Is there such thing as 1st integral, 2nd integral etc. or just the indefinite integral?

Question

If there is, then how do you write its notation?
^_^

Answer 1

If you are thinking in relation to first derivative, second derivative, etc., then formulae such as double and triple integrals would be somewhat of a correspondence. For a double integral, the notation would be (integral sign)(integral sign) f(x,y) dx dy, but you can use a function in one variable.

However, the analogy to derivatives breaks down when applying the indefinite integral more than once as you need to specify the constant term after each integration. Usually this is done with given constraints, such as integrating a function of acceleration to determine distance travelled. The double integral et. al. is usually definite and produces a returned value.

Hope that helps.

Answer 2

It all depends on what dimensional space you are working in. Most 1st and 2nd year calculus courses only deal in R^2, and R^3 (some Professors go into R^4), but one can integrate as many times as they want depending on what dimension ones function is in. It is nearly impossible to visualize what is happening beyond 4 dimensions, but it is possible. I am a have just started work on my PHD and I regularly intergrate n-times where n = 0, 1, 2, 3, .... It is actually possible to continue integrating any function as many times as you like, but there is little or no point in doing so.

Answer 3

yes. you can have multiple depths of integrations. They are noted by integral signs beside each other. When you do this, you must remember to note what each is taking integral with respect to. possibly CAL 2 but I know you see them in CAL 3. Integeration rules are all the same for these, its just your variables that may change. People say 2nd & 3rd Integral, but you are just integrating multiple times to similar variables.

Answer 4

?(3x - 2)^20 dx is pretty a lot in the variety u^n du u = 3x - 2, du = 3dx. that is perfect if I had a three accessible contained in the market, so i'm purely gonna positioned on there, yet I actually have were given to balance it out, so I''ll placed a a million/3 out the front. (a million/3)? (3x - 2)^20 3dx now i have u^du, i am going to now do u^(n + a million)/(n + a million) (a million/3)(3x - 2)^21 / 21 simplify... (a million/sixty 3)(3x - 2)^21 + C

Answer 5

there is a double integral .. and a three times integral ... when you integrate the function with respect to more than one variable