The Fundamental Theorem of Calculus.

1 The Fundamental Theorem of two main concepts of calculus are integration and differentiation. TheFundamental Theorem of calculus (FTC) says that these two concepts are es-sentially inverse to one Fundamental Theorem states that ifFhas a continuous derivative on aninterval [a,b], then baF (t)dt=F(b) F(a).This form allows one to compute integrals by finding FTC says that integration undoes differentiation (up to a constant whichis irrevocably lost when taking derivatives), in the sense thatF(x) = xaddt(F(t))dt+CwhereC=F(a).The second part of the Fundamental Theorem says that differentiation undoesintegration, in the sense thatf(x) =ddx xaf(t)dt,wherefis a continuous function on an open interval Letf(x) =11+x4+a, and letFbe an antiderivative off, so thatF = thatFhas exactly one critical 1. Clemson calculus Letf(x) = 2x1 1 + 20xf(x) by parts to get 20xf(x)dx= 2012x2f (x)dx= 20x22 1 + rest is a straightforward integral after substituting forx3.

2 The solu-tion What function is defined by the equationf(x) = x0f(t)dt+ 1? Letfbe such thatxsin( x) = x20f(t) (4). (x) = x0f(t) (x2) =xsin( x). Take thederivative of both sides to get 2xf(x) = sin( x) + xcos( x) and pluginx= 2 to findf(4) = /2. Kenneth Roblee, Troy Ifa,b,c,dare polynomials, show that x1a(x)c(x)dx x1b(x)d(x)dx x1a(x)d(x)dx x1b(x)c(x)dxis divisible by (x 1) the expression in question byF(x).Fis a know (x 1)4dividesFif and only ifF (1) = 0. Check this bydifferentiation. (Larson ).6. Suppose thatfis differentiable, and thatf (x) is strictly increasing forx 0. Iff(0) = 0,prove thatf(x)/xis strictly increasing forx > (MCMC 2005 ) Suppose thatf: [0, ) [0, ) is a differentiablefunction with the property that the area under the curvey=f(x) fromx=atox=bis equal to the arclength of the curvey=f(x) fromx=atox=b. Given thatf(0) = 5/4, and thatf(x) has a minimum value onthe interval (0, ), find that minimum area under the curvey=f(x) fromx=atox=bis baf(t)dt,and the arclength of the curvey=f(x) fromx=atox=bis ba 1 + (f (t)) , baf(t)dt= ba 1 + (f (t))2dtfor all nonnegativeaandb.]]

3 In particular, we can write x0f(t)dt= x0 1 + (f (t))2dtfor all nonnegativex. Both sides of the above equation define a functionofx, and since they are equal, their derivatives are equal; their derivativesare given by the Second Fundamental Theorem of calculus :ddx( x0f(t)dt)=ddx( x0 1 + (f (t))2dt), ,f(x) = 1 + (f (x)) , we are looking for a functionywhich satisifies the differential equationy= 1 + (y ) equation is separable:y= 1 + (y )2 y2= 1 + (y )2(1) (y )2=y2 1(2) y = y2 1(3) dy y2 1=dx.(4)Integrating both sides yields dy y2 1= dx ln y+ y2 1 =x+C(where the first integral is evaluated using the trig substitutiony= sec and the two arbitrary constants of integration are combined into one con-stant on the right hand side). Next, sincef(0) = 5/4 is positive, we candrop the absolute value, and solve fory:ln(y+ y2 1)=x+C y+ y2 1 =ex+C=Aex(whereA=eC) y2 1 =Aex y(5) y2 1 = (Aex y)2=A2e2x 2 Ayex+y2(6) 1 =A2e2x 2 Ayex(7) 2 Ayex=A2e2x+ 1(8) y=A2e2x+ 12 Aex=A2ex+12Ae x.

4 (9)Usingf(0) = 5/4, we find54=A2+12A A=12or gives two possible functions:y=14ex+e xory=ex+14e latter has a minimum atx= ln 2, which is not positive, so wereject that function. The former has a minimum atx= ln 2, and theyvalue is 1. Note: One could also deduce from the differential equationy = y2 1 that at the minimum value, sincey = 0, they-value mustbe (MCMC 2006 ) Letf(t) andf (t) be differentiable on [a,x] and for eachxsuppose there is a numbercxsuch thata < cx< xand xaf(t)dt=f(cx)(x a).Assume thatf (a)6= 0. Then prove thatlimx acx ax a= (x) = xaf(t) Taylor s expansion ofF(x), we haveF(x) =F(a) + (x a)F (a) +(x a)22F ( x),where xlies strictly betweenaandx, and asxgoes toa, xalso goes toa. We also haveF(a) = 0,F (x) =f(x), andF (x) =f (x). Thus,F(x) = 0 + (x a)f(a) +(x a)22f ( x).By definition,f(cx) =1x aF(x) =f(a) +x a2f ( x).Therefore,f(cx) f(a)x a=12f ( x).On the other hand we can writef(cx) f(a)x aas a productf(cx) f(a)cx a cx ax taking the limits of these asxgoes toa, we getlimx a12f ( x) = limx af(cx) f(a)cx a cx ax gives12f (a) = limx af(cx) f(a)cx alimx acx ax other words,12f (a) =f (a) limx acx ax showslimx acx ax a=12.