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Chapter 5 | Integration
antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting. Proof Let P ={ x i }, i =0, 1,…, n be a regular partition of ⎡ ⎣ a , b ⎤ ⎦ . Then, we can write F ( b )− F ( a ) = F ( x n )− F ( x 0 ) = ⎡ ⎣ F ( x n )− F ( x n −1 ) ⎤ ⎦ + ⎡ ⎣ F ( x n −1 )− F ( x n −2 ) ⎤ ⎦ +…+ ⎡ ⎣ F ( x 1 )− F ( x 0 ) ⎤ ⎦
= ∑ i =1 n
⎡ ⎣ F ( x i )− F ( x i −1 ) ⎤ ⎦ .
Now, we know F is an antiderivative of f over ⎡ ⎣ a , b ⎤ for i =0, 1,…, n we can find c i in [ x i −1 , x i ] such that F ( x i )− F ( x i −1 ) = F ′( c i ⎞
⎦ , so by the Mean Value Theorem (see The Mean Value Theorem )
⎠ ( x i − x i −1 ) = f ( c i )Δ x .
Then, substituting into the previous equation, we have
F ( b )− F ( a ) = ∑ i =1 n
f ( c i )Δ x .
Taking the limit of both sides as n →∞, we obtain
→∞ ∑ i =1 n
F ( b )− F ( a ) = lim n
f ( c i )Δ x
b
= ∫
f ( x ) dx .
a
□
Example 5.20 Evaluating an Integral with the Fundamental Theorem of Calculus Use The Fundamental Theorem of Calculus, Part 2 to evaluate ∫ −2 2 ⎛ ⎝ t 2 −4 ⎞ ⎠ dt . Solution Recall the power rule for Antiderivatives : If y = x n , ∫ x n dx = x n +1 n +1 + C . Use this rule to find the antiderivative of the function and then apply the theorem. We have
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