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Displacement is a fundamental concept in calculus and physics that helps us understand how much a particle has moved, irrespective of the path taken. It represents the net change in position of a particle from one point in time to another. Mathematically, for a velocity function \( v(t) \), the displacement over a time interval \([a, b]\) is calculated by integrating the velocity function over that interval. This gives us:\[ \text{Displacement} = \int_{a}^{b} v(t) \, dt \]For example, if we have a velocity function like \( v(t) = t - \sqrt{t} \) for \( 0 \leq t \leq 4 \), the displacement is found by evaluating the integral from 0 to 4. The result represents how far the particle has moved along its path, taking into account direction — positive forward movement and negative backward movement cancel each other out.

Integral calculus is a branch of calculus focused on the accumulation of quantities and the spaces under and between curves. The main tool used in integral calculus is the process of integration, which is essentially the reverse operation of differentiation. While differentiation gives us the rate of change of a quantity, integration allows us to sum up infinitesimally small changes to find the overall change.To find the displacement or the distance traveled by a particle, we use definite integrals. This involves:

• Setting up an integral with bounds that correspond to the time interval during which you're examining the motion.
• Integrating the velocity function over that interval to compute the net change in position (displacement) or the total path length (distance).

For instance, if the velocity function is given as \( v(t) = \frac{1}{\sqrt{t+1}} \) over the interval \([0, 3]\), then the definite integral provides the total displacement: \[ \int_{0}^{3} \frac{1}{\sqrt{t+1}} \, dt = 2 \]This example demonstrates how integration sums up changes across the entire interval to give a total measure of change in position.

Velocity is an essential concept in understanding motion. It describes not just the speed of an object, but also its direction. In calculus, velocity is often described as a function \( v(t) \) that shows how the velocity varies over time.The sign of the velocity function over a given interval is important. It tells us whether the particle is moving forward (positive velocity) or backward (negative velocity). This influences both displacement and distance calculations:

• Displacement: Considers direction. Integrating the velocity function directly gives the net change in position.
• Distance: Ignores direction and considers the total path length. It requires integrating the absolute value of velocity to ensure all changes contribute positively.

Consider the velocity function \( v(t) = t - \sqrt{t} \). On the interval \( 0 \leq t \leq 4 \), the function changes sign, meaning there are parts of the interval where the particle moves forward and parts where it moves backward. Therefore, finding the total distance requires integrating \( |v(t)| \) for separate subintervals where the particle's motion does not change direction.