If f is continuous on ℝ , prove that ∫a^b f(-x) d x=∫-b^-2 f(x) d x

Name: Problem 83, Chapter 5: Calculus Early Transcendentals
Uploaded: 2021-12-08T02:38:42-08:00
Duration: 5 min 18 s
Channel: Bobby Barnes
Description: If f is continuous on ℝ , prove that ∫a^b f(-x) d x=∫-b^-2 f(x) d x

If f is continuous on ℝ , prove that ∫a^b f(-x)...

Key Concepts

Continuity of Functions

Continuity of the function over the interval of integration is an essential concept that guarantees the integrability of the function. When a function is continuous on an interval, it assures that the definite integral exists and that the application of substitution and other integral properties are valid. This makes it possible to confidently manipulate the integral expressions without concern for discontinuities or undefined behavior.

Properties of Definite Integrals

Understanding the properties of definite integrals is crucial when performing variable substitutions and manipulating integration limits. Key properties include the effect of reversing the limits (which introduces a negative sign) and the additivity over intervals. These properties are fundamental in equating the transformed integral with another integral that has its limits correctly adjusted to match the change in variable.

Substitution Rule

This is a technique used to change the variable of integration in a definite integral. By substituting u = -x, for instance, one can transform an integral of the form ? f(-x) dx into an equivalent form involving f(u) with adjusted limits of integration. This technique relies on accurately computing the new integration bounds and the differential, ensuring that the integral’s value remains unchanged after the transformation.

Transcript

00:01 So if we want to prove this, let's just go ahead and start with the left -hand side and see if we can work our way to the right -hand side.

00:07 So the left -hand side is going to be equal to the integral from a to b of f -negative x, d -x.

00:15 Well, if we want to make it look like what we have over here, we just have f of x.

00:20 So let's just go ahead and say u is equal to this negative x right here.

00:27 And now, taking the derivative, we would have d -u over d -x is equal to, negative 1, and then we can multiply that dx over like this, giving us du is equal to negative 1, negative dx.

00:45 So if we make this negative on the outside and on the inside like this, now this negative x, dx become d .u, so let's go ahead and write that.

00:56 So it would be negative integral of, so it would be f of u, and then we have du out here.

01:05 And, our bounds, well, when x is equal to a, u is going to be equal to negative a, so this is negative a.

01:18 And then when x is equal to b, we have u is equal to negative b, so that's negative b.

01:27 Now, we don't have this negative out here, and then you notice that our bounds are switched.

01:31 So we also have this property that says when we have a negative on the outside, we can use that to switch our bounds.

01:38 So, i'm going to go ahead and do this.

01:43 So this is going to be equal to the integral of negative b to negative a of f of u, d, since, and i guess the important part is that it is continuous since f is continuous.

02:07 All right, now at this point, u is just a dummy variable, so we can just replace this with whatever we want.

02:14 And so then this is just going to be equal to negative b, negative a, f of x, dx, since u is just a dummy variable.

02:33 And then you can see how we started on the left -hand side, and we got to the right -hand side.

02:39 So that's the first part of the question.

02:42 Now, the second part, they want us to draw a diagram given f of x, positive that has this property.

02:50 So let me actually pick this up and scoot it down.

02:57 All right.

02:58 So we have f x greater than equal to zero and zero less than a less than b.

03:15 So let's go ahead and just draw what this looks like...

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