使用梯形法計算一二次函數(shù)的數(shù)值積分

$\int_{a}^f(x)dx$

we can partition the integration interval $[a,b]$ into smaller subintervals,

and approximate the area under the curve for each subinterval by the area of

the trapezoid created by linearly interpolating between the two function values

at each end of the subinterval:

The blue line represents the function $f(x)$ and the red line

is the linear interpolation. By subdividing the interval $[a,b]$, the area under $f(x)$ can thus be approximated as the sum of the areas of all

the resulting trapezoids.

If we denote by $x_{i}$ ($i=0,\ldots,n,$ with $x_{0}=a$ and

$x_{n}=b$) the abscissas where the function is sampled, then

$$

\int_{a}f(x)dx\approx\frac{1}{2}\sum_{i=1}{n}\left(x_{i}-x_{i-1}\right)\left(f(x_{i})+f(x_{i-1})\right).

$$

The common case of using equally spaced abscissas with spacing $h=(b-a)/n$ reads simply

$$

\int_{a}f(x)dx\approx\frac{h}{2}\sum_{i=1}{n}\left(f(x_{i})+f(x_{i-1})\right).

$$

具體計算只需要這個公式

積分

道理很簡單，就是把積分區(qū)間分割為很多小塊，用梯形替代，其實還是局部用直線近似代替曲線的思想。這里對此一元二次函數(shù)積分，并與python模塊積分制對比(精確值為4.5)用以驗證。

$$

\int_a^b (x^2 - 3x + 2) dx

$$

from scipy import integrate

def f(x):

return x*x - 3*x + 2

def trape(f,a,b,n=100):

f = np.vectorize(f) # can apply on vector

h = float(b - a)/n

arr = f(np.linspace(a,b,n+1))

return (h/2.)*(2*arr.sum() - arr[0] - arr[-1])

def quad(f,a,b):

return integrate.quad(f,a,b)[0] # compare with python library result

a, b = -1, 2

print trape(f,a,b)

print quad(f,a,b)

4.50045

4.5

總結(jié)

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