The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus Through the examples we looked at for the area under graphs of functions, we were led to an interesting observation: there seems to be a relationship between the process of integration, which is just a fancy way of performing sums, and the process of differentiation. In fact, this observation is the one basic fact which underlies almost all of our work in this course. Consequently, we will give it a name which indicates its importance: The Fundamental Theorem of Calculus. What does the Fundamental Theorem mean? Before we jump in and tell you about the theorem, we will try and give you an intuitive feel for it through a demonstration. We have seen already that the definite integral of a positive function can be interpreted as the area under the graph of the function. But what about functions which are negative? There's a pretty simple explanation in that case as well. Remember that the definite integral is given by a sum where the points are formed by breaking the interval into n pieces of width . When the function f was positive, we could interpret each term as the area of a very thin rectangle. However, remember that area is always considered to be positive and so if the function f is negative, the term represents the additive inverse of the area of a small rectangle. Know More About Define Mean Math.Tutorvista.com

Page No. :- 1/5

This means that when we perform the sum , we are actually computing the additive inverse of the area of all the rectangles and so the area A is given by In other words, the definite integral measures the additive inverse of the area between the graph and the x axis. Now when the function f has some regions where it is positive and others where it is negative, we may compute the definite integral by integrating over the regions where the function is positive and add that to the integral over the region where the function is negative. This leads to the observation that where is the area bounded by the region where the function is positive and is the area bounded by the region where the function is negative.So in this most general case, the definite integral can still be thought of as measuring area, but it does so by measuring some areas as negative and others as positive. With this interpretation, we can try to convey graphically the main idea behind the Fundamental Theorem of Calculus. Below, you will see the graph of a function f on the left. The graph on the right is the function Now when you drag the ball on the graph to the left, you will see a region of area which contributes to the definite integral. Yellow area contributes positively, while orange area contributes negatively. Notice that as you first drag the ball to the right away from the origin, you are in a region where is positive. Consequently, as you increase x , the new area which is exposed is weighted positively in the definite integral. This means that the function is increasing in that region. When you reach the point at which , you have uncovered all of the positive area. This produces a maximum value for . Increasing the value of x even further causes the value of the function to become negative. Then the new area which is exposed is weighted negatively and this causes the function to decrease. Learn More Standard Deviation Definition Math.Tutorvista.com

Page No. :- 2/5

Rate of Change Problems Rate of Change Problems The rate of change in algebra is the ratio that shows the relation between the two variables in equation. In general, the coefficient of x is called as the rate of change of an equation. For example, in y=3x+4, the rate of change is 3. 3 is the coefficient of x. In this equation, the constant variable is 4. Problem 1: A rectangular water tank (see figure below) is being filled at the constant rate of 20 liters / second. The base of the tank has dimensions w = 1 meter and L = 2 meters. What is the rate of change of the height of water in the tank?(express the answer in cm / sec). Solution to Problem 1: The volume V of water in the tank is given by. V = w*L*H We know the rate of change of the volume dV/dt = 20 liter /sec. We need to find the rate of change of the height H of water dH/dt. V and H are functions of time. We can differentiate both side of the above formula to obtain

Math.Tutorvista.com

Page No. :- 3/5

dV/dt = W*L*dH/dt note W and L do not change with time and are therefore considered as constants in the above operation of differentiation. We now find a formula for dH/dt as follows. dH/dt = dV/dt / W*L We need to convert liters into cubic cm and meters into cm as follows 1 liter = 1 cubic decimeter = 1000 cubic centimeters = 1000 cm 3 and 1 meter = 100 centimeter. We now evaluate the rate of change of the height H of water. dH/dt = dV/dt / W*L = ( 20*1000 cm 3 / sec ) / (100 cm * 200 cm) = 1 cm / sec.