Integral calculus
      The definite integral
      Physical applications to the definite integral
         Describing motion of objects using velocity - time graphs
      Evaluating the area under the graph of a function using the definition of the definite integral, examples
Physical applications to the definite integral
Describing motion of objects using velocity - time graphs
In kinematics motion is defined as change in position relative to some fixed point or object.
The distance the object travels in a period of time is its speed. Velocity is speed in a given direction.
The displacement is the vector form of the distance involving direction and magnitude of the change in position of the object, meaning it is the straight line distance between the initial and final positions of the body.
Acceleration is the increase in speed or velocity over a period of time.
Example:   When an object travels with a constant velocity v over a period of time t, the displacement d, for the object we calculate using the formula  d = v t.
Then, the constant velocity v is represented in the coordinate system (t, v) as a line parallel to the t-axis passing through the point (0, v) while the displacement d is given by the area of the rectangle with the base t and height v, as is shown in the left figure below.
 
Example:   If an object travels with constant acceleration a then its velocity v is changing by a constant amount each unit of time t.
Thus, the velocity v = a t  is represented as the line through the origin with slope equal to the acceleration, as shows the velocity-time graph in the right picture above. 
When an object travels with constant acceleration a then displacement d traveled in a certain amount of time t is the same as object travels with constant velocity v = (a t)/2, that is, equals the half of the final velocity at the end of the time interval.
Therefore, the length of the path an object travels with constant acceleration in a given time interval is represented in the velocity-time graph by the area of the rectangle of the base t and height  (at)/2
that correspond to the area of the triangle of the base t with height  at, shown in the above figure.
Evaluating the area under the graph of a function using the definition of the definite integral, examples
Example:   Evaluate the area under the graph of the quadratic function over the interval [0, x], where x is any value of the argument, using the definition of the definite integral.
Solution:   Let's use the partition of the interval [0, x] into n equal subintervals, so that  Dx = x / n and calculate the lower sum s of inscribed rectangles, as is shown in the left figure below. According to definition
The area under the graph of  f(x) = x2 over the interval [0, x] amounts 2/3 of the area of the triangle OPxP.
 
Example:   Evaluate the area under the graph of the rectangular or equilateral hyperbola f (x) = 1/x  over the interval [1, x], using the definition of the definite integral.
Solution:   Let's divide the interval [1, x] into n subintervals of different lengths by partition points that form
the geometric sequence,   1,  qq2, . . . ,  qn - 1qn = x  (q > 1)
and calculate the lower and upper sums, s and S, as is shown in the right figure above.
Note that all rectangles of the sum s have the same area 1 - 1/q, while all rectangles of the sum S have the same area  q - 1.
Calculus contents E
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