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Substitutions Involving e^x or ln(x) - Concept
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Because the derivatives of e^x and ln(x) are e^x and 1/x, respectively, the integrals of the latter two functions are just the former. If the variable parts of these functions are not just x, but a polynomial of x, then the substitution method must be used to get the correct answer.
I want to talk a little bit more about the method of substitution as it applies to functions involving e to the x or natural log. But first of all, let's go over these derivative formulas really quick.
The derivative of e to the x is e to the x and that's where we get the integral formula for e to the x, right? All we have to do is reverse this. The integral of e to the x is e to the x plus c. And which one of these do we use for the anti-derivative? We always use this formula for the anti-derivative of 1 over x because the domain of this of this formula is all real numbers except 0. The domain of this formula is only the positive numbers. This this is only true in the positive numbers. So it's not as good a formula to use for your anti-derivative. So that's why we we say that the integral of 1 over x dx is the natural log of the absolute value of x+c.
Now, every derivative formula can be turned into an integral formula. We remember that. One of the key problems with these formulas is that people sometimes misuse them. And I want to show you an example of how that can happen right now. For example integrating e to the 3 plus point 4x, and this is not exactly the same as e to the x. But some people might be tempted to just say, well, using this formula it would be e to the 3 plus point 4x+c. But I want to show you that this is not, this is not correct. The derivative, you differentiate the answer you should get the original function that you were integrating. So the derivative of that is going to be, you take the derivative of e to anything and you get e to that power, e to the 3 plus 0.4x times the derivative of 3 plus 0.4x and that's 0.4. Okay, this is not what we started out within the integral. So this is not true. Okay, so you can't just use this formula directly.
Same thing here. This looks a lot like, it's of the same form as you know 1 over x dx. I have 1 over 2-5x dx, right? I can write this as 1 over 2-5x. And so shouldn't that be the natural log of the absolute value of 2-5x+c? Well, you'll see it's not. Let's differentiate the answer and see if we get the original, the original function back again.
The derivative of natural log of the absolute value of 2-5x+c is, well first of all the derivative of natural log of the absolute value of something is just from this formula back here. Right? It's just 1 over that something. But here we'll have to use the chain rule. So it's 1 over 2-5x times the derivative of 2-5x which is -5. And the derivative of the +c is just 0. Alright. Now I can see that this is not the same. I started with 1 over 2-5x and I ended up with 1 over 2-5x times -5. So this is not right.
Okay, so both of these, both of these examples I just did are incorrect. The way they integrate these is using the method of substitution. I'll show you this in upcoming examples.
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