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Integral Calculus Formulas

Integration is the reverse process of differentiation. So, we may call it as Inverse Differentiation. Integration is the process to find a function with its derivative. Basic integration formulas on different functions are very useful and important. This article deals with the concept of integral calculus formulas with concepts and examples. Integral calculus is the branch of mathematics dealing with the formulas for integration, and classification of integral formulas. The student will take benefits from this concrete article. Let us learn the concept and the integral calculus formulas.

Concept of the Integral Calculus

Integration is the algebraic method to find the integral for a function at any point on the graph. The integral comes from not only to determine the inverse process of taking the derivative.

But also for solving the area problem as well. 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.

Given a function, f(x), is an anti-derivative of f(x), is any function F(x) such that

\(F’(x)=f(x)\)

If F(x) is any anti-derivative of f(x), then the most general anti-derivative of f(x) will be its indefinite integral.

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

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

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

Integral Calculus Formulas

Some Properties

  1. \(\displaystyle \int{{k\,f\left( x \right)\,dx}} = k\int{{f\left( x \right)\,dx}}\) where k is any number. So, we factor the multiplicative constants out of indefinite integrals.
  1. \(\displaystyle \int{{ – f\left( x \right)\,dx}} = – \int{{f\left( x \right)\,dx}}\) This is really the first property with k=−1k=−1 and so no proof of this property will be given.
  1. \(\displaystyle \int{{f\left( x \right) \pm g\left( x \right)\,dx}} = \int{{f\left( x \right)\,dx}} \pm \int{{g\left( x \right)\,dx}}\) It means the integral of a sum or difference of functions is the sum or difference of their individual integrals. This is applicable and useful for many functions as we need.

Formula for Integral Calculus

  1. \(\int x^n\; dx = {1\over n+1}x^{n+1}+C \\\)
  2. \(\hbox{ unless n=-1 } \\\)
  3. \(\int e^x \;dx= e^x+C \\\)
  4. \(\int {1\over x} \;dx= \ln x+C \\\)
  5. \(\int \sin x\;dx=-\cos x+C \\\)
  6. \(\int \cos x\;dx= \sin x + C\\\)
  7. \(\int \sec^2 x\;dx=\tan x+C \\\)
  8. \(\int {1\over 1+x^2} \; dx=\arctan x+C\)
  9. \(\int a^x \;dx= {a^x\over \ln a}+C \\\)
  10. \(\int \log_a x\;dx={1\over \ln a}\cdot{1\over x}+C \\\)
  11. \(\int { 1 \over \sqrt{1-x^2 }} \; dx=\arcsin x+C\\\)
  12. \(\int { 1 \over x\sqrt{x^2-1 }} \; dx=\hbox{ arcsec}\, x+C\)

Solved Examples for Integral Calculus Formulas

Q.1: Evaluate the following: \(\int 4x^5-3×17\sqrt{x}+{3\over x}\;dx\)

Solution:

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

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

Where C is integral constant.

Example-2: Evaluate the following indefinite integral: \(\int{{{x^4} + 3x – 9\,dx}}\)

Solution: It is asking for the most general anti-derivative. Therefore,

The indefinite integral is,

\(\int{{{x^4} + 3x – 9\,dx}}\)

\(= \frac{1}{5}{x^5} + \frac{3}{2}{x^2} – 9x + c\)

Where c is integral constant.

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