Q1 Use integration by substitution to solve the integral below.

\[ \int 4e^{-7x} \, dx \]

Identify the substitution

Let:

\[ u = -7x \]

Compute \(du\):

\[ du = -7dx \]

and dx:

\[ dx = -\frac{1}{7}du \]

Rewrite the integral using \(u\)

Substitute \(u\) and \(dx\):

\[ \int 4e^u \cdot -\frac{1}{7}du \]

This simplifies to:

\[ -\frac{4}{7} \int e^u \, du \]

Integrate with respect to \(u\)

The integral of \(e^u\) with respect to \(u\) is \(e^u\):

\[ -\frac{4}{7}e^u + c \]

Substitute for \(x\)

Replace \(u\) with the original function of \(x\):

\[ -\frac{4}{7}e^{-7x} + c \]

Answer

The integral of \(4e^{-7x}\) with respect to \(x\) is:

\[ F(x) = -\frac{4}{7}e^{-7x} + c \]


Q2 Contamination Rate

Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac{dN}{dt} = -\frac{3150}{t^4} - 220.\) bacteria per cubic centimeter per day, where t is the number of days since treatment began. Find a dt t4 function N( t ) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter.

We integrate the rate of change: \[ \frac{dN}{dt} = -\frac{3150}{t^4} - 220. \]

We integrate term by term:

First Term Integral:

\[ \int -\frac{3150}{t^4} \, dt = -3150 \int t^{-4} \, dt. \] Using the power rule for integration, we get: \[ -3150 \left( \frac{t^{-3}}{-3} \right) = 1050t^{-3} + c \]

Second term integral:

\[ \int -220 \, dt = -220t + c \]

Answer

We combine the results: \[ F(t) = 1050t^{-3} - 220t + c \]


Q3 Area of Shaded Region

To find the area of the yellow shaded triangular region below the line using integration, we need to set up the integral with the given function and limits:

To find the area, we integrate the function from \(x = 4\) to \(x = 8\).

\[ \text{Area} = \int_{4}^{8} f(x) \, dx = \int_{4}^{8} (2x - 9) \, dx \]

Evaluate the Integral:

\[ \int (2x - 9) \, dx = x^2 - 9x + C \]

Substitute the Limits:

\[ \text{Area} = \left[ (8)^2 - 9 \times 8 + C \right] - \left[ (4)^2 - 9 \times 4 + C \right] \] \[ = (64 - 72 + C) - (16 - 36 + C) \] \[ = -8 - (-20) \] \[ = 12 \text{ square units} \]

Answer

The area of the yellow shaded triangular region below the line \(f(x) = 2x - 9\), between \(x = 4\) and \(x = 8\), is 12 square units.


Q4 Area of Bounded Area

Plot the functions
# Create a sequence of x values
x <- seq(-10, 10, by = 0.1)

# Calculate y values
y1_values <- x^2 - 2 * x - 2
y2_values <- x + 2

# Set limits for the plot
x_lim <- range(x)
y_lim <- c(-10, 10) 

# Plot the first function
plot(x, y1_values, type = 'l', col = 'red', ylim = y_lim, xlim = x_lim, xlab = '', ylab = '', main = 'Graphs of y = x^2 - 2x - 2 and y = x + 2', axes = FALSE)

# Add the second function to the plot
lines(x, y2_values, type = 'l', col = 'blue')

# Add axes at the origin
axis(side = 1, at = seq(floor(x_lim[1]), ceiling(x_lim[2]), by = 1), pos = 0)
axis(side = 2, at = seq(y_lim[1], y_lim[2], by = 1), pos = 0)

# Add legend
legend('bottomright', legend = c('y = x^2 - 2x - 2', 'y = x + 2'), col = c('red', 'blue'), lty = 1)

Calculate the Intersection Points

The intersection point are the limits of the integration. For this calculation, I used the uniroot function which identifies the roots by determining where the difference between the functions is zero over a given interval.

# Equations
f1 <- function(x) { x^2 - 2 * x - 2 }
f2 <- function(x) { x + 2 }

# Find the intersections by solving f1(x) - f2(x) = 0
solved_function <- function(x) { f1(x) - f2(x) }

# Find the first intersection point
root1 <- round(uniroot(solved_function, interval = c(-10, 0))$root)

# Find the second intersection point
root2 <- round(uniroot(solved_function, interval = c(0, 10))$root)

# Return the rounded roots
c(root1, root2)
## [1] -1  4
Setup the Integral

To evaluate the integral between the curves for the functions \(y = x^2 - 2x - 2\) and \(y = x + 2\) between the intersection points \(x = -1\) and \(x = 4\), we compute the integral. This represents the area between the two curves from \(x = -1\) to \(x = 4\).

\[ \int_{-1}^{4} ((x + 2) - (x^2 - 2x - 2)) \, dx \]

