MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and theFundamental Theorem of Calculus (pp. 207-233, Gootman)Chapter Goals:• Understand the statement of the Fun damental Theorem of Calculus.• Learn how to compute the antiderivative of some basic functions.• Learn how to use the substitution m ethod to compute the antiderivative of morecomplex functions.• Learn how to solve area and distance traveled problems by means of antiderivatives.Assignments:Assignment 22 Assignment 23 Assignment 24 (Review)So far we have learned about the idea of the integral, and what is meant by computing the definite integral ofa function f (x) over the interval [a, b]. As in the case of derivatives, we now s tudy procedures for computingthe definite integral of a function f(x) over the interval [a, b] that are easier than computing limits of Riemannsums. As with derivatives, however, the definition is important because it is only through the definition thatwe can understand why the integral gives the an swers to particular problems.◮ Idea of the Fundamental Theorem of Calculus:The easiest procedure for computing defi nite integrals is not by computing a limit of a Riemann sum, butby relating integrals to (anti)derivatives. This relationship is so important in Calculus that the theorem thatdescribes the relationships is called the Fundamental Theorem of Calculus.◮ Computing some antiderivatives:In previous chapters we were given a function f (x) and we found the der ivative f′(x). In this section, we will dothe reverse. We will be given a function f (x) that is the derivative of another function F (x) and will computeF (x). In other words find a function F (x) such that F′(x) = f (x). F (x) is called an a ntiderivative of f(x).For example (x3)′= 3x2so an antiderivative of f (x) = 3x2is F (x) = x3.Note that F (x) = x3+ 2 is also an antiderivative of f(x) = 3x2because (x3+ 2)′= 3x2. In general, if F (x) isan antiderivative of f (x), then so is F (x) + C where C is any constant. This leads to the following notation.Definition of the indefinite integral:The indefinite integral of f(x), denoted byZf(x) dxwithout limits of integration, is the general antiderivative of f (x).For example, it is easy to check thatZ3t2dt = t3+ c, where c is any constant.Recall that the power rule for derivatives gives us (xn)′= nxn−1. We multiply by n and subtract 1 from theexponent. Since antiderivatives are th e reverse of derivatives, to compute an the antiderivative we first increasethe power by 1, then divide by the new power.105The formulas below can be verified by differentiating the righthand side of each expression.Some basic indefinite integrals:1.Zxndx =1n + 1xn+1+ C n 6= −1 2.Zexdx = ex+ Cn = −1 in formula 1 leads to division by zero, bu t for this sp ecial case we may use (ln(x))′=1x:3.Z1xdx = ln |x| + CRules for indefinite integrals:A.Zc f(x) dx = cZf(x) dx B.Z(f(x) ±g(x)) dx =Zf(x) dx±Zg(x) dxExample 1:Evaluate the indefinite integralZ(t3+ 3t2+ 4t + 9) dt.Example 2: Evaluate the indefinite integralZ6√tdt.Warning:We do not h ave simple derivative ru les for products an d q uotients, so we should not expectsimple integral rules for products and quotients.Example 3:Evaluate the indefinite integralZt3(t + 2) dt.106Example 4: Evaluate the indefinite integralZx2+ 9x2dx.We now have some experience computing antiderivatives. We will now see how antiderivatives give us an elegantmethod for finding areas under curves.Example 5:Find a formula for A(x) =Zx1(4t + 2) dt, that is, evaluate th edefinite integral of the function f (t) = 4t + 2 over the interval [1, x] inside [1, 10].(Hint: think of this definite integral as an area.) Find the values A(5), A(10), A(1).What is the derivative of A(x) with respect to x?ty01x106A(x)f(t) = 4t + 2Observations: There are two important things to notice about the function A(x) analyzed in Example 1:A(1) =Z11(4t + 2) dt = 0 A′(x) =ddxZx1(4t + 2) dt|{z }A(x)= 4x + 2.Notice what the last equality says: The instantaneous rate of change of the area under the cur ve y = 4t + 2 att = x is sim ply equal to the value of the curve evaluated at t = x.Why? A(x) measures the area of some geometric figure. As x increases, the w idth of the figure increases, andso the area increases. A′(x) measures the rate of increase of the figure. Now, as x increases, the right wall ofthe figure sweeps out ad ditional area, so the rate at which the area increases shou ld be equ al to the height ofthe right wall.The following pages will make this idea more precise.107Idea: Suppose that for any function f (t) it were true that the area function A(x) =Zxaf(t) dt satisfiesA(a) =Zaaf(t) dt = 0 A′(x) =ddxZxaf(t) dt= f (x).Moreover, suppose that F (x) is any andtiderivative of f(x) (i.e., F′(x) = f (x) = A′(x).) By The ConstantFunction Theorem (Chapter 6), there exists a constant value c such that F (x) = A(x) + c.All these facts put together help us easily evaluateZbaf(t) dt. Indeed,Zbaf(t) dt:::::::::= A(b) = A(b) − 0= A(b) − A(a):::::::::::= [A(b) + c] − [A(a) + c]= F (b) − F (a):::::::::::The above ‘speculations’ are actually true for any:::::::::::continuous function on the interval over which we areintegrating. These results are stated in the following theorem, which is divided into two parts:The Fundamental Theorem of Calculus:PART I: Let f (t) be a continuous function on the interval [a, b].Then the function A(x), defined by the formulaA(x) =Zxaf(t) dtfor all x in the interval [a, b], is an antiderivative of f(x), that isA′(x) =ddxZxaf(t) dt= f (x)for all x in the interval [a, b].PART II:Let F (x) be any antiderivative of f (x) on [a, b], so thatF′(x) = f (x)for all x in the interval [a, b]. ThenZbaf(x) dx = F (b) − F (a).Special notations: The above theorem tells us that evaluating a definite integral is a two-step process:find any antiderivative F (x) of the fun ction f (x) and then compute the difference F (b) −F (a). A notation hasbeen devised to separate the two steps of this process: F (x)bastands for the difference F (b) − F (a). ThusZbaf(x) dx = F (x)ba= F (b) − F (a).108◮ Some properties of definite integrals:1.Zaaf(x) dx = 0 2.Zbak f(x) dx = kZbaf(x) dx3.Zba(f(x) ± g(x)) dx =Zbaf(x) dx±Zbag(x) dx4.Zbaf(x) dx +Zcbf(x) dx =Zcaf(x) dx 5.Zbaf(x) dx = −Zabf(x) dx6. If m ≤ f(x) ≤ M on [a, b] then m(b −a) ≤Zbaf(x) dx ≤ M(b − a)Geometric