Chapter 4: Integration
Section 4.3: Fundamental Theorem of Calculus and the Indefinite Integral
Fundamental Theorem of Calculus
The basic mechanism for evaluating a definite integral is the Fundamental Theorem of Calculus (FTC), formally stated as Theorem 4.3.1.
Theorem 4.3.1: Fundamental Theorem of Calculus
f is continuous on a,b
For f, F is any antiderivative on a,b
gx≡∫axft ⅆt, with x∈a,b
g is continuous on a,b, differentiable on a,b, and g′x=fx
The first conclusion of the FTC is that a definite integral, the limit of a Riemann sum, can be evaluated by subtracting the endpoint values of any antiderivative. That is the meaning of the symbols in the first conclusion of the theorem, where F is any antiderivative of the integrand f, and the value of the definite integral is Fb−Fa.
The second conclusion of the FTC is that a definite integral with a varying endpoint defines a function whose derivative is the integrand. In other words, the integral in hypothesis (3) defines an antiderivative, so that integration (the limit of a Riemann sum) and differentiation are inverse operations.
Because of the FTC, and only because of the FTC, the integral sign is used to denote the operation of finding an antiderivative. Thus, the antiderivative is identified with the indefinite integral, denoted by ∫f ⅆx.
For the functions fk in Table 4.3.1, Maple computes their definite integrals on a,b by taking the limit of a right Riemann sum. The results are consistent with finding an antiderivative Fk and evaluating Fkb−Fka, that is, by using the FTC.
Limit of right Riemann sum
f1x = x
limn→∞∑k=1nf1a+k h h = −12⁢a2+12⁢b2
f2x = x2
limn→∞∑k=1nf2a+k h h = −13⁢a3+13⁢b3
f3x = x3
limn→∞∑k=1nf3a+k h h = −14⁢a4+14⁢b4
f4x = x
limn→∞∑k=1nf4a+k h h = −23⁢a3/2+23⁢b3/2
f5x = sin⁡x
simplifylimn→∞∑k=1nf5a+k h h = −cos⁡b+cos⁡a
f6x = cos⁡x
simplifylimn→∞∑k=1nf6a+k h h = sin⁡b−sin⁡a
f7x = ⅇx
simplifylimn→∞∑k=1nf7a+k h h = ⅇb−ⅇa
Table 4.3.1 Evaluation of ∫abfkx ⅆx via Riemann sum is consistent with FTC
The first four functions all yield to the "Power rule for antidifferentiation,"
that is, to the rule "add 1 to the power and divide by the new power." (In symbols, this is the third entry in Table 3.10.1.) Antiderivatives for the exponential and trig functions can also be found in Table 3.10.1.
If F is an antiderivative of f, a common notation for the difference Fb−Fa that arises in the evaluation of the definite integral ∫abfx ⅆx is F |ab. (Sometimes, the stroke is replaced by "]", the closing square bracket.) Although such notation can sometimes be written in Maple, it is not part of the code structure used when a definite integral is evaluated by Maple.
From Riemann Sum to Antiderivative: A Sketch
The following is a sketch of a demonstration as to why the limit of a Riemann sum should result in the difference in the endpoint values of an antiderivative of the integrand. The essential link is the Mean Value theorem stated in the form fx+h=fx+f′c h for some c between x and x+h.
Consider a right Riemann sum for the derivative f′x on the interval a,b. Apply the Mean Value theorem to each subinterval xk,xk−1,k=1,…,n, selecting as the point to evaluate f′ as the ck for which the equality
holds, k=1,…,n. In the Riemann sum, replace each term h f′ck with fxk−fxk−1 so that the Riemann sum ∑k=1nh f′ck becomes
which collapses to fb−fa.
In the limit as n→∞ and h→0, the need to pick the ck to satisfy the Mean Value theorem disappears since each subinterval shrinks to zero length, and there must always be an appropriate ck in every such shrinking subinterval.
In the definite integral (which is just the limit of the Riemann sum) where the integrand is the derivative f′, the value of the definite integral is the difference fb−fa, the difference in the endpoint values of the antiderivative.
Calculate the area bounded by the x-axis, the graph of y=x3 and the vertical lines x=−1, x=2; then use the FTC to evaluate the definite integral ∫−12x3 ⅆx. Are these two quantities the same?
Use the FTC to evaluate the definite integral ∫−135 x3−7 x2+4 ⅆx.
Obtain an explicit rule for the function gx=∫0x11+x2 ⅆx; then show g′x is the integrand evaluated at x.
Use Maple to obtain an explicit rule for the function Gx=∫0xsinu ⅆu; then show G′x is the integrand evaluated at x.
If wx=∫ayxft ⅆt, show that w′x=fyx y′x.
If wx=∫yxaft ⅆt, show that w′x=−fyxy′x
Obtain ddx∫0xsinht ⅆt.
Evaluate the indefinite integral ∫5 sec2x−3 sinx ⅆx.
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