Calculate the Area (Compute the Integral)

We evaluate the integral of the difference between the functions \(y = x^2 - 2x - 2\) and \(y = x + 2\) from \(x = -1\) to \(x = 4\):

\[ \int_{-1}^{4} [(x + 2) - (x^2 - 2x - 2)] \, dx \]

This simplifies to:

\[ \int_{-1}^{4} [-x^2 + 3x + 4] \, dx \]

To compute this integral, we’ll find the antiderivative

\[ \int [-x^2 + 3x + 4] \, dx = \left[-\frac{1}{3}x^3 + \frac{3}{2}x^2 + 4x\right] \]

Evaluate the antiderivative from \(x = -1\) to \(x = 4\):

\[ \left[-\frac{1}{3}(4)^3 + \frac{3}{2}(4)^2 + 4(4)\right] - \left[-\frac{1}{3}(-1)^3 + \frac{3}{2}(-1)^2 + 4(-1)\right] \]

After performing the calculations for each term:

\[ \left[-\frac{1}{3}(64) + 24 + 16\right] - \left[\frac{1}{3} + 1.5 - 4\right] \]

This simplifies to:

\[ \left[-\frac{64}{3} + 24 + 16\right] + \left[-\frac{1}{3} - 1.5 + 4\right] \]

Combine like terms:

\[ \left[-\frac{64}{3} + 24 + 16\right] + \left[-\frac{1}{3} - 1.5 + 4\right] \]

\[ = -\frac{64}{3} + 40 - \frac{1}{3} - 1.5 + 4 \]

\[ = -\frac{65}{3} + 42.5 \]

\[ = -21.6667 + 42.5 = 20.8333 \]

Answer

The area between the curves from \(x = -1\) to \(x = 4\) is \(20.8333\) square units. We confirm using pracma R module.

library(pracma)

# Define the  function
integrand <- function(x) { (x + 2) - (x^2 - 2*x - 2) }

# Calculate the integral from x = -1 to x = 4
area <- integral(integrand, -1, 4)
area
## [1] 20.83333


Q6 Use integration by parts to solve the integral below.

\[ \int \ln(9x) \cdot x^6 \, dx \]

Identify \(u\) and \(dv\)

Let:

\[ u = \ln(9x) \]

and

\[ dv = x^6 \, dx \]

Compute \(du\):

\[ du = \frac{1}{x} \, dx \]

and \(v\) (the integral of \(dv\)):

\[ v = \frac{x^7}{7} \]

Apply the integration by parts formula

\[ \int u \, dv = uv - \int v \, du \]

Substitute \(u\), \(du\), and \(v\) into the formula:

\[ \int \ln(9x) \cdot x^6 \, dx = \ln(9x) \cdot \frac{x^7}{7} - \int \frac{x^7}{7} \cdot \frac{1}{x} \, dx \]

Simplify the integral

Simplify the integral on the right:

\[ \int \frac{x^7}{7} \cdot \frac{1}{x} \, dx = \frac{1}{7} \int x^6 \, dx = \frac{x^7}{49} \]

Integrate and combine terms

\[ \int \ln(9x) \cdot x^6 \, dx = \ln(9x) \cdot \frac{x^7}{7} - \frac{x^7}{49} + C \]

Answer

The integral of \(\ln(9x) \cdot x^6\) with respect to \(x\) is:

\[ F(x) = \ln(9x) \cdot \frac{x^7}{7} - \frac{x^7}{49} + C \]


Q7 Check for PDF

For the function \(f(x) = \frac{1}{6x}\) to be a probability density function (PDF) over a certain interval, it needs to satisfy two conditions:

PDF Conditions
  1. The function must be non-negative for all \(x\) in the interval.
  2. The integral of the function over the interval must equal 1.
Check for Positive Values

Since \(x\) is in the interval \([1, e^6]\), \(x\) is always positive. Therefore, \(\frac{1}{6x}\) is also always positive on this interval.

Check Integral Equal 1

The integral \(\int_{1}^{e^6} \frac{1}{6x} \, dx\) should result in 1 for \(f(x)\) to be a probability density function.

The integral of \(\frac{1}{x}\) with respect to \(x\) is \(\ln(x)\). We can write the integral as:

\[ \int_{1}^{e^6} \frac{1}{6x} \, dx = \frac{1}{6} \int_{1}^{e^6} \frac{1}{x} \, dx = \frac{1}{6} \left[ \ln(x) \right]_{1}^{e^6} \]

Sustitute the Limits:

\[ \frac{1}{6} \left[ \ln(e^6) - \ln(1) \right] = \frac{1}{6} [6 - 0] = 1 \]

Answer

Since the integral over the interval \([1, e^6]\) equals 1 and the function is non-negative over the interval, \(f(x) = \frac{1}{6x}\) is a probability density function.