Connection to Areas

Key Questions

What is an integral?

In mathematics, we talk about two types of integrals. Definite integrals and indefinite integrals.

Generally, an integral assigns numbers to functions in a way that can describe displacement, area, volume and even probability.

Definite Integrals

This type of integral relates to numerical values. It is used in pure mathematics, applied mathematics, statistics, science and many more. However, the very basic concept of a definite integral describes areas.

The definite integral of a function $f$ $f$ over an interval $[a, b]$ $[a,b]$ represents the area defined by the function and the x-axis from point $a$ $a$ to point $b$ $b$ , as seen below.

![https://en.wikipedia.org/wiki/Numerical_integration]()

The symbol used to represent this area $S$ $S$ and integral, respectively, is

$S = \int_{a}^{b} f (x) d x$ $S=int_a^b f(x)"d"x$ , where

$⋄ f is called the integrand$ $diamond f " is called the integrand"$
$⋄ a and b are the lower and upper bounds$ $diamond a and b " are the lower and upper bounds"$
$⋄ x is a dummy variable$ $diamond x " is a dummy variable"$

You might be wondering what $d x$ $"d"x$ means. Formally, it doesn't mean anything but rather it tells you which variable you are differentiating with respect to or in our case, tells you the variable of integration.

When we say the area defined by the function $f$ $f$ with the x-axis, we mean the net area. The net area is not the same as absolute area.

If the graph of the function is above the x-axis, then it is said that the net area is positive. If it is below, the net area is negative. This might be harder to grasp at first. This is visualised below:

![https://en.wikipedia.org/wiki/Integral]()

For example, say we are tasked with finding the net area under the curve $f (x) = x^{2}$ $f(x)=x^2$ from $0$ $0$ to $1$ $1$ :

In our case,

$S = \int_{0}^{1} x^{2} d x$ $S = int_0^1 x^2 "d"x$

To not complicate this answer, here's a video describing it in greater detail:

As such, it has been proven that $S = \frac{1}{3}$ $S=1/3$ .

We can make a general case here; for every $n \neq - 1$ $n!=-1$ ,

$\int_{0}^{τ} x^{n} d x = \frac{τ^{n + 1}}{n + 1}$ $int_0^tau x^n "d"x = tau^(n+1)/(n+1)$

At the same time, the video describes Riemann sums. These are used to compute integrals. Generally, the Riemann sum of a function $ϕ$ $phi$ is

$\int_{a}^{b} ϕ (x) d x = lim n \to \infty b n \sum i = a n ϕ (x_{i}) Δ x_{i}$ $int_a^b phi(x)dx = lim_(n->oo) sum_(i=an)^(bn) phi(x_i)Deltax_i$

where $Δ x_{i} = x_{i} - x_{i - 1}$ $Deltax_i = x_i-x_(i-1)$ and $x_{i}$ $x_i$ , as mentioned in the video above, represent some "marks" on the x-axis. One possible solution is by letting $x_{i} = i / n$ $x_i=i"/"n$ . Then $Δ x_{i} = 1 / n$ $Deltax_i = 1"/"n$ . While this is usually simpler, it might not be easiest or fastest way to compute integrals.

If we remember the general case formed earlier, about the integral of $x^{n}$ $x^n$ from $0$ $0$ to $τ$ $tau$ ; well, this is called the Power Rule. There are many, many ways different formulas for integrals, which I won't cover in this answer. This is just a very general idea of what integrals are.

Indefinite Integrals

These are represented as integrals with bounds. Let $I$ $I$ be the indefinite integral of a function $f$ $f$ .

$I = \int f (x) d x$ $I=int f(x)"d"x$

You can think of indefinite integrals as generalisations of definite ones.

Instead of being defined by areas, volumes or something else, indefinite integrals correlate to derivatives. The indefinite integral of a function $f$ $f$ is also called the antiderivative and is often noted as $F (x)$ $F(x)$ .

The Fundamental Theorem of Calculus bridges the gap between a function, its derivative and its indefinite integral. Basically, it says that $F$ $F$ is defined as the function which, when differentiated, gives $f$ $f$ :

$F' (x) = f (x)$ $F'(x) = f(x)$

Now, say we want to find the antiderivative of the function $f (x) = x^{2}$ $f(x)=x^2$ .

$F (x) = \int f (x) d x = \int x^{2} d x$ $F(x) = int f(x)"d"x=intx^2"d"x$

Using our former definition, what function do we have to differentiate to get $x^{2}$ $x^2$ ? The Power Rule for derivatives states that, if $f (x) = x^{n}$ $f(x)=x^n$ , then $f' (x) = n x^{n - 1}$ $f'(x) = nx^(n-1)$ . As such, if we assume $F (x)$ $F(x)$ to be an algebraic function of the type

$F (x) = constant \cdot x^{exponent}$ $F(x) = "constant"*x^"exponent"$ , we have:

$F (x) = c x^{k} \Rightarrow F' (x) = c k x^{k - 1}$ $F(x) = cx^k => F'(x) = ckx^(k-1)$

Except that this is not complete. Remember that, when differentiating a constant with respect to a variable, it practically dissapears, hence the true form of $F (x)$ $F(x)$ is

$F (x) = {constant}_{1} \cdot x^{exponent} + {constant}_{2}$ $F(x) = "constant"_1*x^"exponent"+"constant"_2$

Let $α$ $alpha$ and $C$ $C$ be the two constants and $k$ $k$ the exponent.

