Finding the area under curves has various applications in physics, allowing us to calculate important quantities like displacement, velocity, and acceleration. Let's consider an example to understand how integrals can be used in physics.

Suppose an object is moving along a straight line with a time-dependent velocity. We can represent the velocity of the object as a function v(t), where t is the time. To find the displacement of the object during a given time interval, we need to calculate the area under the velocity curve between the initial and final times.

For instance, if v(t) = 3t, we integrate this function with respect to time over the interval [1, 4] to find the area under the curve. The integral of 3t with respect to t is (3/2)t^2. Evaluating this integral between 1 and 4 gives us [(3/2)(4)^2 - (3/2)(1)^2] = [(3/2)(16 - 1)] = 22.5. Therefore, the displacement of the object during this time interval is 22.5 units.

By using integrals to calculate such areas, we can determine displacements, velocities, and accelerations for objects with complex motion.