I am sure that you must be familiar with the formulae for the areas of different geometrical objects like a square, rectangle, triangle etc. But do you know how to evaluate the areas under various complex curves using the known basic areas? You’ll most likely think about dividing the given area into some basic shapes that you are aware of and add up your areas to approximate the final result. This process in mathematics is actually known as integration and is studied under integral calculus.

We also have various general integration formulae which may be utilized straightaway to yield the resultant areas under the curve as well. So let us now look at these formulae and understand integration better!

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## Integral Calculus

As the name should hint itself, the process of Integration is actually the reverse/inverse of the process of Differentiation. It is represented by the symbolÂ âˆ«, for example, $$ \int (\frac{1}{x}) dx = log_e x + c $$

where,

- (\(\frac{1}{x}\)) – the integrand
*dx*– denotes that x is the variable with respect to which the integrand has to be integrated*log*– the resultant function_{e}x*c*– the constant of integration

You may note that,

\(\frac{d}{dx}(log_e x) = \frac{1}{x}\)

Comparing with the example for integration, you can see it for yourself how the integration and differentiation of a function are interconnected. Besides, from the formulae of differentiation, we also have –

\(\frac{d}{dx}(log_e x + c) = \frac{1}{x}\)

Thus, you always need to attach a constant of integration with the result (unless provided with some initial or boundary conditions), because the differentiationÂ of such a function also gives the same result.

Since you must already be aware of the formulae for derivatives of various known functions, we can use those results in the opposite way here now. Even if you don’t remember the exact formulae, you may verify the result yourself by differentiating the integrand and looking for yourself. Thus, building upon this simple connection with differentiation, we can understand the following basic formulae for integration.

*Source: wikimedia*

## Basic Integral Calculus Formulae

- Change of the variable of integration doesn’t affect the result, $$\int f(x)dx = \int f(z)dz = \int (f(u)du$$
- \( \int x^n dx = \frac{x^{n+1}}{n+1} + c \text{(n â‰ -1)}Â \)
- \( \text{Case of n = -1 of above formula – } \int \frac{dx}{x} = log_e x + cÂ \)
- \( \int e^x dx = e^x + c $$ $$ \Rightarrow \int a^x dx = \frac{a^x}{log_e a} + cÂ \)

Irrespective of the integral you may evaluate, the following properties always hold –

**Product with a constant**$$ \int cf(x) dx = c\int f(x)dx + k \text{(where c – known constant, k – constant of integration)}$$**Additive PropertyÂ**$$ \int (f(x) \pm g(x))dx = \int f(x)dx \pm \int g(x)dx $$

For certain specific recurring formats of the integrands, given below are some standard formulae which may be easily verified by differentiation. To apply them, sometimes some manipulation needs to be done to convert the given problems in the standard format.

## Standard Integral Calculus Formulae

$$Â \Rightarrow \int \frac{dx}{x^2 – a^2} = \frac{1}{2a}log|\frac{x – a}{x + a}| + c $$ $$Â \Rightarrow \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}log|\frac{a + x}{a – x}| + c $$ $$ \Rightarrow \int \frac{dx}{\sqrt{x^2 + a^2}} = log |x + \sqrt{x^2 + a^2}| + c $$ $$ \Rightarrow \int \frac{dx}{\sqrt{x^2 – a^2}} = log (x + \sqrt{x^2 – a^2}) + c $$ $$ \Rightarrow \int \sqrt{x^2 + a^2} dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}log(x + \sqrt{x^2 + a^2}) + c $$ $$ \Rightarrow \int \sqrt{x^2 – a^2} dx = \frac{x}{2}\sqrt{x^2 – a^2} – \frac{a^2}{2}log(x + \sqrt{x^2 – a^2}) + c $$ $$ \Rightarrow \int e^x{f(x) + f'(x)} dx = e^x f(x) + c $$

Sometimes, or rather most of the times, the integrands are not in such a simple form. For eg – You know the integration of e^{x}, but what about the integration of e^{3x}? You’ll need to use the method of substitution for such cases.

## The Method of Substitution (Change of Variable)

This method is used to reduce a seemingly complex integrand to a known simple form, for which the integration formula is known already. With enough practice and a good understanding of the integration formulae, you’ll understand yourself what substitutions to make.

