 # Integration and Evaluation of Simple Algebraic Functions

Welcome to class!

In today’s class, we will be talking about Integration and evaluation of simple algebraic functions. Enjoy the class!

### Integration It is the reverse of differentiation. When we integrate, we have to add a constant (c).

To integrate a term, increase its power by 1 and divide by this figure. In other words: From our knowledge of derivatives, we can immediately write down a number of antiderivatives. Here is a list of those most often used: #### Methods of integration

##### 1. Integration by parts:

This technique for turning one integral into another is called integration by parts. We can rewrite this as If we let u = f(x) and v = g(x) then du = f ′ (x) dx and dv = g ′ (x) dx and To use this technique, we need to identify likely candidates for u = f(x) and dv = g ′ (x) dx.

Example:

Let u = ln x                            ,                du = 1/x dx

let dv = x dx                         ,                 v = x 2 /2

therefore, ###### 2. Partial fractions: If the integrand (the expression after the integral sign) is in the form of an algebraic fraction and the integral cannot be evaluated by simple methods, the fraction needs to be expressed in partial fractions before integration takes place.

The steps needed to decompose an algebraic fraction into its partial fractions results from a consideration of the reverse process − addition (or subtraction).

Consider the following addition of algebraic fractions: We want to go the other way around i.e. if we were to start with the expression and try to find the fractions whose sum gives this result, then the two fractions obtained, i.e. are called the partial fractions of So, if we needed to integrate this fraction, we could simplify our integral in the following way: We integrate the two fractions using what we learned in Basic Logarithmic Form: Where K is the constant. ##### 3. Substitution:

The substitution method (also called u−substitution) is used when an integral contains some function and its derivative.

In this case, we can set u equal to the function and rewrite the integral in terms of the new variable u. This makes the integral easier to solve.

Do not forget to express the final answer in terms of the original variable x. Substitution for integrals corresponds to the chain rule for derivatives.

Suppose that F(u) is an antiderivative of f(u): If u = u(x) is a differentiable function and using the chain rule, we have Integrating both sides, Therefore, This is the substitution rule formula for indefinite integrals.

Note that the integral on the left is expressed in terms of the variable x. The integral on the right is in terms of u.

Example:

Find the integral ∫(3x+2)5dx.

Solution

We make the substitution u = 3x+2.

Then

du = d(3x+2) = 3dx.

So, the differential dx is given by

dx = du / 3

Therefore, #### Area under a curve In this case, we find the area by simply finding the integral:

$Area={\int }_{a}^{b}f\left(X\right)dx$

Example:

Find the area bounded by y = x2 − 4, the x-axis and the lines x = −1 and x = 2.

Solution #### Simpson’s rule

In Simpson’s Rule, we will use parabolas to approximate each part of the curve. This proves to be very efficient since it’s generally more accurate than the other numerical methods we’ve seen. We divide the area into n equal segments of width Δx. The approximate area is given by the following: Note: In Simpson’s Rule, n must be EVEN.

We hope you enjoyed the class.

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