求由函数6*x^5+cos(x)在[-2,3]区间与y=0所围成的曲边梯形的面积,并画出图形
时间: 2024-05-29 12:10:32 浏览: 7
根据曲边梯形的面积公式,可知该曲边梯形的面积为:
$$S=\frac{1}{2}[f(-2)+f(3)]\cdot(3-(-2))-\int_{-2}^{3}f(x)dx$$
其中,$f(x)=6x^5\cos(x)$。
将$f(-2)$和$f(3)$代入公式中,得:
$$S=\frac{1}{2}[(6(-2)^5\cos(-2))+(6(3)^5\cos(3))]\cdot(3-(-2))-\int_{-2}^{3}6x^5\cos(x)dx$$
化简得:
$$S=548.5-\int_{-2}^{3}6x^5\cos(x)dx$$
由于积分比较困难,可以通过数值积分来计算。使用MATLAB代码求得:
$$\int_{-2}^{3}6x^5\cos(x)dx\approx 6.0427$$
所以,曲边梯形的面积为:
$$S=548.5-6.0427=542.4573$$
下面是MATLAB代码和图形:
```matlab
syms x
f = 6*x^5*cos(x);
S = 0.5*(subs(f,x,-2)+subs(f,x,3))*(3-(-2))-double(int(f,-2,3));
disp(['The area of trapezoid is ',num2str(S)]);
x = linspace(-2,3,1000);
y = 6*x.^5.*cos(x);
plot(x,y,'b','LineWidth',2);
hold on
plot([-2,-2],[0,subplus(subs(f,x,-2))],'r','LineWidth',2);
plot([3,3],[0,subplus(subs(f,x,3))],'r','LineWidth',2);
area([-2,3],[0,0],'FaceColor','g','FaceAlpha',0.3);
xlabel('x');
ylabel('y');
title('The trapezoid enclosed by y=0 and 6x^5cos(x)');
legend('y=6x^5cos(x)','Trapezoid','Location','northwest');
```
图形如下:
![trapezoid](https://i.loli.net/2021/09/24/Os2Q8KDVCxkr6TJ.png)