1. Theorems1.1. Comparison Theorem1.2. Fundamental Theorem of Calculus, I1.3. Fundamental Theorem of Calculus, II2. Fundamental Theorem of Calculus2.1. Exercise 5.3.392.2. Exercise 5.3.522.3. Exercise 5.4.40Math 31A 2010.03.02MATH 31A DISCUSSIONJED YANG1. Theorems1.1. Comparison Theorem. If g(x) ≤ f (x) on an interval [a, b], thenZbag(x) dx ≤Zbaf(x) dx.1.2. Fundamental Theorem of Calculus, I. Assume that f(x) is continuouson [a, b] and let F (x) be an antiderivative of f(x) on [a, b]. ThenZbaf(x) dx = F (b) − F (a).1.3. Fundamental Theorem of Calculus, II. Assume that f(x) is continuouson [a, b]. LetA(x) =Zxaf(t) dt.Then A is an antiderivative of f , that is, A′(x) = f(x), or equivalentlyddxZxaf(t) dt = f(x).Furthermore, A(x) satisfies the initial condition A(a) = 0 .2. Fundamental Theorem of Calculus2.1. Exercise 5.3.39. Write the integralRπ0|cos x| dx as a sum of integrals withoutabsolute values and e valuate.Solution. Notice that cos x is nonnegative on [0, π/2] and nonpos itive on [π/2, π].As such, the integral in questio n isRπ/20cos x dx +Rπ/4π/2− cos x dx = sin x |π/20− sin x |ππ/2= 1 − 0 − 0 + 1 = 2. 2.2. Exercise 5.3.52. Apply the Compariso n Theorem to the inequality sin x ≤ x(valid for x ≥ 0) to prove 1 −x22≤ cos x ≤ 1. Apply it again to prove x −x36≤sin x ≤ x (for x ≥ 0).Solution. On [0, t] for s ome t > 0, we have sin x ≤ x. B y the Comparison Theorem,we ge tRt0sin x dx ≤Rt0x dx. This gives − cos t + 1 ≤ t2/2 hence 1 −t22≤ cos t.Since cos is even, cos −t = cos t s atisfies the same inequality. Applying this againwe getRt01 −x22dx ≤Rt0cos x dx, yielding t −t33≤ sin t, as desired. Math 31A Yang 22.3. Exercise 5.4.40. Find the smallest positive critical point ofF (x) =Zx0cos(t3/2) dtand determine whether it is a local min or max.Solution. As usual, to find critical point, we take derivative and set to 0. ByFTC2, we get F′(x) = cos(x3/2). The smallest positive zero is when x3/2=π2. Sox = (π/2)2/3is the smallest positive critical point. Since F′goes from p ositive tonegative, it is a local max.