积分是一种找到函数f(x)的方法,该函数的导数Df(x)等于给定函数。这就是为什么它也被称为反导数。通常对积分进行计算,以找到可提供有关面积,位移,体积的信息的函数,该信息由于收集了无法单独测量的小数据而出现。在这里,我们可以通过替换方法或形式为f(p)p’(x)的积分来定义积分。
- 代入积分法
- f(p)p’(x)形式的积分
代入积分法
只要给定函数f(x)乘以给定函数f(x)’的导数,即形式为∫g(f(x)f(x)’)dx,就可以使用替换法积分。当要被集成的函数不是在一个标准的形式,它有时可通过合适的取代被变换为积形式。积分
∫ f {g (x)} g’ (x) dx can be converted to
∫f (θ) dθ by substituting g (x) by θ, so that if
∫f (θ) dθ = F(θ) + c, Then
∫f{g (x)} g’ (x) dx = F{g(x)} + c
This is a direct consequence of chain rule, for
d/dx [F {g(x)} + c] = d/dθ [F(θ) + c] . dθ/dx = f {g(x)} g’(x)
没有明确的替代公式。对被积物形式的敏锐观察将有助于选择要进行替代的函数。但是,必须确保与上述情况一样,如此选择的函数的导数必须与dx一起出现。有时可能仅需调整常数即可。变量viz的任何符号。除了给定积分的变量外,还可以选择s,t,u,v,w,x,y,z进行替换。但是,积分结束后,应将原始变量放回原处。
例子
示例1:整合∫2x.cos(x 2 )dx
解决方案:
Let, I = ∫ 2x. cos (x2) dx …………. (i)
Substituting x2 = t ………… (ii)
by differentiate the above equation
2x dx = dt ………… (iii)
put the equ (ii) and (iii) in equ (i), we get
= ∫ cos t dt
Integrate the above equation then, we get
= sin t + c
Put the value of t in the above equ
= sin (x2) + c
Hence, I = sin (x2) + c
例2:对∫sin(x 3 )进行积分。 3x 2 dx
解决方案:
Let, I = ∫ sin (x3). 3x2 dx …………. (i)
Substituting x3 = t ………… (ii)
by differentiate the above equation
3x2 dx = dt ………… (iii)
put the equ (ii) and (iii) in equ (i), we get
= ∫ sin t dt
Integrate the above equation then, we get
= – cos t + c
Put the value of t in the above equ
= – cos (x3) + c
Hence, I = – cos (x3) + c
例3:整合∫2x cos(x 2 − 5)dx
解决方案:
Let, I = ∫ 2x cos(x2 − 5) dx ……… (i)
Put, x2 – 5 = t ……… (ii)
differentiate the above equ
2x dx = dt ……… (iii)
put the equ (ii) and (iii) in equ (i), we get
= ∫ cos (t) dt
Integrate the above equ then, we get
= sin t + c
Put the value of t in above equ
= sin (x2 – 5) + c
Hence, I = sin (x2 – 5) + c
例4:整合∫x/(x 2 +1)dx
解决方案:
Let, I = ∫ x / (x2+1) dx
we rearrange the above equ
= (1/2) ∫ 2x / (x2+1) dx (i)
put , x2 + 1 = t (ii)
2x dx = dt (iii)
put the equ (ii) and (iii) in equ (i), we get
= (1/2) ∫ 1/t dt
Integrate the above equ then, we get
= (1/2) log t + c
now, put the value of t in the above equ
= (1/2) log (x2 +1) + c
Hence, I = (1/2) log (x2 + 1) + c
示例5:整合∫(2x + 3)(x 2 + 3x) 2 dx
解决方案:
Let, I = ∫ (2x + 3) (x2 + 3x)2 dx (i)
Substitute x2 + 3x = t (ii)
differenitate the above equ
2x + 3 dx = dt (iii)
put the equ (ii) and (iii) in equ (i), we get
= ∫ t2 dt
Integrate the above equ then, we get
= t3/3 + c
now, put the value of t in the above equ
= (x2 + 3x)3 / 3 + c
Hence, I = (x2 + 3x)3/3 + c
f(p)p’(x)形式的积分
f (p). p'(x) where p is a function of x
I = ∫ f(p). p'(x) dx
Let p(x) = t
p'(x) dx = dt
I = ∫f(t) dt
例子
示例1:∫cos(x 2 )2x dx
解决方案:
∫cos(x2) 2x dx
Here, f = cos, g(x) = x2, g'(x) = 2x
So put,
x2 = t ….. (1)
differentiate the above equ w.r.t x
2x dx = dt ….. (2)
so the equ is,
∫cost dt ….. (3)
from integrate the equ (3)
sin t + c …… (4)
then put the value t in equ (4)
The final answer is = sin(x2) + c
示例2:整合∫cos(x 3 )。 3x 2 dx
解决方案:
Let, I = ∫ cos (x3). 3x2 dx (i)
Here, f = cos, g(x) = x3, g'(x) = 3x2
Substituting x3 = t (ii)
by differentiate the above equation
3x2 dx = dt (iii)
put the equ (ii) and (iii) in equ (i), we get
= ∫ cos t dt
Integrate the above equation then, we get
= sin t + c
Put the value of t in the above equ
= sin (x3) + c
Hence, I = sin (x3) + c
例3:整合∫2x sin(x 2 − 5)dx
解决方案:
Let, I = ∫ 2x cos(x2 − 5) dx (i)
Here, f= cos, g(x) = x2 – 5, g'(x) = 2x
Put, x2 – 5 = t (ii)
differentiate the above equ
2x dx = dt (iii)
Put the equ (ii) and (iii) in equ (i), we get
= ∫ cos (t) dt
Integrate the above equ then, we get
= – sin t + c
Put the value of t in above equ
= – sin (x2 – 5) + c
Hence, I = – sin (x2 – 5) + c