For a given complex integralÂ âˆ«F(h(x))dx, you may make the substitution h(x) = z (your new variable of integration. Followed by this step, you’ll also have to change the variable of integration. Thus, dx will have to be changed to dz. From h(x) = z; you may arrive at h'(x)dx = dz.

Substituting this in the place of dz, and proceeding with the new variable, you may now be able to successfully integrate the resulting function. Let us look at a couple of important applications below for a better idea of this process –

**â‡’ \(\int f(ax) dx = ?\)**

SubstituteÂ *z = ax. *Then on differentiating, we haveÂ *dz = adx*. Use this in the integral to get –

\(\int f(ax) dx = \int f(z) \frac{dz}{a}\)

\(= \frac{1}{a}\int f(z) dz\)

Thus if you actually know how to solve the integral \(\int f(x)dx\), the integral \(\int f(ax)dx\) with a = a real number, can be easily evaluated using the change in the variable of integration. Obviously in the last step here,Â \(\int f(z)dz\) =Â \(\int f(x)dx\).

**â‡’ \(\int \frac{f'(x)}{f(x)} dx = ?\)**

Substitute *f(x) = z* in this case. On differentiating, we haveÂ *f'(x)dx = dz*. Put this back in the integral to get –

\(\int \frac{f'(x)}{f(x)} dx = \int \frac{dz}{z}\)

Using the formulae for integration,

\( = log_e z + c\)

Re-substituting for z in the result,

\( = log_e f(x) + c\)

These two results arrived at, above are very useful in evaluating simple integrals and find great application in problems.

## Solved Examples for You on Integral Calculus

**Question:Â **Solve $$ \int 4e^{-7x}dx $$

**Solution:** Using the result derived at in theÂ *Variable Substitution methodÂ *above, and the basic formulae for integration, we can find the result as – $$Â \int 4e^{-7x}dx = 4\int e^{-7x}dx$$ $$ = 4.\frac{e^{-7x}}{-7} $$ $$ = -\frac{4}{7}e^{-7x} $$

**Question:** Solve $$ \int 8x^2(3x^3 – 1)^{16} dx $$

**Solution:** Substitute as follows –

3x^{3} – 1 = z

9x^{2}dx = dz

x^{2}dx = dz/9

Using this substitution, the integral can be changed, and eventually solved as – $$\int 8x^2(3x^3 – 1)^{16} dx = \int 8.\frac{dz}{9}.z^{16} $$ $$ = \int \frac{8}{9}z^{16} dz $$ $$ = \frac{8}{9}\frac{z^{17}}{17} $$ $$ = \frac{8}{17}(3x^3 – 1)^{17} $$

**Question:** Solve $$\int \frac{dx}{1 – 6x – 3x^2} $$

**Solution:Â **We can rewrite the given integrand in the format of a standard integral as follows – $$\int \frac{dx}{1 – 6x – 3x^2} = \int \frac{dx}{1 – 3(2x + x^2)}$$ $$ = \int \frac{dx}{1 – 3(2x + x^2 + 1 – 1)}$$ $$ =Â \int \frac{dx}{4 – 3(x + 1)^2}$$ $$ = \int \frac{1}{3}\frac{dx}{\frac{4}{3} – (x + 1)^2} $$

Now, we may proceed by substitution of variable as –

x + 1 = z

dx = dz

Then, the integral can be evaluated as – $$\int \frac{1}{3}\frac{dx}{\frac{4}{3} – (x + 1)^2} = \int \frac{1}{3}\frac{dz}{\frac{4}{3} – z^2}$$ $$ \text{Using the necessary standard formula:} $$ $$ = \frac{1}{3}\frac{1}{2.\frac{3}{4}}log|\frac{\frac{4}{3} + z}{\frac{4}{3} – z}| + c $$ $$ = \frac{1}{8}log|\frac{4 + 3z}{4 – 3z}| + c $$ $$ = \frac{1}{8}log|\frac{4 +3(x + 1)}{4 – 3(x + 1)}| + c $$ $$ = \frac{1}{8}log|\frac{7 + 3x}{1 + 3x}| + c $$

This concludes our discussion on the topic of introduction to integral calculus.

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