∫f(x)dx
∫u^ndu → 1/(n+1)u^(n+1)+C
Also known as indefinite integrals, learning about antiderivatives served as a basis to integral calculus. Antiderivatives are basically the undone version of a derivative (hence the name). Antiderivatives are useful in concepts such as motion, as it allows you to find different information of a function depending on the equation you have. For example, the position equation, giving the current position of an object in question (the object being graphed) is the original function, f(x). It’s derivative, f’(x), is the function for velocity. The derivative of the velocity function, f’’(x), gives the object’s acceleration. If you only had the acceleration or velocity equations, you would be able to attain the other functions using antiderivatives. Furthermore, antiderivatives lead us to the other main kind of calculus: Integral calculus.
The antiderivative of f(x) is F(x). ∫f(x)dx represents the antiderivative of f. While in derivatives you bring an exponent down and multiply it to the number it’s attached to and then subtract one from the exponent, antiderivatives is the opposite as you first add one to the exponent and then divide the number it’s attached to by the new exponent. For indefinite integrals, you would also add +C, as the integral is indefinite and thus approximate. C is the constant of integration.
In the example photo here, the equation given to integrate is ∫3x^2dx. To solve this, one would add one to the exponent of 2, attaining a grand total of three. You would then divide the now 3x^3 by 3, the new denominator, which cancels out the 3x and makes it just x. You now have x^3, but--oh my--a pesky dx is spoiling your happiness! You just want it to go, go far away while still representing an indeterminate value, so you say “C ya later” and replace it by adding +C. You now have x^3+C (the ∫ has integrated away). So, F(x)=x^3+C.
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You can also solve for C if you are truly daring, giving you a more accurate equation. These are called “Anti-derivatives and initial conditions”, meaning you are solving for the antiderivative equation at a specific point.
In this example, we are solving for the antiderivative of the equation dy/dx=-x/y, and are looking to solve for the specific equation at the points (2, -1). You start with separating the variables, so y’s and x’s are each on the same side. After doing that, I got ∫y^1dy=∫-x^1dx. You would then integrate each side. y^1+1=y^2, divided by the new exponent=1/2y^2. Meanwhile, -x^1+1=-x^2 divided by 2. The dy and dx are integrated, and dx will become +C, giving you 1/2y^2=-1/2x^2+C (Note: there should be a -½x in the photo above, but I neglected it. 1000 apologies.). The +½ and -½ cancel, and you are left with y^2=-x^2+C, but that is not enough, oh no it is not. You want more. You want Y to be single and all alone on its side of the equation. So, to get rid of the ^2, you must take the square root of y^2--but what you do to one side of the equation must be done to the other. So, you are now stuck with √y^2=√-x^2+C. But since the ^2 and √ cancel each other out, y is successfully forced into solitude, it’s only companion in the world ^2 having been cancelled out. Meanwhile, √-x^2+C are all partying it up together under the square root because C is there too and is not cancelled by the square root. And so, you are now left with y=-√-x^2+C. Congratulations. But you are still not done yet. Now, you have the initial conditions to worry about--(2,-1). As 2 is the x value and -1 the y, plug them in accordingly and you get -1=-√2^2+C. (Note: While you typically want two answers, a positive and negative answer, for square roots, you are only looking for the negative one in this case due to y’s original value in the original equation. As the original equation was -x/y, y’s value was y^-1, a negative value. Thus, you want the negative answer/it will be -√2^2+C). You are now going to solve for C. 2^2=4; -1=√4+C. You need to cancel the square root in order to get to C, so you cancel both sides of the equation by squaring both sides.
(1^2=-√(4+C)^2)= 1=-4+C. Now, you separate the variables. 1+4=5=C. C=5. You rewrite the equation with 5, and you now have Y=-√x^2+5. Woot. |
Antiderivatives relate to my physics class as I have had to cancel out equations or work backwards at times in order to solve things; antiderivatives are a form of working backwards in my opinion in order to attain a solution. Furthermore, as antiderivatives can be used to attain a velocity equation from an acceleration one, or a position equation from a velocity equation, I feel I could apply antiderivative techniques in order to solve for these things in physics, be it to derive equations in order to do so, etc.
Advice: Please keep up with your work, it will make everything that much easier. That is all. |