In mathematics, we usually need to find the derivative of some mathematical functions. It gives the rate of change of one variable with respect to others. Integration is the opposite process of differentiation. The fundamental use of integration is to get back the function whose derivatives are known. So, it is like an anti-derivative procedure. Thus, integrals are computed by viewing an integration as an inverse operation to differentiation. In this topic, the student will learn the Integration concepts as well as some integration formula with examples. Let us learn it!

**Integration Formula**

**Concept of integration:**

Integration is the algebraic method to find the integral for a function at any point on the graph. Finding the integral of some function with respect to some variable x means finding the area to the x-axis from the curve. Therefore, the integral is also called the anti-derivative because integrating is the reverse process of differentiating.

The integral comes from not only to determine the inverse process of taking the derivative. But also for solving the area problem as well. Similar to the process of differentiation for finding the slope at any point on the graph, this process of integration will be used to find the area of the curve up to any point on the graph.

The integral of the function of x from range a to b will be the sum of the rectangles to the curve at each interval of change in x as the number of rectangles goes to infinity.

The notation, which we have stuck with for historical reasons, is as peculiar as the notation for derivatives:

The integral of a function f(x) with respect to x is written as:

\(\int f(x)\;dx\)

Also, integration is considered as almost an inverse to the operation of differentiation means that if,

\({d\over dx}f(x)=g(x)\)

then

\(\int g(x)\;dx=f(x)+C\)

The extra C called the constant of integration, which is really necessary. This is because that after all differentiation kills off constants, which is why integration and differentiation are not exactly inverse operations of each other.

Since integration is almost the inverse operation of differentiation, the recollection of formulas and processes for differentiation is possible. So, many differentiation formulae will be used to provide the corresponding formula for the integration.

Definite integrals are the special kind of integration, where both endpoints are fixed. So, it always represents some bounded region, for computation.

**Some properties of Integration:**

And since the derivative of a sum is the sum of the derivatives, the integral of a sum is the sum of the integrals:

\(\int f(x)+g(x)\;dx=\int f(x)\;dx+\int g(x)\;dx\)

And, likewise, constants â€˜go throughâ€™ the integral sign:

\(\int c\cdot f(x)\;dx=c\cdot \int f(x)\;dx\)

**Formula for Integration:**

- \(\int x^n\; dx = {1\over n+1}x^{n+1}+C \)
- \(\hbox{ unless n=-1 } \)
- \(\int e^x \;dx = e^x+C \)
- \(\int {1\over x} \;dx= \ln x+C \)
- \(\int \sin x\;dx=-\cos x+C \)
- \(\int \cos x\;dx= \sin x + C\)
- \(\int \sec^2 x\;dx=\tan x+C \)
- \(\int {1\over 1+x^2} \; dx=\arctan x+C \)
- \(\int a^x \;dx= {a^x\over \ln a}+C \)
- \(\int \log_a x\;dx={1\over \ln a}\cdot{1\over x}+C \)
- \(\int { 1 \over \sqrt{1-x^2 }} \; dx=\arcsin x+C\)
- \(\int { 1 \over x\sqrt{x^2-1 }} \; dx=\hbox{ arcsec}\, x+C \)

**Solved Examples**

Q.1: Evaluate:

\(\int 5\cdot 2^x-{ 23 \over x\sqrt{x^2-1 }}+5x^2\;dx\)

Solution:

\(\int 5\cdot 2^x-{ 23 \over x\sqrt{x^2-1 }}+5x^2\;dx \)

= \({5\cdot 2^x\over \ln 2}-23\,\hbox{arcsec}\,x+{5x^3\over 3}+C \)

Q.2: Evaluate:

\(\int 4x^5-3×17\sqrt{x}+{3\over x}\;dx \)

Solution:

\(\int 4x^5-3×17\sqrt{x}+{3\over x}\;dx \)

= \({4x^6\over 6}-{3x^2\over 2}{ 17x^{3/2} \over 3/2 }+3\ln x+C \)

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26