In calculus, an antiderivative or indefinite integral, G, of a function g, is the function that can be differentiated to obtain the original function, that is, G’ = f.
G(x) = $\int g(x)dx$
$\int u^{5}\frac{du}{4} = \frac{u^{6}}{24} + C = \frac{(x^{4}+2)^{6}}{24} + C.$
$\int u^{-1}du = ln|u| + C = ln|lnx| + C.$
$\int \frac{x}{1+x^{2}}dx=\sqrt{1+x^{2}} + C.$
$\int xe^{-x^{2}}dx = \frac{-1}{2}e^{-x^{2}} + C.$
$\frac{d}{dx}(1+x^{2})^{1/2} = \frac{1}{2}(1+x^{2})^{-1/2}(2x)$
Let F and G tow different antiderivatives of f(x). If F’ = G’, then F(x) = G(x) + C. They are unique up to a constant.
Proof. If F’ = G’ ⇒ (F-G)’ = F’ - G’ = 0 ⇒ F-G is constant ⇒ F(x) = G(x) + C.
$\frac{dy}{dx}=f(x)$. The solution is going to be the antiderivate or the integral, y = $\int f(x)dx$
Another way would be, $\frac{dy}{dx}=f(x) ⇒ dy = f(x)dx ⇒ y = \int dy = \int f(x)dx$
Example. $(\frac{d}{dx}+x)y=0$.
$\frac{dy}{dx}=-xy$ ⇒ $\frac{dy}{y}=-xdx$ ⇒ $\int \frac{dy}{y}=-\int xdx ⇒ |lny| = \frac{-x^2}{2}+C ⇒ $
$e^{ln|y|}= e^{\frac{-x^2}{2}+C} ⇒ |y| = Ae^{\frac{-x^2}{2}} ⇒ y = ±Ae^{\frac{-x^2}{2}}$ where A = e^{c}. We have missed the solution y = 0 because we have divided by y.
$\frac{dy}{dx}=f(x)g(y) ⇒ \frac{dy}{g(y)}=f(x)dx ⇒ H(y) = \int \frac{dy}{g(y)};~ F(x) = \int f(x)dx $. Finally, we know that H(y) = F(x) + C, where C is a constant. Previously y is defined implicitly, you may want to do it an explicit equation, y = H^{-1}(F(x) + C).
Let’s solve the following geometrical problem. Let f(x) be a function such that the slope of the tangent to the curve at any point is twice the slope the ray from the origin.
$\frac{dy}{dx}=2\frac{y}{x} ⇒ \frac{dy}{y} = \frac{2dx}{x} ⇒ \int \frac{dy}{y} = \int \frac{2dx}{x} ⇒ ln(y) = 2ln(x) + C ⇒ e^{ln(y)} = e^{2ln(x) + C} ⇒ y = Ax^2~ where~ A=e^c$
For the case when a < 0, it is enough to realize that $\int \frac{dy}{y} = \int \frac{2dx}{x} ⇒ ln|y| = 2ln|x| + C$. Besides, y = ax^{2} ⇒ $\frac{dy}{dx}=2ax=\frac{2ax^2}{x}=\frac{2y}{x}$ and it works for a ≥ 0 or a < 0. However, there is a unresolved problem when x = 0.
A definite integral of a function is defined as the area of the region bounded by the function's graph between two points, say a and b. The integral of a real-valued function f(x) on an interval [a, b] is written as $\int_{a}^{b} f(x)dx$ (Figure 1.a.).
To compute the area, we can divide the region under the function into a series of rectangles corresponding to function values and multiplies by the step width to find the sum. Of course, we need to rectify this sum (it is just an approximation) by taking a limit, and as the rectangles get infinitesimally thin, we achieve the desired result (Figure 1.b.).
