Find the area between the graph of the given function and the x -axis over the given interval, i

Find the area between the graph of the given function and...

Key Concepts

Substitution Method in Integral Calculus

The substitution method is a technique used to simplify integrals by making a change of variable. This method transforms the integrand into a simpler or more standard form, making it easier to evaluate. It is especially useful when dealing with composite functions or when the derivative of one part of the integrand appears elsewhere in the integral.

Absolute Value in Integration

The area between a curve and the x-axis is found by integrating the absolute value of the function, rather than the function itself, because parts of the graph may lie below the axis. This ensures that all contributions to the area are counted positively, which is particularly important for functions that change sign over the interval of integration.

Improper Integrals

Improper integrals involve integrals with one or both limits of integration being infinite or where the integrand becomes unbounded within the interval. To evaluate them, one typically replaces the infinite limit with a variable and takes the limit as that variable approaches infinity. This concept is essential when determining convergence or divergence and correctly computing areas in such cases.

Odd Functions and Symmetry

An odd function is defined by the property that f(-x) = -f(x) for all x in its domain. When integrating an odd function over an interval symmetric about the origin, the contributions from the negative and positive sides cancel each other out, resulting in a net integral of zero. Recognizing this property is crucial when evaluating signed areas versus actual areas under the curve.

Transcript

00:01 We want to find the improper integral of negative infinity to infinity of x over 1 plus x squared squared.

00:12 And one thing you might notice is that this here, x over 1 plus x squared squared squared, is an odd function.

00:37 And a property that we have for odd functions on symmetric intervals, because negative infinity to infinity can be thought of as a symmetric interval, is that this is equal to zero.

00:53 But since we're using improper intervals, let's actually do it the long way to show that this is actually equal to zero.

01:03 But this is just kind of as like a side note.

01:05 If we have an odd function on a symmetric interval, then we can just go ahead and say that the area for this will be zero right away.

01:16 So hopefully whatever we do here matches this answer.

01:21 So the first thing we want to do is to split this up so we can write this as a improper interval.

01:30 And since we have two pieces that are going to be improper, what we're going to want to do is to break it up into the following two intervals.

01:39 So first we'll write negative infinity, and we can choose any number that we want.

01:45 I'm just going to choose zero.

01:50 X over 1 plus x squared dx plus 0 to infinity of x over 1 plus x squared dx, and that's just by using the additivity over domains for integrals and then we can write each of these individual intervals as an improper integral so this is going to be the limit as so the limit as the approach is infinity of the interval of negative b or so actually maybe what i should do instead is say a approaches negative infinity or a to zero x over 1 plus x squared squared dx plus the limit as b approaches infinity of 0 to 0 to 0 to infinity of 0 to infinity integrated and then we have 1 plus x squared squared dx so we have both of these so since they're the same integral let's go ahead and just do one off onto the side and then we'll get the answer for both so the integral of just x over 1 plus x squared squared dx so looking at this you might think we want to do a substitution, and the thing that you might want to choose to be your u is the 1 plus x squared, or 1 plus x squared.

04:23 So i'm going to say let u equal 1 plus x squared, because when i take the derivative of this with respect to x, then find the differential we end up with du is equal to 2x dx.

04:41 And we have an x here.

04:46 And a dx here, so that's that part.

04:49 So if we just divide by 2, we end up with d u over 2 is equal to x dx.

04:58 So let's go ahead and make our change of variables.

05:03 So x dx is going to become d u over 2.

05:08 So d u over 2, and then 1 over is going to be something squared.

05:18 And the thing that we're squaring is going to be u because 1 plus x squared is our u and this here we can rewrite u to the negative second power and i'm also going to just pull this one -half out front of the integral and d now integrating this as power rules well one -half you now to the negative first power divided by its new power plus some constant c and then plugging in our u and then re -recipricating this here we'll end up with negative one -half let me go ahead and write what u is going to be first so it'll be two one plus x squared over 1 plus c so this is what this integral here ends up becoming and so we don't need to write the c here since when we end up taking the bounds and subtracting it the constants would just cancel out so we don't need to write that again so we have the limit as a approaches negative infinity of one half, one plus x squared, evaluated from a to zero, plus the integral of the limit as b approaches infinity of one half times one plus x squared.

07:16 And then write that a little bit better.

07:19 So 1 plus x squared and move that 1 over...

07:28 Not stretch it.

07:30 We just write the 1 again.

07:32 Over 1...

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