##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# Simple Substitutions - Concept

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

The substitution method is useful on some indefinite integrals that are not as simple as they look. These include functions such as e^(5-2x) or the square root of [x-3], in which the variable part of the function is more complex than just an x. These **simple substitutions** require use of the substitution method to solve.

When you're solving indefinite integrals, there are some deceptively simple situations that require the method of substitution. Let's take a look at an example.

First of all, recall the power rule for antiderivativeS. The integral of x to the n is 1 over n plus 1 x to the n plus 1 plus c. Now, when I compare 2 integrals, this really simple integral root x and this slightly more complicated one but still fairly simple root 4x+3. Now when I integrate root x, I can just immediately convert that to a power of x and use the power rule from up here. The question is can I do that for root 4, 4x+3? I will get this answer right? Just two thirds 4x+3 to the 3 halves plus c. Just following the exact same pattern.

Now I want to check this answer by differentiation to see if it's correct. So only take this answer and differentiate it. And I have two thirds 4x+3 to the 3 halves plus c. Okay, the two thirds will come on in front and then fourth, 4x+3 to the 3 halves, the 3 halves will come out and I have 4x+3 to the one half, right? This power minus 1. But I still have to multiply by 4, the derivative of the inside part. Have to use the chain rule. And of course the derivative of the +c is 0. Now when I look at this I get two thirds and 3 halves. This is just 1, then I have a 4 here. 4 times the square root of 4x+3.

Now let's take a look at what we started with. The integral of root 4x+3. We didn't have a 4 in the front. So this method has not worked. In general, it's not going to work unless you have just a single variable raised to a power, right? The fact that I have a 4x+3, this is not the same as x. I have to use the method of substitution in a case like this even though it's a very simple integral.

So integral 2 requires substitution. Let's take a crack at it. So I have the integral of root 4x+3 dx. Remember when I use the method of substitution, I look for a composite function, and this is a composite function. I want to substitute for the inside part. So w=4x+3 and I want to calculate a dw, so the dw is going to be 4 times dx. Now, I have a dx. I don't have a 4 times dx. So what I need to do is multiply both sides by one quarter. So a quarter dw is going to equal dx, and this is what I'm going to replace my dx with. One quarter dw.

So I get the integral of the square root of w times one quarter dw. Now I can switch this to w to the one half and pull the one quarter up in front. So one quarter w to the one half dw. And now I can use the power rule. Here I do have just a simple variable raised to a power. So it's one quarter, I have to raise the exponent by 1. So one half becomes 3 halves and then dividing by 3 halves is the same as multiplying by 2 thirds. This gives me one sixth, right? The twos cancel. I get a one half times one third. one sixth w to the 3 halves plus c.

And the last thing to do is resubstitute for w. I need to put a 4x+3 in there. So this is one sixth sorry. 4x+3, 2 3 halves plus c. That's my final answer. Not exactly the same as the 2 thirds that I thought it was before. This is the correct answer. One sixth 4x+3 to the 3 halves plus c.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

###### Get Peer Support on User Forum

Peer helping is a great way to learn. Join your peers to ask & answer questions and share ideas.

##### Concept (1)

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete