Indefinite Integrals

If you are given the speed of your bike as a function of time – would it be possible for you to calculate the distance traveled within certain time periods by using just that function? Well, integrals are the property which will help you to find it very easily. Other interesting examples could be to find the time period of milk getting poured into your glass or to find how much area your bed covers inside of your room. Exciting huh? Let’s find out more about indefinite integrals:

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How Integration is Done?

Integration is a method of adding things up. It joins slices and makes them whole. However, here we are going to use it to find the area of any random curve. Like take this curve for instance –

\(y = f(x)\)

Now, if I ask you to find the area of this curve then you might not be able to find it. But, what if we divide this curve into small pieces? This should make it a little simple.

But what if we divide it into even smaller pieces, it will be much better, right?

          

Similarly, if we keep on decreasing the size of the pieces, it will take the shape of the given curve, hence making the calculated area a lot more accurate. But how do we add them up now, this is all so small! Well, we don’t have to.

There is another way of calculating this area and for that, all we have to do is find the antiderivative of the given function. Yes, you read it right, Integration is the inverse of Differentiation. So, we know that to find the derivative of f(x), we write

\(d/dx (f(x)) = g(x)\)

So to find the antiderivative, we will write –

\( \displaystyle \int g(x).dx = f(x)+c\)

Here, c is the constant of integration. Now you must be wondering why this c? Let me give you an example to help you with this.

If \(f(x) = x^2-3\) , then \(f’(x) = 2x\). But now, when you’ll take the antiderivative of 2x, you’ll get \(x^2\) only. This is a problem. We started with \(x^2-3\) but we now have \(x^2\), to counter that we introduce a constant of integration ‘c’, this makes our answer, \(x^2+c\) which is nothing but the family of all possible \(x^2\) curves. Of course, we can get back \(x^2-3\), but more on that later.

Area under the curve

For now, we wanted to calculate the area of the curve \(y = f(x)\), which will be

\( \displaystyle \int f(x).dx = h(x)+c \)

Formulae for Indefinite Integrals

Now that we already taken care of the concept of Integration, let’s take a quick look at some of the basic indefinite integrals formulae –

\( \displaystyle \int x^n dx = \frac{x^{n+1}}{(n+1)} + c\)

\( \displaystyle \int \sin{x} dx = – \cos{x} + c\)

\( \displaystyle \int \cos{x} dx = \sin{x} + c\)

\( \displaystyle \int \frac{1}{x} dx = \ln{x} + c\)

\( \displaystyle \int e^x dx = e^x + c\)

You can download Integrals Cheat Sheet by clicking on the download button below

Solved Examples for You

Q: Find the integration of \(x^3\).

Sol:  \( \displaystyle \int x^3 dx \)

\(= \frac{x^3}{3+1} + c\)
\(= \frac{x^4}{4} + c\)

Q: Find the integration of \(3 \cos{x} + e^x \).

Sol: \( \displaystyle \int (3 \cos{x} + e^x) dx \)

\(= 3 \sin{x} + e^x + c \)

Solved Questions for You

Question 1: What does an indefinite integral mean?

Answer: An indefinite integral refers to a function which takes the anti-derivative of another function. We visually represent it as an integral symbol, a function, and after that a dx at the end.

Question 2: Why is it called indefinite integral?

Answer: The reason that we call it the indefinite integral is because there is a remarkable link between the definite integral and the indefinite integral.

Question 3: How do you integrate fractions?

Answer: In order to integrate fractions we begin by multiplying or dividing the top and bottom of the fraction with a number. At times it will be of assistance if you split a fraction up before making an attempt to integrate it. You can use the method of partial fractions for it.

Question 4: What are integrals used for?

Answer: Integrals may be of use for computing the area of a two-dimensional region which has a curved boundary. We may also use it for computing the volume of a three-dimensional object having a curved boundary. We can calculate the area of a two-dimensional region by making use of the aforesaid definite integral.