Let’s try to calculate $\int_{0}^{b} x^2dx$ where a = 0 and b is arbitrary. We divide the segment [0, b] into n pieces (Figure 1.c.). The sum of the areas of all rectangles is equal to $b^3(\frac{1^2+2^2+3^2+···n^2}{n^3})$
To try to understand ${1^2+2^2+3^2+···n^2}$, we can observe that it is basically a pyramid, the base of which is n x n blocks. On top of it, we put a second layer of n-1 x n-1 blocks (Figure 1.d). And we keep on piling up layer after layer. At the top of the pyramid, there is just one block of stone. Underneath our staircase pyramid, there is an ordinary pyramid, and its volume is $\frac{1}{3}(base~ ·~ height)=\frac{1}{3}n^2~ ·~ n$ On the outside, there is also another ordinary pyramid, with base and height (n+1)·(n+1), (n+1) respectively.
So our initial quantity ${1^2+2^2+3^2+···n^2}$ is to be found between these two pyramids: $\frac{1}{3}n^3 < 1^2+2^2+3^2+···n^2 < \frac{1}{3}(n+1)^3$ (Figure 1.e.) If we divide by n^{3}, $\frac{1}{3} < \frac{1^2+2^2+3^2+···n^2}{n^3} < \frac{1}{3}(1+\frac{1}{n})^3$ Therefore, our total area $\int_{0}^{b} x^2dx=\frac{1}{3}b^3.$
A Riemann sum is another approximation of an integral by a finite sum. The interval is partitioned into n pieces and we pick any height of f in each interval (e.g. f(c_{i}), Figure 1.f.) The sum is defined as $\sum_{i=1}^n f(c_i)\Delta x,~ where~\Delta x=\frac{b-a}{n}$ Please notice that in the limit, as the rectangles get thinner, the difference between the area covered by the rectangles and the area under the curve will get smaller, and finally it does not matter, we get $\int_{a}^{b} f(x)dx$
How much money do we borrow in the whole year? $\sum_{i=1}^{365} f(\frac{i}{365})\Delta t$ → [It is going to be very close] $\int_{0}^{1} f(t)dt$
How much money do you actually owe? Obviously, and regretfully, the interest on our debt is compounded continuously. If P is our principal (we start with a debt of P), then after time t you owe Pe^{rt} where r is the interest rate.
So, you borrow these amounts $f(\frac{i}{365})\Delta t$, but when you borrow these quantities, the amount of time left in the year is 1 -i/365, which is the amount of time this incremental debt will accumulate interest, that is, $(f(\frac{i}{365})\Delta t)e^{r(1-\frac{i}{365})}$. An at the end of the year, $\sum_{i=1}^{365} (f(\frac{i}{365})\Delta t)e^{r(1-\frac{i}{365})}$ that is basically $\int_{0}^{1} e^{r(1-t)}f(t)dt$
The Fundamental Theorem of Calculus states roughly that the integral of a function f over an interval is equal to the change of any antiderivate F (F’(x) = f(x)) between the ends of the intervals, i.e., $\int_{a}^{b} f(x)dx = F(b)-F(a)=F(x) \bigg|_{a}^{b}$
Examples:
For some distance s(t), the velocity function is v(t)=s’(t). Therefore, given a velocity function, v(t) -it is what is measured in the speedometer- $\int_{a}^{b} v(t)dt$ =s(b)-s(a) -the actual distance traveled, what is measured in the odometer, that is the instrument on your car dashboard that displays how many miles or kilometers your vehicle has traveled.
Let’s say that you checked your speedometer every single second ($\Delta t = 1 second$), the Riemman Sum is equal $\sum_{i=1}^n v(t_i)\Delta t≈\int_{a}^{b} v(t)dt$ =s(b)-s(a)
Let’s notice that * $\int_{0}^{2π} sinxdx = (-cosx)\bigg|_{0}^{2π}=-cos2π-(-cos0) = 0$
Definite integrals can be interpreted as the signed area of the region in the plane that is bounded by the graph bounded by its graph. Be careful, areas above the horizontal -x- axis of the plane are positive while areas below the horizontal axis are negative.