The region R1 is bounded by the graph of y=tanx and the x -axis on the interval [0, π/ 3] . The

The region R1 is bounded by the graph of y=tanx and the x...

Key Concepts

Definite Integration

Definite integration is a fundamental concept in calculus used to compute the net area under a curve between two specified bounds. It relies on finding an antiderivative of the function and evaluating it at the limits of integration, as stated in the Fundamental Theorem of Calculus. This process allows one to quantify the accumulation of quantities, such as area, over an interval.

Area Under a Curve

Calculating the area under a curve is a common application of integration. When a function is integrated over an interval, the result represents the total area between the curve and the x-axis, considering regions above the axis as positive and those below as negative. This technique is essential in problems where understanding the space covered by a graph is necessary.

Trigonometric Functions

Trigonometric functions, such as tangent and secant, are periodic functions that arise frequently in mathematics, physics, and engineering. Their properties, including their behavior and asymptotic characteristics over specific intervals, play an important role in both theoretical and applied problems.

Integration Techniques for Trigonometric Functions

Integrating functions like tan x and sec x often requires special techniques and knowledge of standard integral formulas. The antiderivative of tan x involves the natural logarithm of the cosine function, while the antiderivative of sec x also typically produces logarithmic expressions after manipulation. Mastery of these techniques is crucial for effectively solving problems that involve areas bounded by these curves.

Comparison of Areas

The process of comparing areas bounded by different functions involves computing the definite integrals for each region and then evaluating which yields a larger numerical result. This comparison can reveal insights about the relative behaviors of different functions over distinct intervals, and it is a practical application of both integration and the properties of the functions involved.

Transcript

00:02 So for the region, under the graph of a certain function, between that and the x -axis, what you actually do is you take the integral, as you may or may not have heard, the integral for the given bounds, so the interval specified 0 to pi over 3 in this case, of the function, but specifically the absolute value of the function.

00:28 If you're just looking at specifically the area, because if we're looking at the area, then what goes under the curve should rather add instead of subtracting from the result.

00:41 So what you need to do then is be able to split this integral into the parts that are positive and the parts that are negative.

00:51 But the thing is, as you may notice, and you should already know how to check this, but basically for the interval from 0 to pi over 3, tan x is, in fact, always positive, and secant x is always positive as well.

01:23 And you can check this using various ways, such as the unit circle or just looking at the graphs.

01:29 But that makes things easier, basically, because then the absolute values can just be removed if they're positive so we'll start with r1 our 1 font r1 is integral from 0 to power 3 of tan x dx and for that you'll need the integral of tan x and by use substitution that would be equal to it would be equal to one of absolute value of c x plus c so basically r1 is going to be equal to you're going to sub the two values the two bounds in.

02:16 You can get long absolute value of secant.

02:18 Sorry, there should have another absolute value on there.

02:21 Sequent pi over three minus long absolute value, secant zero.

02:32 And before i, i'm just going to do the same thing for the next part, the other region.

02:41 And r2 is going to therefore be just equal to the integral of just function since it's always positive anyway and this is a known integral that you may have been asked to prove before so i won't show that here but the integral is equal to the it's equal to the law of absolute value of secant x plus tan x so here i'm just going to draw a line to split down split up the work basically r2 is going to be equal to one has the value secant pi 3.

03:31 So yeah, we're going to first sub in pi over 3 because that's the upper bound plus 10 by over 3 and then subtract and we're going to sub the lower bound 0.

03:43 So we have minus the lineups value so you can 0 plus 10 0.

03:49 Now for each of these regions we're just going to have to sub the different values in.

03:55 Really what we just need to know, we can just write them all down...

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