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| Fundamental Theorem of Calculus |
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We have already defined, for a continuous function f(x) on a closed interval [a, b]
 as the area of the region bounded by the curve y = f(x), X-axis and x= a and x = b. |
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| In other words, area of the shaded region is a function of x. |
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| The function A(x) is shown in figure below. |
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| This area function A(x) is the anti derivative of f(x). That is f(x) = A'(x) |
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| We state fundamental theorems of integral calculus without proof as they are beyond syllabus. |
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| Let f(x) be a continuous function on the closed interval [a, b]. |
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| Let the area function A(x) be defined by |
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| then |
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| Let f(x) be a continuous function defined on an interval [a,b]. |
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| between the limits a and b. This statement is also known as 'fundamental theorem of calculus'. |
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| We call b, the upper limit of x and a, the lower limit. |
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| If in place of F(x) we take F(x)+c as the value of the integral, we have |
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| = [F(b) + c] - [F(a) + c] = F(b) + c - F(a) - c = F(b) - F(a) |
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| Thus, the value of a definite integral is unique. It does not depend on the constant c and so in the evaluation of a definite integral the constant of integration does not play any role. |
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| Note: From the above two theorem, we infer the following |
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 (Anti derivative of the function f(x) at b) |
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| - (Anti derivative of the function f(x) at a) |
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| (ii) The fundamental theorem of integral calculus shows a close relationship between differentiation and integration |
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| (iii) These theorems give an alternate method evaluating definite integral, without calculating the limit of a sum. |
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| Example: |
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| Evaluate the definite integral of the following |
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| Solution: |
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