac{7}{5}]\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\approx\frac{27}{20}[(11sin(17cos((-4))))+2(11sin(17cos((-\frac{13}{10}))))+(11sin(17cos((\frac{7}{5}))))]\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\approx-10.84\end{align*}$

报告错误

例子问题2:梯形的金额

$\begin{align*}&\text{Calculate the trapezoidal sums integral approximation of :}\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\\&\text{Using points over }2\text{ intervals.}\end{align*}$

可能的答案:

-1.68

-0.60

-5.20

-0.90

正确答案:

-5.20

解释：

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{0}^{2}(6cos(20e^{(x)}))dx\\&\text{So the interval is }[0,2]\text{ the subintervals have length }\frac{2-(0)}{2}=1\\&\text{and the x-values are:}\\&[0,1,2]\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\approx\frac{1}{2}[(6cos(20e^{((0))}))+2(6cos(20e^{((1))}))+(6cos(20e^{((2))}))]\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\approx-5.20\end{align*}$

报告错误

例子问题3:梯形的金额

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\end{align*}$

可能的答案:

35.68

4.20

114.16

6.37

正确答案:

35.68

解释：

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\\&\text{So the interval is }[-1,8.9]\text{ the subintervals have length }\frac{8.9-(-1)}{3}=\frac{33}{10}\\&\text{and the x-values are:}\\&[-1,\frac{23}{10},\frac{28}{5},\frac{89}{10}]\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\approx\frac{33}{20}[(5sin(9e^{((-1))}))+2(5sin(9e^{((\frac{23}{10}))}))+2(5sin(9e^{((\frac{28}{5}))}))+(5sin(9e^{((\frac{89}{10}))}))]\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\approx35.68\end{align*}$

报告错误

问题4:梯形的金额

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

可能的答案:

-93.67

-33.45

-138.84

-16.73

正确答案:

-16.73

解释：

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{1}^{6.4}(-6cos(19x^{2}))dx\\&\text{So the interval is }[1,6.4]\text{ the subintervals have length }\frac{6.4-(1)}{3}=\frac{9}{5}\\&\text{and the x-values are:}\\&[1,\frac{14}{5},\frac{23}{5},\frac{32}{5}]\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\approx\frac{9}{10}[(-6cos(19(1)^{2}))+2(-6cos(19(\frac{14}{5})^{2}))+2(-6cos(19(\frac{23}{5})^{2}))+(-6cos(19(\frac{32}{5})^{2}))]\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\approx-16.73\end{align*}$

报告错误

例子问题1:梯形的金额

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

可能的答案:

3.13

5.51

7.65

27.55

正确答案:

27.55

解释：

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{2}^{6.5}(17cos(2sin(2x)))dx\\&\text{So the interval is }[2,6.5]\text{ the subintervals have length }\frac{6.5-(2)}{3}=\frac{3}{2}\\&\text{and the x-values are:}\\&[2,\frac{7}{2},5,\frac{13}{2}]\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx\frac{3}{4}[(17cos(2sin(2(2))))+2(17cos(2sin(2(\frac{7}{2}))))+2(17cos(2sin(2(5))))+(17cos(2sin(2(\frac{13}{2}))))]\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx27.55\end{align*}$

报告错误

例子问题6:梯形的金额

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\\&\text{Using }2\text{ intervals.}\end{align*}$

可能的答案:

-2160.41

-744.97

-130.70

-88.69

正确答案:

-744.97

解释：

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{0}^{4.4}(-196cos(3x)^{2})dx\\&\text{So the interval is }[0,4.4]\text{ the subintervals have length }\frac{4.4-(0)}{2}=\frac{11}{5}\\&\text{and the x-values are:}\\&[0,\frac{11}{5},\frac{22}{5}]\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\approx\frac{11}{10}[(-196cos(3(0))^{2})+2(-196cos(3(\frac{11}{5}))^{2})+(-196cos(3(\frac{22}{5}))^{2})]\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\approx-744.97\end{align*}$

报告错误

示例问题7:梯形的金额

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\end{align*}$

可能的答案:

