AP Calculus BC Unit 8 Applications of Integration

Unit 8: Applications of Integration ● Average value: ○ Used to find the average of all the instantaneous values of a function ○ ■ Use the formula above to calculate average value. ○ Example: find the average value of f(x) = 𝑥 3 from x = 1 to x = 3 ■ All you do is plug in the values into the formula → 3 1 ∫ 3−1 1 𝑥 3 𝑑𝑥 ■ Then solve for the integral → (½)[(3)4 /4 - (1)4 /4]= 10 ● Connecting position, velocity, and acceleration ○ The integral of acceleration represents an object’s velocity over an interval of time ○ The integral of velocity represents an object’s displacement over an interval of time. ○ The integral of speed represents the object’s total distance travelled in that time interval ● Applying integrals into context ○ Remember that the integral of a function represents the accumulation of the rate of change (AKA net change) of that function over the time interval ● Area between two curves as functions of x ■ 1) To find the area between the curves of two different functions in respect to x, it’s easiest to sketch the two curves if possible. This just makes it easier to understand how to approach the problem. ■ 2) Then, find the points of intersection between the curves, and this will determine the limits of your integral. ● This can be done by setting the two equations equal to each other, since both equal ‘y’, and thus, they must be equal to each other → 5x - 𝑥 2 =x, and then solve for x → 4x = 𝑥 2 → x=0, x=4 ○ Thus the limits of this integral are from x=0 to x=4 𝑏 ■ 3) set up the integral → ∫𝑎 [𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑡ℎ𝑎𝑡′𝑠 "𝑜𝑛 𝑡𝑜𝑝"] − [𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑡ℎ𝑎𝑡′𝑠 "𝑏𝑒𝑙𝑜𝑤"]𝑑𝑥 ● In this example, y = 5x-𝑥 2 is the function that’s “on top”, and y = x bounds the area at the bottom of the shaded area 4 ○ ∫0 4 5𝑥 − 𝑥 2 − 𝑥 𝑑𝑥= ∫0 4𝑥 − 𝑥 2 𝑑𝑥 ■ 4) integrate, and solve for the area. ● Area between curves as a function of y ○ If the equation of the functions are given in terms of y, then the same steps are used to find the area between the curves, with some changes ■ The limits of the integral will change if the functions are in terms of y. Rather than setting the limits with x values, the limits will be set using y values. ○ As you can see in the picture above, the functions are in terms of y. The limits of the integral would be from y = -2 to y = 4, because that’s where the two curves intersect. ■ Also, when writing the integral, rather than being in terms of 𝑏 ∫𝑎 𝑏 𝑓(𝑥) 𝑑𝑥, it’ll now be written as: ∫𝑎 𝑓(𝑦) 𝑑𝑦, and to determine which function is written first in the formula, rather than the function “on top” going first, you now determine the function that’s “to the right” ● In this example, x= y + 1 is the function that’s binding the right side of the area between the curves, so the integral would be set up 4 like this: ∫−2 1 4 𝑦 + 1 − (2 𝑦 2 − 3)𝑑𝑦= ∫−2 1 − 2 𝑦 2 + 𝑦 + 4 𝑑𝑦 ● Area between curves with more than 2 intersection points. ○ If there are more than 2 intersection points between two curves, then multiple integrals will need to be set up. ■ In the picture above, the curves have 3 intersection points. The blue curve (2𝑥 3 − 𝑥 2 − 5𝑥)is the curve on top from x = -2 to x = 0, but then it becomes the curve on the bottom from x = 0 to x = 2. Thus, 2 integrals need to be set up and added to each other to account for the change: ● 0 ∫−2 2 2𝑥 3 − 𝑥 2 − 5𝑥 − (𝑥 2 + 3𝑥) 𝑑𝑥+ ∫0 𝑥 2 + 3𝑥 − (2𝑥 3 − 𝑥 2 − 5𝑥) 𝑑𝑥 ○ There needs to be two integrals because the function that’s “on the top” changes at x = 0, and thus, the integrals have to change, or else the wrong area would be