Area Find the area of the region bounded above by y=2 cosx and below by y=secx,-π/ 4 ≤x ≤π/ 4

Name: Problem 41, Chapter 8: Thomas Calculus
Uploaded: 2020-06-10T20:53:59-08:00
Duration: 7 min 18 s
Channel: David Ku
Description: Area Find the area of the region bounded above by y=2 cosx and below by y=secx,-π/ 4 ≤x ≤π/ 4

Area Find the area of the region bounded above by y=2 cosx...

Key Concepts

Trigonometric Functions

Trigonometric functions such as cosine, sine, and secant are fundamental to many problems in calculus, especially when they are used as representations of curves. Knowledge of their properties, graphs, and identities is essential for correctly setting up integrals, evaluating them, and understanding the behavior of the functions within the given interval.

Area Between Curves

The area between two curves is determined by integrating the difference between the top function and the bottom function over an interval. This method relies on subtracting one function from the other to obtain the vertical distance at each point, then summing these small areas using integration. Understanding this concept is crucial for solving problems involving regions bounded by different curves.

Definite Integral

A definite integral is used to calculate the accumulation of quantities, such as area under a curve, over a specified interval. In the context of finding areas between curves, the definite integral provides a framework to sum the infinitesimally small contributions of the vertical difference between two functions over a given interval.

Transcript

00:01 In this video, we will be going over how to find the area underneath two curves.

00:06 So we are given in this problem, y equals 2 cosine x and y equals ccon x.

00:11 Y equals 2 cosinex is represented by this red line here, while the blue liner is represented by y equals cconics.

00:19 Our goal in this video is trying to find this area, which would be represented by y equals 2 cosine x, and subtracted by the area underneath y equals cconx represented here, so that we are left with the area in between both the curves.

00:39 So first we have this integral of negative pi over 4 to pi before our given bounds of 2 cosine x d x.

00:51 Our first step we can do is we can take out that two out front since it's just a coefficient.

00:59 And we're left with just cosine x d x.

01:04 And we know that the integral of cosine x is just sine of x.

01:12 And we have to rewrite our bounds here.

01:15 Of negative pi over 4 to pi over 4.

01:18 And now we can plug these in so that we have 2 times sine of pi over 4, subtracted by sine of negative pi over 4.

01:32 If we know our unit circle, we know that sine pi over 4 is just root 2 over 2, subtracted by sign of negative pi over 4, which would just be the negative root 2 over 2.

01:44 And if we combine these terms, we get that 2 times 2 root over 2 is the final sum of that of the two terms.

01:56 And we can simplify this even further, where we can eliminate these two twos.

02:01 So we are left with 2 -2 over, sorry, 2 root 2 .2.

02:06 And we want to approximate this to decimals, which would equal 2 .8 -28.

02:16 Now we want to find the integral of sicken -x, using the same bounds of sickenx dx.

02:28 So right off the bat, we don't know what the integral of sikin -x is, but we can multiply it by something so that we get into a form where we can u sub and use that technique.

02:41 So what i'm going to do is keep these bounds the same, but we're going to multiply the numerator and the denominator by sicken x plus tangent x.

02:57 And we would pretend there's a one there, and we multiply both.

03:04 And now we can multiply this through so that we get something that can be u subbed.

03:13 So our product would be secan x, secan squared x, plus secan x, tangent x...

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