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▭\:\longdivision{▭} | \times \twostack{▭}{▭} | + \twostack{▭}{▭} | - \twostack{▭}{▭} | \left( | \right) | \times | \square\frac{\square}{\square} |
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- \twostack{▭}{▭} | \lt | 7 | 8 | 9 | \div | AC |
+ \twostack{▭}{▭} | \gt | 4 | 5 | 6 | \times | \square\frac{\square}{\square} |
\times \twostack{▭}{▭} | \left( | 1 | 2 | 3 | - | x |
▭\:\longdivision{▭} | \right) | . | 0 | = | + | y |
Imagine watching a movie in reverse: a shattered vase reassembling itself, footsteps returning to their source, rain rising back to the clouds etc. In calculus, this reversal has a name: the antiderivative. While derivatives break down motion and change into their smallest parts, antiderivatives do the opposite, they build up the original function from its changes. It’s the process of going back to the origin of anything mathematically, figuring out the original path from the speed or the total growth from the rate. Antiderivatives are fundamental not just in mathematics, but in understanding the world when viewed through the lens of restoration.
The concept of antiderivatives, or integration, has roots that reach back to ancient Mathematics. Early Mathematicians like Archimedes used geometric methods to find areas under curves, laying the groundwork for integral calculus long before the formal notation existed. However, the true development of antiderivatives as we know them today, began in the 17th century with the work of Isaac Newton and Gottfried Wilhelm Leibniz.
Newton approached the subject through motion and physical applications.Leibniz on the other hand focused on the formal, symbolic manipulation of integrals and their structure. He introduced the familiar integral sign ($\int$), derived from the Latin word 'summa' which means summation or total.
Both scientists recognized a powerful link between differentiation and integration which is now famously known as the Fundamental Theorem of Calculus.
Antiderivative is essentially the reverse of a derivative. If the derivative of a function gives us the rate at which something is changing, then the antiderivative allows us to recover the original function from that rate of change. Theoritically, if F(x) has a derivative F'(x) = f(x), then F(x) is called antiderivative of f(x).
Mathematically, this can be represented as:
$\frac{d}{dx}\left(F\left(x\right)\right)=f\left(x\right)$
$\int f\left(x\right)dx=F\left(x\right)+c$
Here, the symbol $\int$ denotes integration (finding the antiderivative), and c is an arbitrary constant known as the constant of integration. This constant appears because when differentiating a constant, it disappears — so when reversing the process, we include c to account for all possible original functions.
In simpler terms, if you know how fast something is changing, the antiderivative helps you figure out the total amount or original value that was changing.
Geometrically, an antiderivative represents the area under a curve of a given function. If f(x) is a function that describes a rate (like speed), then the antiderivative of f(x) gives the total accumulated value (like distance) from that rate over a given interval.
This concept is often visualized using a graph. Imagine plotting f(x) on a coordinate plane. The definite integral of f(x) from a to b is written as :
$\int _a^bf\left(x\right)dx$
This gives the net area between the curve and the x-axis from x=a to x=b. If the curve lies above the x-axis, this area is positive. If it lies below, the area is negative. The antiderivative function F(x) on the orher hand tells us how this area accumulates as we move along the x-axis.
So, while derivatives give the slope of a curve (how steep it is at a point), antiderivatives give the area under the curve, a powerful way to understand accumulation.
Power rule :
$\int x^n=\frac{x^{n+1}}{n+1}$
Example 1 : If $f\left(x\right)=x^5$, the
$\frac{d}{dx}\left(f\left(x\right)\right)=\frac{x^6}{6}$
Constant Rule :
$\int c=x+c$
Example 2 : If $f\left(x\right)=5$ , then,
$\int f(x)=5x+c$
Constant Multiple Rule :
$\int cf\left(x\right)=c\int f\left(x\right)$
Example 3 : If $f\left(x\right)=4x^7$, then,
$\int f\left(x\right)=\int 4x^7$
$\int 4x^7=4\cdot \frac{x^8}{8}=\frac{x^8}{2}+c$
Sum Rule :
$\int \left(f\left(x\right)+g\left(x\right)\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx$
Example 4 : If $f\left(x\right)=x^3+2x^2+7$, then,
$\int f\left(x\right)=\int x^4+\int 2x^2+\int 7+c$
$\int f\left(x\right)=\frac{x^4}{4}+\frac{x^3}{2}+7x+c$
Difference rule :
$\int \left(f\left(x\right)-g\left(x\right)\right)dx=\int f\left(x\right)dx-\int g\left(x\right)dx$
Example 5 : If $f\left(x\right)=x^3-2x^2$, find $\int f\left(x\right)$.
