Evaluate. (Be sure to check by differentiating!) ∫(x^3+1)^4 x^2 d x

Name: Problem 4, Chapter 4: Calculus and its applications
Uploaded: 2021-04-05T19:46:53-08:00
Duration: 4 min 32 s
Channel: Lucas Finney
Description: Evaluate. (Be sure to check by differentiating!) ∫(x^3+1)^4 x^2 d x

Evaluate. (Be sure to check by differentiating!) ∫(x^3+1)^4...

Key Concepts

Chain Rule

In differentiation, the chain rule is used for finding the derivative of a composite function. When checking an antiderivative by differentiating, the chain rule ensures that the inner function’s derivative is correctly applied. This is essential in verifying integrals solved by substitution, as it confirms the integration was correctly performed by reversing the effect of the chain rule.

Verification through Differentiation

Once an antiderivative is obtained, differentiating it should yield the original integrand if the integration was performed correctly. This process of checking by differentiating leverages the fundamental connection between derivatives and integrals, ensuring that the antiderivative is indeed the correct primitive function.

U-Substitution

This is a method in integration where a substitution is made to simplify the integral into an easier form to integrate. By setting a part of the integrand to a new variable, often denoted as u, the integral can be transformed into a simpler integral in terms of u. This method is particularly useful when the derivative of the chosen substitution appears elsewhere in the integrand.

Transcript

00:01 All right, for this problem, we want to try to evaluate the integral of x cubed plus 1 to the power of 4 times x squared.

00:08 Now, this is obviously not an integral that we can evaluate directly, so we need to see if we can use one of our other techniques to allow us to integrate this.

00:17 The first technique that we should check, it's always going to see if we can do u substitution.

00:22 Which for u substitution, we want to see if we can write our function in the form, our integrating, in the form of f of g of x times g prime of x d x.

00:46 If we can do that, then we can make our use of substitution work out here.

00:52 So if we look at this, we have this x cubed plus one, which is then being put to the power of four, then we have the x squared out front.

01:02 So if we look at this, well, it looks like, we could say that our g of x, our internal function, is x cubed plus one.

01:11 If we look at what g prime of x is, well, g prime of x is going to be 3x squared.

01:20 So, it's not perfectly in that form that we're looking for, but we should be able to see, well, you know, we end up getting a scalar multiple times x squared.

01:32 So if we have this constant multiplying what, you know, multiplying our x squared, we can always bring that outside of our integral.

01:41 So if we look at doing our u substitution then, you know, it would make sense to say, well, u equals x cubed plus one, which would mean that du would equal 3x squared dx, or dx is equal to du over 3x squared, which then means when we substitute that in, we will get, i'll write this over the side here, so i have more room, we do that substitution.

02:13 We'll get that of that is equal to, and we have u to the power of 4, and we have our x squared, and then we'll have du over 3x squared.

02:24 So we'll have that the x squareds multiply together to 1, and we'll have that our integral simplifies down to you to the power of 4, the u over 3...

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