## Calculus Tutorial

*Integration by Parts*

#### Intro

The integration by parts formula S udv = uv – S vdu is derived from the product rule of differentiation d (uv) = vdu + udv. From the derivative form we will be making the function or equation back to its original function though it is in product form. I will be giving example(s) how you will deal with this.

#### Sample Problem

#### Solution

The first step in solving this particular problem is that we first identify what is u and what is dv. Therefore,

S xe^x dx

u = x. dv = e^x dx

du = dx. v = e^x

If I will be going to make e^x as your u and dv = x dx, it will make your equation very long, thus you are prolonging integrating. Applying the formula

S udv = uv – S vdu

S xe^x dx = xe^x – S e^x dx

The integral of S e^x dx = e^x + C. Therefore,

S xe^x dx = xe^x – e^x + C

Factoring e^x,

S xe^x dx = e^x (x – 1) + C

This will be your final answer. Though you can consider the upper part as your answer but much appreciated if it is in the factored form. To check if the integrated function is correct, find its derivative using the product rule of differentiation.

y = e^x (x – 1) + C

The product rule is

d(uv) = vdu + udv

Say that u = e^x v = x – 1

du = e^x. dv = 1

Applying the formula,

y’ = e^x (x – 1) + e^x(1)

We remove the grouping symbol by using the DPMA or Distributive Property of Multiplication over Addition/Subtraction,

y’ = xe^x – e^x + e^x

And by cancellation or combining of similar terms,

y’ = xe^x

As you can see it is the same as in the given, therefore e^x(x – 1) + C is the correct answer for this particular problem.

# About The Author

Mathematics Instructor/Tutor |

I am a tutor for almost 5 years and I am teaching Mathematics and Science. A graduate of Bachelor of Science in Electronics Engineering and a Licensed Elecronics Technician. I am currently taking Continuing Professional Education and soon taking Licensure Examination for Teachers. |