Let’s return to a problem from the Trapezoidal Rule section. What is the purpose of definite integrals? They give the exact area beneath a curve. However, they have little to do with the limits of Chapter 3, so don’t worry. The a and b in definite integrals are called the limits of integration. From that number, subtract the result of plugging the lower bound into the antiderivative. Once you’ve done that, plug the upper bound, b, into the antiderivative. Translation: In order to evaluate the definite integral, find the antiderivative of f(x). If f(x) is a continuous function on with antiderivative g(x), then Remember, you can’t spell Fundamental Theorem without “fun”! These are, indeed, two giant differences, but you’ll be surprised by how much they actually have in common with our previous integrals, which we will now refer to by their proper name, indefinite integrals. These are slightly different from the integrals we’ve been dealing with for two reasons: (1) they have boundaries, and (2) their answers are not functions with a “+ C” tacked on to the end-their answers are numbers. The first part of the Fundamental Theorem deals with definite integrals.
I love them both equally, as I would my own children. Some even refer to one as the Fundamental Theorem and the other as the Second Fundamental Theorem. Mathematicians can’t seem to agree which is the more important part and, therefore, number them differently. In fact, the fundamental theorem has two major parts. In this theorem lies the fabled connection between the antiderivative and the area beneath a curve. Today, interesting.” This is quite accurate, if not a little understated.
He once said something I remember to this day: “Fundamental theorems are like the beginning of the world. This tells us this: when we evaluate f at n (somewhat) equally spaced points in, the average value of these samples is f ( c ) as n → ∞.I once had a Korean professor in college named Dr. Lim n → ∞ 1 b - a ∑ i = 1 n f ( c i ) Δ x = 1 b - a ∫ a b f ( x ) d x = f ( c ). = 1 b - a ∑ i = 1 n f ( c i ) Δ x (where Δ x = ( b - a ) / n ) = 1 b - a ∑ i = 1 n f ( c i ) b - a n = ∑ i = 1 n f ( c i ) 1 n ( b - a ) ( b - a ) Multiply this last expression by 1 in the form of ( b - a ) ( b - a ): The average of the numbers f ( c 1 ), f ( c 2 ), …, f ( c n ) is:ġ n ( f ( c 1 ) + f ( c 2 ) + ⋯ + f ( c n ) ) = 1 n ∑ i = 1 n f ( c i ). Next, partition the interval into n equally spaced subintervals, a = x 1 < x 2 < ⋯ < x n + 1 = b and choose any c i in. First, recognize that the Mean Value Theorem can be rewritten asį ( c ) = 1 b - a ∫ a b f ( x ) d x ,įor some value of c in. The value f ( c ) is the average value in another sense. This proves the second part of the Fundamental Theorem of Calculus. Consequently, it does not matter what value of C we use, and we might as well let C = 0. This means that G ( b ) - G ( a ) = ( F ( b ) + C ) - ( F ( a ) + C ) = F ( b ) - F ( a ), and the formula we’ve just found holds for any antiderivative. Furthermore, Theorem 5.1.1 told us that any other antiderivative G differs from F by a constant: G ( x ) = F ( x ) + C. We now see how indefinite integrals and definite integrals are related: we can evaluate a definite integral using antiderivatives. = - ∫ c a f ( t ) d t + ∫ c b f ( t ) d t = ∫ a c f ( t ) d t + ∫ c b f ( t ) d t Using the properties of the definite integral found in Theorem 5.2.1, we know First, let F ( x ) = ∫ c x f ( t ) d t. Suppose we want to compute ∫ a b f ( t ) d t. Consider a function f defined on an open interval containing a, b and c. We have done more than found a complicated way of computing an antiderivative.
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