found. ● Volumes with cross sections: ○ The picture above demonstrates what is meant by a “cross-section”. Basically, finding the volumes of solids using cross sections is done by adding up numerous rectangles (or whatever shape the problem asks for) together. ○ Square cross sections: 𝑏 ■ ∫𝑎 (𝑓(𝑥))2 𝑑𝑥 ● The function is squared because the height of the volume is f(x), since squares have the same length for all 4 sides. Integrating this allows you to “add up” all of the square cross sections within the given interval. ■ For the problem above, the integral would be set up like this: 6 ∫4 6 (√𝑥)2 𝑑𝑥= ∫4 𝑥 𝑑𝑥. After setting up the integral, integrate to solve for the volume. ■ *note that in this example, it explicitly said that the cross sections were perpendicular to the x axis. Thus, the function was written in terms of x. However, there are problems that ask for the volume of a solid with cross sections that are perpendicular to the y axis, in which case you’d need to write the function in terms of y. ○ Rectangular cross sections: 𝑏 ■ ∫𝑎 𝑓(𝑥)ℎ 𝑑𝑥 ● The only difference for a rectangular cross section is that because the height of the solid isn’t the same as its base, you can’t square the function. Instead, the height will be given, so you just multiply the height of the solid to the function. ■ For ex., if the problem above asked for rectangular cross sections, and said that the height of the solid was h=4, the integral would be set up like this: 6 ∫4 4√𝑥 𝑑𝑥, and then you can integrate as you normally would. ○ *If the problem gives you two curves, and asks you to find the volume of a solid who’s base is the area between the two curves, the integral would be set up like: 𝑏 ∫𝑎 (𝑓(𝑥) − 𝑔(𝑥))ℎ 𝑑𝑥 ■ Just like finding the area between two curves, f(x) is the curve that’s “on top”, or if it’s in terms of y, “to the right” ● Disk Method: ○ This method is used to find the volume of solids whose cross sections are circular disks. ○ If you take the curve y=𝑒 𝑥 and revolve it around the x-axis, then you can find the volume of the generated solid by adding up the volumes of all of the disks. ■ 1) Each disk has an area of 𝜋𝑟2. To find the area of the disks, you can plug the equation of the function into ‘r’ → 𝜋(𝑒𝑥 )2. The volume of the disk is equal to the area of the circle times the thickness of each disk, which is ‘dx’ → volume of each disk = 𝜋(𝑒𝑥 )2 dx ■ 2) Then, to add up all of the volumes of each disk, integrate this equation using the given limits. In this case, the graph shows you that 3 the limits of the solid are from x=0 to x=3 → ∫0 𝜋(𝑒 𝑥 )2 𝑑𝑥. ■ 3) Integrate the function to find the volume of the solid. ○ You can also use the Disk Method for a function that’s being revolved around the y-axis. The only difference is that the function being integrated should be in terms of y. ● Washer’s Method: ○ This is used to find the volume of a solid whose cross sections are ring-shaped (look at image below) ○ The formula for Washer’s Method is in the picture above. ■ R(x) represents the equation of the curve that accounts for the outside of the ring, and r(x) represents the inner circle of the ring. ● Arc Length / Length of a Curve: ○ If the function is continuous and differentiable on [a,b], then the formula above can be used to find arc length. ○ Ex. (taken from Princeton book): 1: Find the length of the curve 𝑦 = 𝑥 3/2 from x = 0 to x = 4. 3 ■ 1) find f’(x) → f’(x) = 2 𝑥 1/2 4 ■ 2) plug in the values into the formula → ∫0 4 ■ 3) integrate → ∫0 3 √1 + ( 𝑥 1/2 )2 𝑑𝑥 2 9 √1 + 𝑥 𝑑𝑥 (use u-substitution to finish integrating) 4 ○ Arc length of a planar curve can give you the distance travelled by an object.