Solution :
$\int f\left(x\right)dx=\int x^3dx-\int 2x^2dx+c$
$\int f\left(x\right)dx=\frac{x^4}{4}-\frac{2x^3}{3}+c$
Integration by Substitution (Reverse Chain Rule) :
$\int f\left(g\left(x\right)\right)⋅g′\left(x\right)dx=\int f\left(u\right)du=F\left(u\right)+c$
where $u=g\left(x\right)$
Example 6 : If $f\left(x\right)=x\cdot \left(x^2+1\right)^4dx$, find $\int f\left(x\right)dx$.
Solution :
We know that $\int f\left(g\left(x\right)\right)⋅g′\left(x\right)dx=\int f\left(u\right)du=F\left(u\right)+c$ if $u=g\left(x\right)$
Here, let $u=x^2+1$
So, $\frac{d}{dx}\left(u\right)=2x$
$du=2xdx$
$\frac{du}{2}=xdx$
$\int x\cdot \left(x^2+1\right)^4dx$=$\int u^4\cdot \frac{du}{2}$=$\frac{1}{2}\int u^4du$
$\frac{1}{2}\int u^4du$=$\frac{1}{2}\cdot \frac{u^5}{5}+c=\frac{u^5}{10}+c$
Substitute back $u=x^2+1$
$=\frac{\left(x^2+1\right)^5}{10}+c$
Final answer,
$\int x\cdot \left(x^2+1\right)^4dx$=$\frac{\left(x^2+1\right)^5}{10}+c$
Integration by parts :
$\int udv=u⋅v−\int vdu$
Example 7 : Find $\int x⋅e^xdx$ given $\int e^xdx=e^x$
Solution :
$\int udv=u⋅v−\int vdu$
Let u=x (as it is simpler when differentiated) and $dv=e^xdx$
So, $dv=e^xdx$ and $v=\int e^xdx=e^x$
$\int x⋅e^xdx=x⋅e^x−\int e^x⋅dx=x⋅e^x−e^x+c$
$\int x⋅e^xdx=e^x\left(x-1\right)+c$
$\int \frac{1}{\sqrt{a^2-x^2}}dx=sin^{-1}\frac{x}{a}+c$
$\int \frac{1}{a^2+x^2}dx=\frac{1}{a}tan^{-1}\left(\frac{x}{a}\right)+c$
$\int \frac{1}{x\sqrt{x^2-a^2}}dx=\frac{1}{a}sec^{-1}\left(\frac{x}{a}\right)+c$
Example 8 : Find $\int \frac{x}{x^2+4}dx$.
Solution : Using substitution method,
$u=x^2+4$ and so, $du=2xdx$
$\int \frac{x}{x^2+4}dx=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}ln\left(u\right)+c$
Putting $u=x^2+4$ back in the equation,
$\int \frac{x}{x^2+4}dx=\frac{1}{2}ln\left(x^2+4\right)+c$
Example 9 : Find $\int \sqrt{x}dx$.
Solution :
Here, we would use power rule.
$\int x^ndx=\frac{x^{n+1}}{n+1}+c$
Here, n=$\frac{1}{2}$
So, $\int x^ndx=\frac{x^{n+1}}{n+1}+c=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c=\frac{2x^{\frac{3}{2}}}{3}+c$
Example 10 : Find $\int \left(\frac{1}{\sqrt{4-x^2}}\right)dx$.
Solution : We know that :
$\int \frac{1}{\sqrt{a^2-x^2}}dx=sin^{-1}\frac{x}{a}+c$
Here, a=2. So,
$\int \left(\frac{1}{\sqrt{4-x^2}}\right)dx=sin^{-1}\frac{x}{2}+c$
Example 11 : Find $\int \left(x^2+1\right)^3⋅2xdx$
Solution : Here, we would use substitution.
Let $u=x^2+1$ and so $du=2xdx$
$\int \left(x^2+1\right)^3⋅2xdx=\int u^3du=\int u^3du$
Substituting back $u=x^2+1$
$\int \left(x^2+1\right)^3⋅2xdx=\frac{\left(x^2+1\right)^4}{4}+c$
Antiderivatives, also known as indefinite integrals, play a vital role in interpreting and solving real-world problems involving accumulation, area, and reverse rates of change. Here are some key areas where antiderivatives are applied:
Physics: In kinematics, if acceleration is known as a function of time, its antiderivative gives velocity. Similarly, taking the antiderivative of velocity gives the displacement of the body.
Environmental Science: Antiderivatives are used to calculate the total accumulation of quantities like rainfall, pollution, or growth over time when the rate of change is known.
Biology : If a population growth rate is known, taking its antiderivative helps find the actual population function over time.
Engineering : In mechanics, when force is variable, the work done is the antiderivative of the force over displacement.
Business and Inventory : If the rate of sales or stock depletion is known, antiderivatives help estimate total sales over a time interval.
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