$F (x) = α x^{k} + C \Rightarrow F' (x) = α k x^{k - 1}$ $F(x) = alphax^k + C => F'(x) = alphakx^(k-1)$

Since $F' (x) = f (x)$ $F'(x) = f(x)$ , we can conclude that

$α k x^{k - 1} = x^{2} \Rightarrow {\begin{matrix} k - 1 = 2 α k = 1 \end{matrix} \Leftrightarrow {\begin{matrix} k = 3 α = \frac{1}{3} \end{matrix}$ $alphakx^(k-1)=x^2=>{(k-1=2),(alphak=1) :}<=>{(k=3),(alpha=1/3) :}$

$∴ \int x^{2} d x = \frac{x^{3}}{3} + C$ $:. int x^2"d"x = x^3/3+C$

Analogously, if we define $f (x) = x^{n}$ $f(x) = x^n$ for $n \neq - 1$ $n!=-1$ , then

$F (x) = \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $F(x) = int x^n "d"x = x^(n+1)/(n+1)+C$

$C$ $C$ is called the constant of integrations and simply is a random constant. It's not a particular value. It's just there for the sake of correctness.

Bridging the gap between definite and indefinite integrals

Our previous antiderivative of $x^{n}$ $x^n$ does resemble something which we got earlier while talking about definite integrals. We see that, if we allow $C$ $C$ to be $0$ $0$ , then

$F_{(C = 0)} (τ) = \frac{τ^{n + 1}}{n + 1}$ $F_((C=0))(tau) = tau^(n+1)/(n+1)$

But we know that this is also equal to $\int_{0}^{τ} x^{n} d x$ $int_0^tau x^n "d"x$ :

$F_{(C = 0)} (τ) = \int_{0}^{τ} x^{n} d x$ $F_((C=0))(tau) = int_0^tau x^n "d"x$

This is where the connection between definite integrals and indefinite integrals is visible, stated formally below:

If $F (x)$ $F(x)$ is the antiderivative of a function $f (x)$ $f(x)$ and we let $C = 0$ $C=0$ , then

$\int_{a}^{b} f (x) d x = F (b) - F (a)$ $int_a^b f(x) "d"x = F(b)-F(a)$

I hope this answer wasn't too intimidating.

Hammer · 1 · Jun 16 2018
How are derivatives related to area?

Answer:

See below.

Explanation:

Derivatives aren't directly related or associated with area.

Derivatives describe the $s l o p e$ $slope$ of a tangent line in relation to another line, at a given point.

However, derivatives can relate to the rate of change in an objects area. This is usually figured through [implicit differentiation.](https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-advanced/ab-implicit-diff/v/implicit-differentiation-1)

This finds the rate at which an area is increasing/decreasing, by finding the derivative of the appropriate area formula in respect to time.

For explanation sake, here is an example:

Let's say we need to find the rate in which the $a r e a$ $area$ of a circle is increasing, when the rate of the $r a d i u s$ $radius$ increasing is $2$ $2$ m/s with the $r a d i u s$ $radius$ being at $4$ $4$ meters.

We know that the $a r e a$ $area$ of a circle is:

$A = π r^{2}$ $A=πr^2$

Using the chain rule , we can find the derivative of the equation. However, since we are finding the derivative in respect to time, we need to take on a $\frac{d (□)}{d t}$ $(d(square))/dt$ ( $□$ $square$ being the variable in which you are finding the derivative of.

So, here is what it will look like after we find the derivative (See above link if you have questions):

$\frac{d A}{d t} = 2 π r \frac{d r}{d t}$ $(dA)/dt=2πr(dr)/dt$

From here we can simply plug in our values:

$\frac{d A}{d t} = 2 π (4) (2)$ $(dA)/dt=2π(4)(2)$

$\frac{d A}{d t} = 16 π \frac{m^{2}}{sec}$ $(dA)/dt=16π m^2 / sec$

Joseph C. · 1 · Mar 17 2018
What is an anti-derivative?

The anti-derivative (also called the integral) of a function $f$ $f$ is a function $g$ $g$ with the property that its derivative $g'$ $g'$ equals the original function $f$ $f$ .
ie. $\int f (x) d x = g (x) + C \Leftrightarrow g' (x) = f (x)$ $intf(x)dx=g(x)+C iff g'(x)=f(x)$

Trevor Ryan. · 1 · Oct 12 2015

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Connection to Areas

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Limits, Motion, and Areas