Area functions for the same linear function Let f(t)=2 t-2 and consider the two area functions A

Name: Problem 18, Chapter 5: Calculus: Early Transcendentals
Uploaded: 2020-06-24T13:15:34-08:00
Duration: 10 min 45 s
Channel: Bob Monday
Description: Area functions for the same linear function Let f(t)=2 t-2 and consider the two area functions A(x)=∫1^x f(t) d t and F(x)=∫4^x f(t) d t a. Evaluate A(2) and A(3) . Then use geometry to find an expression for A(x), for x ≥1 b. Evaluate F(5) and F(6) . Then use geometry to find an expression for F(x), for x ≥4 c. Show that A(x)-F(x) is a constant and that A^'(x)=F^'(x)=f(x)

Area functions for the same linear function Let f(t)=2 t-2...

Key Concepts

Geometric Interpretation of Integrals

The geometric interpretation of integrals involves visualizing the definite integral as the area under a curve, which can be computed using geometric formulas for simple shapes like triangles and rectangles when the function's graph is linear or can be approximated by straight lines. This approach provides an intuitive means to evaluate integrals by breaking them down into familiar geometric regions, especially when dealing with linear functions.

Definite Integration

Definite integration is a process in calculus that calculates the signed area under the curve of a function between two specified limits. It provides an accumulation of values from the lower limit to the upper limit, considering regions above the x?axis as positive and below the x?axis as negative. This concept allows us to compute areas and solve problems related to accumulation, such as differences between area functions defined on different intervals.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration by stating that if a function is continuous on an interval, then its definite integral over that interval can be evaluated using its antiderivative. Furthermore, the derivative of a function defined by an integral with a variable upper limit equals the original integrand. This theorem underpins the relationship between area functions and rate of change.

Antiderivatives and Constant Differences

Antiderivatives are functions whose derivatives yield the original function. In the context of accumulation or area functions, different choices for the lower limit of integration yield antiderivatives that differ by a constant. This property shows that while integration can be performed with various basepoints, the overall behavior of the resulting functions in terms of their derivatives remains consistent, and any difference between them is a fixed constant.

Transcript

00:01 Okay, so we are given two area functions for the same line, 2t minus t.

00:06 The first one goes from 1 to x, the second one goes from 4 to x, and we were asked to find some information for these area formulas.

00:14 The first one we're asked to find is a of 2.

00:17 So this is the integral from 1 to 2 of 2t minus 2dt.

00:24 Using the fundamental theorem of calculus, we can find the antiderivative of 2t minus t to get t squared minus 2t.

00:32 2t as that is the anti -derivative and then i will simply evaluate from 1 to 2 by substituting in 2 first and i get 2 squared minus 2 times 2 and then i am going to subtract and then i'm going to substitute in 1 so i get 1 squared minus 2 times 1 this leaves me with 4 minus 4 minus 1 minus 2 4 minus 2 4 minus 4 is obviously 0 1 minus 2 is negative 1, and the negative there gives me positive 1.

01:09 Likewise, i can find the area of, or i can substitute 3 into the area function, and that gives me the area from 1 to 3, using the same anti -derivative as the previous problem.

01:23 We still have 2 squared minus 2t from 1 to 3, and then as last time i will substitute in 3, get 3 squared, minus 2 times 3, and then i also substitute in 1.

01:42 So that gives me 9 minus 6, and then again, this is going to be minus 1 with a negative, so really it's just plus 1, and then 9 minus 6 is 3 plus 1 is 4.

01:56 Okay, from here we are now asked to use geometry to find an expression for a of x.

02:02 So obviously we could use the fundamental theme of calculus here, but we need to use geometry.

02:07 So let's draw a picture of this.

02:08 What does this look like? okay, well, the line 2 -t -minus 2 is just a line with the y -erset with the negative 2 and a slope of 2.

02:17 But i'm starting at 1 here.

02:19 So we'll start at 1.

02:21 And we see that this is at 0, or the y -value is 0 when t is 1.

02:28 And then i can use the slope to continue this.

02:31 So if i go to the right 1, i go up 2.

02:35 And out there, i go to the right 1, and now up 2.

02:41 And so this is the line that we are looking at, where this is 2 and 4.

02:50 Okay.

02:52 So we need to find a formula for the area underneath this curve from 1 to any x value.

02:59 So again, i guess i could draw the line this way as well, but our base is here.

03:02 This is where we're starting.

03:04 So the idea is we imagine some x, so we have some x value here.

03:10 This is, again, 1, and we need to find the area of this triangle.

03:16 Okay.

03:19 Let me redraw that.

03:23 So we'll pick this as our x value.

03:25 There we go.

03:26 Don't want to confuse it with being the x value of 2.

03:32 Okay.

03:32 And so then the idea is that we need to find the area of this triangle right here.

03:38 But we know the area of a triangle.

03:40 The area of a triangle is base times height.

03:43 Okay.

03:43 Or one half base times height.

03:45 My apologies.

03:47 So we have one half.

03:48 Now the base of this triangle right here is just the distance from x to 1.

03:54 In this case, x minus 1.

03:56 The height of the triangle is just whatever the y value is when i substitute an x into our on to the equation for our line.

04:07 In this case, that's 2x minus 2.

04:12 And if you were to simplify this out, you would get x squared minus 2x plus 1.

04:25 Okay, and then, so this is just again purely geometric that i just found the area of the triangle to get this formula.

04:33 Now again, we can do a similar thing with f, where we can find, let's write it down here, we have f of x is equal to the integral from 4 to x of 2t minus 2dt.

04:47 Okay, and we can also, we can substitute some values into this, and so we find that f of, in this case 5, is just the area from 4 to 5.

04:59 Of 2t minus 2 d t and again i'm just going to use the fundamental theorem of calculus here since we already have the entire derivative from earlier this time though instead of going at 1 is the starting point we're starting at 4 there going to 5 and we just substitute those in so we get 5 squared minus 2 times 5 minus now we substitute in the 4 squared minus 2 times 4 so we get 25 minus 10 minus 16 plus 8...

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