6.B The Definite Integral

Introduction

Here is some code that approximates the area under \(f(x)=x^2\) on the interval \([a,b]\) using right hand endpoints on 100 subintervals. (How would you change this code to use left hand endpoints instead?)

f=makeFun(x^2 ~ x)
a = 0
b = 3
num = 100
### the approximation of f(x) on interval [a,b]
base = (b-a)/num
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 9.13545

Activities

Drive My Car

A car is driving at a rate of \(v(t)=-t+5\). Here is a plot of the velocity.

  1. Use RStudio to approximate these definite integrals using 1000 subintervals.
  1. \(\displaystyle{\int_0^5 v(x)dx}\)
  2. \(\displaystyle{\int_0^8 v(x)dx}\)
  3. \(\displaystyle{\int_0^{12} v(x)dx}\)

  1. When does the car turn around?

  2. When does the car return to where it started?

Accelerate My Car

A car is traveling at 55 feet per second at time \(t=0\). The plot below shows the car’s acceleration on the interval \([0,18]\).

  1. Is the velocity increasing or decreasing at \(t=2\)? \(t=15\)?
  2. What happens when \(t=6\)?
  3. When else (besides \(t=0\)) is the velocity 55 feet per second?
  4. When is velocity at a global maximum?
  5. When is velocity at a global minimum?

Making Cupcakes

The marginal cost of making cupcakes is given by \[K'(q)=\frac{1}{100}q^2+\frac{1}{5}q+1,\] where \(q\) is in dozens of cupcakes.

  1. Set up a definite integral to calculate \(K(30)-K(0)\), which is the change in costs when we create 30 dozen cupcakes.
  2. Estimate this definite integral using RStudio.
  3. If our fixed costs are $1500, what is the total cost to make 30 dozen cupcakes?

Area Between Curves

Below is a graph of both \(\sin(x)\) and \(\cos(x)\) on the interval \([0,2\pi]\).

slice_plot(sin(x)~x, domain(x=0:2*pi)) %>% 
  slice_plot(cos(x)~x) +
  scale_x_continuous(breaks=seq(0,2*pi,pi/4),labels=c("0", "\u03c0/4", "\u03c0/2", "3\u03c0/4", "\u03c0","5\u03c0/4", "3\u03c0/2", "7\u03c0/2", "2\u03c0"))

  1. Write a sum of integrals which represents the enclosed by the two curves.
    Hint: \(\sin(x) = \cos(x)\) at \(x=\pi/4\) and \(x=5\pi/4\).
  2. Use RStudio to estimate the area between these two curves.

Solutions

Drive My Car

  1. We approximate these definite integrals using 1000 subintervals.
  1. \(\displaystyle{\int_0^5 v(x)dx}\)
f=makeFun(-t+5 ~ t)
a = 0
b = 5
num = 1000
### the approximation of f(x) on interval [a,b]
base = (b-a)/num
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 12.4875
  1. \(\displaystyle{\int_0^8 v(x)dx}\)
f=makeFun(-t+5 ~ t)
a = 0
b = 8
num = 1000
### the approximation of f(x) on interval [a,b]
base = (b-a)/num
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 7.968
  1. \(\displaystyle{\int_0^{12} v(x)dx}\)
f=makeFun(-t+5 ~ t)
a = 0
b = 12
num = 1000
### the approximation of f(x) on interval [a,b]
base = (b-a)/num
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] -12.072

  1. The car turns around at \(t=5\). We have \(v(t) > 0\) for \(t < 5\) and \(v(t) < 0\) for \(t > 5\).

  2. We must look at the signed area. When is the (positive) area above the \(x\)-axis equal to the (negative) area below the \(x\)-axis? This happens at \(t=12\). So this is the time that the car returns to where it started.

Accelerate My Car

See the class slides for a visual explanation of this problem. Here is a descriptive solution.

  1. The car’s velocity is increasing at \(t=2\) because acceleration is positive. The car’s velocity is decreasing at \(t=15\) because acceleration is negative

  2. At \(t=6\), the acceleration is zero. We accelerate up to this point, and we decelerate after this point. So it is a local maximum for velocity.

  3. The velocity is 55 feet per second at \(t=12\) because the area above the \(x\)-axis (positive acceleration so far) equals the area below the \(x\)-axis (negative acceleration so far).

  4. The maximum velocity occurs at \(t=6\)? The car accelerates for the first 6 seconds. Then it decelerates for the next 10 seconds (until \(t=16\)). Then it accelerates again for 2 seconds. However by \(t=18\), the total deceleration (area below the \(x\)-axis) is more than the total acceleration.

  5. The minimum velocity occurs at \(t=12\). This is when we’ve seen the most deceleration, and we’ve decelerated more than we’ve accelerated.

Making Cupcakes

The marginal cost of making cupcakes is given by \[K'(q)=\frac{1}{100}q^2+\frac{1}{5}q+1,\] where \(q\) is in dozens of cupcakes.

  1. By the Fundamental Theorem of Calculus, the definite integral that calculates \(K(30)-K(0)\) is

\[\int_0^{30} K'(q) dq = \int_0^{30} \left( \frac{1}{100}q^2+\frac{1}{5}q+1 \right) dq \]

  1. Here is our RStudio estimate.
f=makeFun(1/100 * q^2 + 1/5*q + 1 ~ q)
a = 0
b = 30
num = 100000
### the approximation of f(x) on interval [a,b]
base = (b-a)/num
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 210.0023
  1. The total costs are approximate \(1500+210\), which is the sum of our fixed costs and our production costs.

Area Between Curves

  1. The sine curve and the cosine curve intersect when \(x=\pi/4\) and \(x=5\pi/4\). The cosine curve is above the sine curve on \([0,\pi/4]\). Then the sine curve is on top on \([\pi/4,5\pi/4]\). Then cosine is back on top on \([5\pi/4, 2\pi\). So the area is \[\int_{\pi/4}^{5\pi/4} ( \cos(x) - \sin(x)) dx + \int_{\pi/4}^{5\pi/4} ( \sin(x) - \cos(x)) dx + \int_{5\pi/4}^{2\pi} ( \cos(x) - \sin(x)) dx.\]

  2. Our RStudio estimates for each part are

f=makeFun(cos(x) - sin(x) ~ x)
a = 0
b = pi /4
num = 100000
### the approximation of f(x) on interval [a,b]
base = (b-a)/num
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 0.4142096
f=makeFun(sin(x) - cos(x) ~ x)
a = pi/4
b = 5* pi /4
num = 100000
### the approximation of f(x) on interval [a,b]
base = (b-a)/num
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 2.828427
f=makeFun(cos(x) - sin(x) ~ x)
a = 5*pi/4
b = 2*pi 
num = 100000
### the approximation of f(x) on interval [a,b]
base = (b-a)/num
points = seq(from=a+base, to=b, by=base)
heights = f(points)
areas = base*heights
sum(areas)
## [1] 2.414225

So our estimate of the area is

0.414 + 2.828 + 2.414
## [1] 5.656