6.C The Indefinite Integral

Activities

Indefinite Integral Practice

Find the indefinite integral.

  1. \(\displaystyle{\int 3x^4\,dx}\)
  2. \(\displaystyle{\int r^5-\frac{7}{r^3}+r^{3/4}\,dr}\)
  3. \(\displaystyle{\int 8t^2-t^{13}\,dt}\)
  4. \(\displaystyle{\int 5e^{4x}\,dx}\)
  5. \(\displaystyle{\int(x+2)^2\,dx}\)
  6. \(\displaystyle{\int 4\sqrt{x}-9x+e\,dx}\)
  7. \(\displaystyle{\int 3^t-7t^4 +\frac{9}{t}\,dt}\)
  8. \(\displaystyle{\int \cos(x)+2e^x +7\,dx}\)
  9. \(\displaystyle{\int \sin(3x)-2\cos(x)dx}\)
  10. \(\displaystyle{\int 7e^{-0.02t}+\frac{4}{t}\,dt}\)

Definite Integral Practice

Evaluate the definite integral using antiderivates.

  1. \(\displaystyle{\int_{-2}^5 x^2 \,dx}\)
  2. \(\displaystyle{\int_0^{\pi/4} \sin(x) + e^{x}\,dx}\)
  3. \(\displaystyle{\int_1^e 4t + \frac{1}{4t} \,dt}\)

Area Between Curves

Find the area enclosed by the curves \(f(x)=5-x^2\) and \(g(x)=x^2-3\) by following the steps below

  1. Find the points of intersection of \(f(x)\) and \(g(x)\).
  2. Which curve is above the other?
  3. Set up the definite integral.
  4. Evaluate (symbolically or with RStudio)

Average Value

The average value of a function \(f(x)\) on interval \([a,b]\) is given by \[ \frac{1}{b-a} \int_a^b f(x) \, dx.\]

Find the average value of function on the given interval.

  1. The average value of \(\sin(x)\) on \([0, \pi]\).
  2. The average value of \(e^{2x}\) on \([0,5]\).
  3. The average value of \(x^2 + 3\) on \([4,6]\)

Modeling Population Growth

The second phase of bacteria growth is called the log phase. The rate of population growth is proportional to the number of bacteria at that time with proportionality constant 0.003. In other words

\[ \frac{dP}{dt} = 0.003 P \]

This is equivalent to the integration formula \[ \int \frac{1}{P} \, dP = \int 0.003 \, dt. \]

  1. Integrate both sides of this equation and find a general equation for \(P\).
  2. Suppose there are 1000 bacteria initially. Find the specific equation for \(P\).

Zombie Apocalypse

The rate \(r(t)\) at which people become zombies during the inevitable zombie apocalypse can be approximated by \[ r(t)=1000 e^{-0.5t}\] where \(t\) is the number of days since the start of the outbreak.

  1. Write the number of infected people \(I(T)\) as an indefinite integral.
  2. How many total people get infected by time \(T\) where \(T\) is a very large number?

Solutions

Indefinite Integral Practice

Find the indefinite integral.

  1. \(\frac{3}{5}x^5 + C\)
  2. \(\frac{1}{6}r^6 + \frac{7}{2}r^{-2} + \frac{4}{7}r^{7/4}+C\)
  3. \(\frac{8}{3}t^3 - \frac{1}{14}t^{14} + C\)
  4. \(\frac{5}{4} e^{4x}+C\)
  5. \(\frac{1}{3}x^3 +2x^2 + 4x + C\)
  6. \(\frac{8}{3} x^{3/2} - \frac{9}{2}x^2 + ex + C\)
  7. \(\frac{1}{\ln 3}3^t - \frac{7}{5}t^5 + 9 \ln t + C\)
  8. \(\sin(x) + 2e^{x} + 7x + C\)
  9. \(-\frac{1}{3} \cos(3x) - 2\sin(x) + C\)
  10. \(-350 e^{-0.02t} + 4\ln t + C\)

Definite Integral Practice

We solve each of these using the Fundamental Theorem of Calculus. If \(F'(x)=f(x)\) then \[ \int_a^b f(x) dx = F(b) - F(a). \]

  1. We have \[\begin{align} \int_{-2}^5 x^2 \,dx &=& \left. \frac{1}{3}x^3 \right|_{-2}^5 = \frac{125}{3} + \frac{8}{3} = \frac{133}{3}. \end{align}\]