-0.56

-1.86

-18.44

-0.29

正确答案:

-1.86

解释：

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\\&\text{So the interval is }[3,7.2]\text{ the subintervals have length }\frac{7.2-(3)}{3}=\frac{7}{5}\\&\text{and the x-values are:}\\&[3,\frac{22}{5},\frac{29}{5},\frac{36}{5}]\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\approx\frac{7}{10}[(11cos(6e^{(4(3))}))+2(11cos(6e^{(4(\frac{22}{5}))}))+2(11cos(6e^{(4(\frac{29}{5}))}))+(11cos(6e^{(4(\frac{36}{5}))}))]\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\approx-1.86\end{align*}$

报告错误

例8:梯形的金额

$\begin{align*}&\text{Calculate the trapezoidal sums integral approximation of :}\\&\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\\&\text{Using points over }2\text{ intervals.}\end{align*}$

可能的答案:

-3.05

-0.38

-12.19

-0.59

正确答案:

-3.05

解释：

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\\&\text{So the interval is }[-5,-3.8]\text{ the subintervals have length }\frac{-3.8-(-5)}{2}=\frac{3}{5}\\&\text{and the x-values are:}\\&[-5,-\frac{22}{5},\frac{11}{5}]\\&\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\approx\frac{3}{10}[(-16sin(9e^{(5(-5))}))+2(-16sin(9e^{(5(-\frac{22}{5}))}))+(-16sin(9e^{(5(\frac{11}{5}))}))]\\&\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\approx-3.05\end{align*}$

报告错误

问题9:梯形的金额

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{-1}^{6.5}(2700sin(3x)^{2})dx\end{align*}$

可能的答案:

91163.20

4849.11

9698.21

53340.17

正确答案:

9698.21

解释：

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-1}^{6.5}(2700sin(3x)^{2})dx\\&\text{So the interval is }[-1,6.5]\text{ the subintervals have length }\frac{6.5-(-1)}{3}=\frac{5}{2}\\&\text{and the x-values are:}\\&[-1,\frac{3}{2},4,\frac{13}{2}]\\&\int_{-1}^{6.5}(2700sin(3x)^{2})dx\approx\frac{5}{4}[(2700sin(3(-1))^{2})+2(2700sin(3(\frac{3}{2}))^{2})+2(2700sin(3(4))^{2})+(2700sin(3(\frac{13}{2}))^{2})]\\&\int_{-1}^{6.5}(2700sin(3x)^{2})dx\approx9698.21\end{align*}$

报告错误

例子问题1:梯形的金额

$\begin{align*}&\text{Calculate the trapezoidal sums integral approximation of :}\\&\int_{2}^{9}(17\cdot 2^ln(3x))dx\\&\text{Using points over }2\text{ intervals.}\end{align*}$

可能的答案:

84.43

3079.98

810.52

5349.44

正确答案:

810.52

解释：

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{2}^{9}(17\cdot 2^ln(3x))dx\\&\text{So the interval is }[2,9]\text{ the subintervals have length }\frac{9-(2)}{2}=\frac{7}{2}\\&\text{and the x-values are:}\\&[2,\frac{11}{2},9]\\&\int_{2}^{9}(17\cdot 2^ln(3x))dx\approx\frac{7}{4}[(17\cdot 2^ln(3(2)))+2(17\cdot 2^ln(3(\frac{11}{2})))+(17\cdot 2^ln(3(9)))]\\&\int_{2}^{9}(17\cdot 2^ln(3x))dx\approx810.52\end{align*}$

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便雅悯
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亚利桑那州立大学，物理学学士。夏威夷大学马诺阿分校，物理学硕士。

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布朗大学，认知科学学士学位。

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AP微积分AB:梯形和

学习AP微积分AB的概念、例题和解释

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例子问题

例子问题1:梯形的金额

例子问题2:梯形的金额

例子问题3:梯形的金额

问题4:梯形的金额

例子问题1:梯形的金额

例子问题6:梯形的金额

示例问题7:梯形的金额

例8:梯形的金额

问题9:梯形的金额

例子问题1:梯形的金额

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