Why Are Definite Integrals Related to Antiderivatives?

[ATCG]

Please help with Anti-derivitives! Need explanation and how to use them. Thank you

 

[StephenPrivitera]

Sounds like fun!


Antiderivatives might be used to solve problems like this:

What function has the derivative f'(x)=3x²?

From your experience you might say f(x)=x³.
That's a good guess, but you will notice that f(x)=x³ + 2 has the same derivative. Remember the derivative of a constant is zero. So this differential equation
f'(x)=3x² actually has an infinite number of solutions. We represent this family of solutions by f(x)=x³+C where C is an arbitrary constant. So now that you can find the antiderivative of 3x², how do you find antiderivatives in general?

Consider the differential equation dy/dx=xⁿ (n does not equal -1). What is the antiderivative of this equation - ie, what is y in terms of x? Consider
y=1/(n+1)*xⁿ⁺¹ - what is the derivative of y with respect to x? It is y'=xⁿ! We express this result in the following form.
dy/dx=xⁿ
dy=xⁿdx ....multiply both sides by dx
[inte] dy= [inte] xⁿdx ...integrate both sides
y=1/(n+1)*xⁿ⁺¹ + C
The integral sign merely tells you to find the antiderivative of the equation. The "dx" and "dy" tell you what variable you are integrating (antidifferentiating) with respect to. To the left of the "dy" and "dx" is the derivative you are trying to undo.
So the left hand side [inte] dy = [inte] 1*dy means what function of y has a first derivative (taken wrt y) equal to 1? Obviously, it is f(y)=y since df/dy=dy/dy=1. The right hand side means what function of x has a derivative of xⁿ? This is the solution to the integral.

Another example,
dy/dx=cosx
dy=cosxdx
[inte] dy= [inte]cosxdx
y=sinx +C
Take the derivative of y to make sure I'm right.

Take a stab at this one:
dy/dx=sec²x
What is y?
________
Technically, the left hand side should be y+C, but this is taken care of in the right hand side since C is entirely arbitrary.

 

[ATCG]

Thanks a lot StephenPrivitera! I understand the anti-derivites now!

 

[Gale]

yeah, that was great help stephen. i haven't even learned integrals yet, but nowi understand them. we've brushed on them in physics and i was pretty clueless, but you've helped me as well.

 

[StephenPrivitera]

No problem. Happy to help anytime, anyday. Calculus is a very interesting topic.

 

[phoenixthoth]

one thing to watch out for in your near future is that you understand why definite integrals are related to antiderivates. the way the notation is set up, it seems like it should be automatic that they are related, but that they are related isn't exactly obvious. the first has to do with area (in some cases) and the other has to do with inverting the operation of differentiation. it took great insight (or luck) to realize the two concepts were related.