- #1
christian0710
- 409
- 9
Hi I'm tying to understand notations in Integration, and would really appreciate some help making sure that my understanding is right.My books writes
Let u and v be functions of x whose domains are an open interval I, and suppose du and dv exist for every x in I.Then it defines
1) ∫(du)= u + C
2) ∫(c*du) = c*∫(du)and
3) ∫(cos(u)du = sin u +CNow i do understand the first 2, but I want to make sure i understand the 3rd rule.
If u is a function of x with the equation u(x)=x^2
Then the derivative
du/dx= 2x
The differential
du=u'(x)dx
Now if it's true that du=u'(x)dx
Then it does make sense that ∫du =u+C because ∫du=∫u'(x)*dx and the integral of the derivative if u is u.
But if u=x^2
Then ∫(cos(x^2)*du = ∫(cos(x^2)*(2x)dx and this is = sin(x^2) + C as the statement above says.
because the derivative of sin(x^2) = cos(x^2)*(2x).
Is this the right interpretation?
Let u and v be functions of x whose domains are an open interval I, and suppose du and dv exist for every x in I.Then it defines
1) ∫(du)= u + C
2) ∫(c*du) = c*∫(du)and
3) ∫(cos(u)du = sin u +CNow i do understand the first 2, but I want to make sure i understand the 3rd rule.
If u is a function of x with the equation u(x)=x^2
Then the derivative
du/dx= 2x
The differential
du=u'(x)dx
Now if it's true that du=u'(x)dx
Then it does make sense that ∫du =u+C because ∫du=∫u'(x)*dx and the integral of the derivative if u is u.
But if u=x^2
Then ∫(cos(x^2)*du = ∫(cos(x^2)*(2x)dx and this is = sin(x^2) + C as the statement above says.
because the derivative of sin(x^2) = cos(x^2)*(2x).
Is this the right interpretation?
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