Sketch the region bounded by the graphs of the equations and find the area of the region. f(x)=√

Sketch the region bounded by the graphs of the equations and...

Key Concepts

Graph Sketching and Function Transformations

Sketching the graphs of given functions aids in visualizing the region of interest. Understanding how functions like the square root function and linear functions are transformed and plotted helps identify which function is above the other, clarifying the setup for the integration process.

Definite Integration

Definite integration is used to compute the exact area of a region bounded by curves. After establishing the bounds from the intersection points, integration enables the calculation of the accumulated area between the graphs over this interval.

Area Between Curves

Calculating the area between two curves is central to this problem. This concept involves integrating the difference between the upper (or outer) function and the lower (or inner) function over the interval formed by the intersection points, ensuring that the area is captured correctly.

Intersection Points

Finding the intersection points of the two curves is a key step because it defines the limits of integration. This involves setting the equations equal to each other and solving for the variable, which lays the groundwork for accurately determining the region's boundaries.

Transcript

00:01 Starting with a graph of these, it would be good for students to be familiar with the parent function.

00:09 So if you were to look at f of x, which is defined as the square root of x, you know, you have that behavior, but then the plus three shifts it up three, which is exactly what i did.

00:22 And then the next piece is that we're looking at g of x.

00:26 I'll write it off to just underneath.

00:28 How about that? is the linear function of up one, right, 2, x, and it's also shifted up 3.

00:37 So it's sort of obvious to me, and i would assume it's obvious to other people as well, that one point of intersection would be at x equals 0, because they share the same y intercept.

00:51 But we don't know the other point of intersection, but you can figure that out by setting them equal to each other.

00:58 And i would hope, let's forgot to write equals, that students would right away cancel out those threes.

01:09 They subtract off.

01:11 Maybe what you would do next is square both sides.

01:16 I don't know if that would be the next step for a lot of people, but it becomes one -fourth x squared.

01:23 And from here, it would actually, i think, makes sense to multiply 4x over and also subtract the x squared over or vice versa.

01:34 It doesn't really matter because when you factor out in x, that's going to be x equals zero.

01:39 And the other point of that intersection would be at x equals four.

01:45 Because if x equals four, four minus four, it is zero, and zero times four, we'd still give me zero.

01:50 That's a zero product property to figure out your upper bound is four.

01:54 If you don't believe me, plug in four for both of these.

01:57 You know, half of four is two plus three is five, and the square to 4 is 2 plus 3 is 5, so they are equal there.

02:07 And then from there, i think it's clear from my problem that the square root of x plus 3 is the upper function, and then you have to subtract off the lower function, which is that 1 .5x plus 3 dx...

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