FINDING DEFINITE INTEGRALS | |
1. THE INTEGRAL FUNCTION | |||
For each value of x belonging to an interval [a,b], we can work out the area bounded by the graph of the function y=f(x) and the X-axis between a and x. Therefore, given a function y=f(x) we can construct another function, such that the area defined earlier can be represented for each value of x. This new function is denoted by A(x). |
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1.- Construct the area function for
the function f(x)=-(x2/10)+2 on the interval [0,4.5]
2.- Analyse the characteristics of the function y=A(x). Is its sign positive or negative? Is it monotone? What relations are there between the properties of f and A?
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3.- What
is the value of F(a)? When is the equation F(x)=A(x) true? What is the
value of F(b)?
4.- Determine which conditions are necessary for the integral function to always be positive. |
2. THE DERIVATIVE OF THE INTEGRAL FUNCTION | ||
Let's select two points on the interval [a,b] which are very close together, x and x+h (where h is a small, positive quantity), and a positive function y=f(x). |
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5.- We
can take the strip in the orange area, found below the graph of f,
between points x and x+h and the rectangles used to evaluate the area
through over estimation (in turquoise) and through under estimation (in red). Find the
relation of order between the areas of the three zones in terms of f, F and
h for each value of x on the interval (a,b).
6.- Deduce and check the following relation for function f 7.- If function f had been monotone increasing which relation would have been obtained? 8.- Examine cases when the function is not monotone and when h<0. |
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9.- Observe what happens when h tends to 0. 10.- Show that the derivative of F(x) is exactly the same as the function f(x). Which basic property of the function f has been used here? |
3. FUNDAMENTAL THEOREM OF CALCULUS | |
If f is a continuous function on the interval [a,b], the integral function F(x) can be differentiated on (a,b) and also its derivative F'(x)=f(x). This is known as the fundamental theorem of calculus. This theorem can also be expressed as follows: the integral function is an antiderivative of f. |
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11.- Look carefully at the window and check the fundamental theorem of calculus in geometric terms. 12.- Look carefully at the following geometric interpretation of the fundamental theorem of calculus: A vertical line of variable height f(x) starts at point a and slides along the X-axis shading in the area between the graph of the function and the X-axis. The shaded area up to the point t=x is . What is the instantaneous rate of change when t=x in the shaded area? The actual change at that moment is what is shaded by the line, i.e. the actual area of the line, which is equal to its length, thus f(x). |
4. BARROW'S RULE | |||
The aim of this section is to focus on the calculus of definite integrals. The definite integral of a function between a and b is represented as F(b), where F is the integral function of f. Furthermore, we know that F is an antiderivative of f. Barrow's rule joins these two concepts together in a practical rule which allows us to find F(b) or the value of the defined integral of f between a and b. |
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13.- If we want to find F(b) we need to know one analytical expression for F. By following the basic rules of integration we can find the antiderivative of f, which we will call G. Therefore, both F and G are antiderivatives of f. Are they the same function?
14.- What are the differences between the graphs of F and G? Does this match up to what you know about the indefinite integral? 15.- Find the constant which makes the antiderivatives of f, F and G different. (Clue: look carefully at what happens when x=a).
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16.- Draw conclusions about the
following relation: If F(x) is the integral function of f and G is any
other antiderivative of f then F(x)=G(x)-G(a) and therefore F(b)=G(b)-G(a).
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Enrique Martínez Arcos | |
Spanish Ministry of Education. Year 2001 | |
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