What is the differentiation and integration of f(x)=1 wrt dx

If $f(x) = 1$, then the derivative of $f(x)$ with respect to $x$ (denoted as $f'(x)$ or $\frac{df}{dx}$) is zero because the function is constant.

$f'(x) = \frac{d}{dx}(1) = 0$

The integral of $f(x) = 1$ with respect to $x$ (denoted as $\int f(x)dx$) is simply $x$ plus a constant of integration (often denoted as $C$).

$\int 1 \,dx = x + C$

So, in summary: $f'(x) = 0$
$\int 1 \,dx = x + C$

where $C$ is the constant of integration.

Change in limits of integration when changing from cartesian to polar coordinates

While changing variables from x, y to r, theta in a double integral, if original limits go from -infinite to +infinite, then what would new limits go to? Please explain step by step.

When changing variables from Cartesian coordinates (x, y) to polar coordinates (r, θ) in a double integral, the limits of integration need to be adjusted accordingly. If the original limits for x and y extend from negative infinity to positive infinity, it typically implies that you are integrating over the entire plane. In polar coordinates, the corresponding region would be described using the radial coordinate (r) and the angular coordinate (θ).

Let's go through the process step by step:

Original double integral:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \,dx\,dy$

Now, let's express the Cartesian coordinates (x, y) in terms of polar coordinates (r, θ):

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

Next, we need to express the differential area element $dx\,dy$ in terms of polar coordinates. The area element in Cartesian coordinates is $dx\,dy$, and in polar coordinates, it becomes $r\,dr\,d\theta$. Therefore:

$dx\,dy = r\,dr\,d\theta$

Now, substitute these expressions into the original integral:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \,dx\,dy = \int_{0}^{2\pi} \int_{0}^{\infty} f(r \cos(\theta), r \sin(\theta)) \,r\,dr\,d\theta$

In polar coordinates, the limits for $r$ typically start from 0 to $\infty$, and the limits for $\theta$ range from 0 to $2\pi$ to cover the entire plane.

So, when changing variables from x, y to r, θ in a double integral where the original limits are from $-\infty$ to $+\infty$, the new limits become:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \,dx\,dy$

becomes

$\int_{0}^{2\pi} \int_{0}^{\infty} f(r \cos(\theta), r \sin(\theta)) \,r\,dr\,d\theta$

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.