Show step by step integration of exp(-x by a)

Certainly! To integrate the function $e^{-\frac{x}{a}}$ with respect to $x$, where $a$ is a constant, you can follow these steps:

1. Recognize the Integral: Identify the function to integrate: $\int e^{-\frac{x}{a}} \, dx$.

2. Use the Power Rule of Integration: Recall the power rule of integration: $\int x^n \, dx = \frac{1}{n+1} x^{n+1} + C$, where $n$ is any constant except -1.

3. Adjust the Exponential Function: The given function is in the form $e^{-\frac{x}{a}}$, which is similar to $e^{kx}$ where $k = -\frac{1}{a}$. The integration of $e^{kx}$ involves multiplying by the reciprocal of the coefficient of $x$ (which is $k$ in this case).

4. Apply the Integration Rule: Apply the integration rule to find the antiderivative of $e^{-\frac{x}{a}}$:

   $$\int e^{-\frac{x}{a}} \, dx = -a e^{-\frac{x}{a}} + C$$

   where $C$ is the constant of integration.

5. Final Answer: The final result of the integration is $-a e^{-\frac{x}{a}} + C$.

So, the step-by-step integration of $e^{-\frac{x}{a}}$ with respect to $x$ is:

$$\int e^{-\frac{x}{a}} \, dx = -a e^{-\frac{x}{a}} + C$$

where $C$ is the constant of integration.

Unit Test 5 - Exploring bivariate numerical data (From Khan Academy)

Q1: Correlation coefficient

Q2: Least squares regression line

Sol 2:

Q3: Scatter plot and line of best fit

Q4: Residual

Q5: Interpretation of slope

Q6: Non linear association

Q7: Scatterplot construction

Q8: Residuals from line of regression

Q9: Coefficient of determination (sq(R))

Q10: Interpretation of scatter plot for equation of line of regression

Sol 10:

Q11: Scatter plot building

Q12: Identifying the association from scatter plot

Derive Jacobian when we are changing from 2D cartesian coordinates to polar coordinates

Here's the derivation of the Jacobian when changing from 2D Cartesian coordinates to polar coordinates:

1. Transformation Equations:

• Cartesian coordinates (x, y) are related to polar coordinates (r, θ) by:
  • x = r cos(θ)
  • y = r sin(θ)

2. Jacobian Matrix:

• The Jacobian matrix represents the partial derivatives of the new coordinates with respect to the old coordinates:

  J = | ∂x/∂r ∂x/∂θ | | ∂y/∂r ∂y/∂θ |

3. Calculating Partial Derivatives:

• ∂x/∂r = cos(θ)
• ∂x/∂θ = -r sin(θ)
• ∂y/∂r = sin(θ)
• ∂y/∂θ = r cos(θ)

4. Filling the Jacobian Matrix:

J = | cos(θ) -r sin(θ) | | sin(θ) r cos(θ) |

5. Evaluating the Determinant:

• The Jacobian, denoted as J, is the determinant of this matrix:

  J = det(J) = cos(θ) * r cos(θ) - (-r sin(θ)) * sin(θ) = r cos^2(θ) + r sin^2(θ) = r (cos^2(θ) + sin^2(θ))

6. Simplifying Using Trigonometric Identity:

• Since cos^2(θ) + sin^2(θ) = 1, we get:

  J = r

Therefore, the Jacobian for the change of variables from Cartesian coordinates to polar coordinates is r.

Significance in Integration:

• When changing variables in a double integral from Cartesian to polar coordinates, the area element dx dy transforms to r dr dθ. This is a crucial step in evaluating integrals over regions that are more naturally expressed in polar coordinates, such as circles and sectors.

Using ChatGPT

Let's go through the steps to calculate the Jacobian for the transformation from Cartesian to polar coordinates.

The transformation from Cartesian coordinates $x$ and $y$ to polar coordinates $r$ and $\theta$ is given by:

$x = r \cos(\theta)$ $y = r \sin(\theta)$

The Jacobian of this transformation is the determinant of the matrix of partial derivatives:

$J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix}$

1. Partial Derivatives: Calculate the partial derivatives:

   $\frac{\partial x}{\partial r} = \cos(\theta)$ $\frac{\partial x}{\partial \theta} = -r \sin(\theta)$ $\frac{\partial y}{\partial r} = \sin(\theta)$ $\frac{\partial y}{\partial \theta} = r \cos(\theta)$

2. Jacobian Determinant: Compute the determinant:

   $J = \cos(\theta) \cdot r \cos(\theta) - (-r \sin(\theta)) \cdot \sin(\theta)$ $J = r \cos^2(\theta) + r \sin^2(\theta)$ $J = r$

So, the Jacobian determinant is $r$ for the given transformation. When changing variables in an integral, you need to multiply by the absolute value of the Jacobian determinant. In polar coordinates, this means you multiply by $r$. The $dx\,dy$ becomes $r\,dr\,d\theta$ in polar coordinates.