  2. We have \[\begin{align} \int_0^{\pi/4} \sin(x) + e^{x}\,dx &= \left. -\cos(x) + e^x \right|_0^{\pi/4} = -\cos(\pi/4) + e^{\pi/4} - (-\cos(0) + e^0) \\ &= -\frac{\sqrt{2}}{2} + e^{\pi/4} - (-1+1) = -\frac{\sqrt{2}}{2} + e^{\pi/4}. \end{align}\]

  3. We have \[\begin{align} \int_1^e 4t + \frac{1}{4t} \,dt &=& \left. 2t^2 + \frac{1}{4} \ln(t) \right|_1^e = 2e^2 + \frac{1}{4} \ln(e^2) - \left( 2e^0 + \frac{1}{4} \ln(1) \right) \\ &=& 2e^2 + \frac{1}{2} - 2 - 0 = 2e^2 + \frac{3}{2} \end{align}\]

Area Between Curves

Find the area enclosed by the curves \(f(x)=5-x^2\) and \(g(x)=x^2-3\) by following the steps below

  1. Setting \(f(x)=g(x)\) gives \(5-x^2=x^2-3\) which means that \(x^2=4\). So the curves intersect at \(x=\pm 2\).
  2. On the interval \([-2, 2]\), function \(f(x)\) is above function \(g(x)\).
  3. The area between these curves is \[ \int_{-2}^{2} ( f(x) -g(x)) dx = \int_{-2}^{2} ( 8 -2x^2) dx \]
  4. Let’s use RStudio. It’s easier!
f=makeFun(8 - 2*x^2 ~ x)
a = -2 
b = 2 
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] 21.33333

Average Value

  1. The average value of \(\sin(x)\) on \([0, \pi]\) is \[ \frac{1}{\pi} \int_0^{\pi} \sin(x) dx = \left. - \frac{1}{\pi} \cos(x) \right|_{0}^{\pi} = - \frac{1}{\pi} (\cos(\pi) - \cos(0)) = \frac{2}{\pi} \]
  2. The average value of \(e^{2x}\) on \([0,5]\) is \[ \frac{1}{5} \int_0^{5} e^{2x} dx = \left. \frac{1}{5} \cdot \frac{1}{2} e^{2x} \right|_{0}^{5} = \frac{1}{10} (e^{10} - 1) \]
  3. The average value of \(x^2 + 3\) on \([4,6]\) is

\[ \frac{1}{6-4} \int_4^{6} (x^2 + 3) dx = \left. \frac{1}{2} \left( \frac{1}{3} x^3 + 3x \right) \right|_{4}^{6} = \frac{1}{2} \left(\frac{6^3}{3} + 18 - \frac{4^3}{3} - 12 \right)= 28.33 \]

Modeling Population Growth

The second phase of bacteria growth is called the . The rate of population growth is proportional to the number of bacteria at that time with proportionality constant 0.003. In other words

\[ \frac{dP}{dt} = 0.003 P \]

This is equivalent to the integration formula \[ \int \frac{1}{P} \, dP = \int 0.003 \, dt. \]

  1. Integrate both sides of this equation and find a general equation for \(P\).

\[ \begin{array}{rcl} \int \frac{1}{P} \, dP &=& \int 0.003 \, dt \\ \ln(P) &=& 0.003t + C \\ P &=& K e^{0.003t} \end{array} \]

where \(K = e^C\).

  1. We know that \(P(t) = K e^{0.003t}\) and that \(P(0)=1000\). This means that \(K=1000\). So our function is \[ P(t) = 1000 e^{0.003t}. \]

Zombie Apocalypse

The rate \(r(t)\) at which people become zombies during the inevitable zombie apocalypse can be approximated by \[ r(t)=1000 e^{-0.5t}\] where \(T\) is the number of days since the start of the outbreak.

  1. The number of infected people is \[I(T) = \int_0^T r(t) \, dt = \int_0^T 1000 e^{-0.5t} \, dt\]

  2. We have \[ \int_0^T 1000 e^{-0.5t} \, dt = \left. \frac{1000}{-0.5} e^{-0.5t} \right|_0^T = - \frac{1000}{0.5} e^{-0.5 T} + \frac{1000}{0.5} e^{0} = 2000 \left( 1 - \frac{1}{e^{0.5 T}}\right). \] For very large \(T\) the second term is basically 0. So we have 2000 zombies by the end of the outbreak. Gosh, that’s not too bad!