Compute $\int_{0}^{t}|s-t|\,ds$: Step-by-Step Guide

In summary, to compute the given integral, we use the interval [0,t] and since s is always less than t, we can rewrite |s-t| as -(s-t). This simplifies the integral to -\int_0^t (t-s) ds. After solving, we get the final answer of t^2/2.
  • #1
Fermat1
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How would I compute the following integral? Let t be in [0,1].

$\int_{0}^{t}|s-t|\,ds$
 
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  • #2
Fermat said:
How would I compute the following integral? Let t be in [0,1].

$\int_{0}^{t}|s-t|\,ds$

Since you are integrating on the interval $[0,t]$ so we have $0\leq s \leq t$.
 
  • #3
Fermat said:
How would I compute the following integral? Let t be in [0,1].

$\int_{0}^{t}|s-t|\,ds$
The integral is for s from 0 to t so s is always less than t and s- t is always negative. |s- t|= -(s- t) so
[tex]\int_0^{t}|s- t| ds= -\int_0^t (t- s) ds= \left[ts- s^2/2\right]_0^t= t^2/2[/tex]
 

FAQ: Compute $\int_{0}^{t}|s-t|\,ds$: Step-by-Step Guide

What is the purpose of computing the integral of |s-t|?

The integral of |s-t| is used to calculate the absolute difference between two variables, s and t. It is commonly used in mathematics and physics to measure the distance between two points.

How do I set up the integral to compute |s-t|?

To compute the integral of |s-t|, you first need to determine the limits of integration, which in this case are 0 and t. Then, you need to write the integrand as |s-t| and include the differential "ds" at the end. This will give you the setup for the integral: ∫0t|s-t|ds

What are the steps to solve the integral of |s-t|?

The steps to solve the integral of |s-t| are as follows:

  1. Set up the integral with the limits of integration and the integrand.
  2. Integrate the absolute value function, which will give you two different cases: when s < t and when s > t.
  3. Evaluate the integral using the limits of integration and the two cases.
  4. Combine the two results to get the final answer.

Can you provide an example of computing the integral of |s-t|?

Sure, let's say we want to compute ∫03|s-2|ds. First, we set up the integral: ∫03|s-2|ds. Then, we integrate the absolute value function, which gives us two cases: ∫02(2-s)ds + ∫23(s-2)ds. Solving these two integrals, we get 2 + 1 = 3. Therefore, the final answer is 3.

What are some real-life applications of computing the integral of |s-t|?

The integral of |s-t| has many real-life applications, such as calculating the displacement of an object, finding the distance traveled by a moving object, or determining the difference between two values in a data set. It is also used in physics to calculate the work done by a force over a certain distance.

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