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The Fundamental Theorem of Calculus.

The two main concepts of calculus are integration and differentiation. The Fundamental Theorem of Calculus (FTC) says that these two concepts are es- sentially inverse to one another.

The fundamental theorem states that if F has a continuous derivative on an interval [a, b], then

abF(t)dt=F(b)-F(a) .

This form allows one to compute integrals by finding anti-derivatives.

The FTC says that integration undoes differentiation (up to a constant which is irrevocably lost when taking derivatives), in the sense that

F(x)=axddt(F(t))dt+C

where C=F(a) .

The second part of the fundamental theorem says that differentiation undoes integration, in the sense that

f(x)=ddxaxf(t)dt,

where f is a continuous function on an open interval containing a and x.

Problems

1. Let f(x)=11+x4+a, and let F be an antiderivative of f, so that F=f. Find a so that F has exactly one critical point.

Solution. a=-1. Clemson Calculus Competition.

2. Let

f(x)=x211+t3dt.

Find

02xf(x)dx.

Solution. Integrate by parts to get

02xf(x)dx=-0212x2f(x)dx=02x221+x3dx.

The rest is a straightforward integral after substituting for x3. The solu- tion is 23.

3. What function is defined by the equation

f(x)=0xf(t)dt+1?

Solution. Larson 6.9.7

4. Let f be such that

xsin(πx)=0x2f(t)dt.

Find f(4) .

Solution. Put F(x)=0xf(t)dt. Then F(x2)=xsin(πx) . Take the derivative of both sides to get 2xf(x)=sin(πx)+πxcos(πx) and plug in x=2 to find f(4)=π/2. Kenneth Roblee, Troy U.

5. If a, b, c, d are polynomials, show that

1xa(x)c(x)dx1xb(x)d(x)dx-1xa(x)d(x)dx1xb(x)c(x)dx

is divisible by (x-1)4.

Solution. Denote the expression in question by F(x) . F is a polynomial. We know (x-1)4 divides F if and only if F(1)=0. Check this by differentiation. (Larson 6.9.3).

6. Suppose that f is differentiable, and that f(x) is strictly increasing for x0. If f(O)=0, prove that f(x)/x is strictly increasing for x>0.

Solution. Larson 6.9.12

7. (MCMC 2005 II.5) Suppose that f:[0, ) [0, ) is a differentiable function with the property that the area under the curve y=f(x) from x=a to x=b is equal to the arclength of the curve y=f(x) from x=a to x=b. Given that f(O)=5/4, and that f(x) has a minimum value on the interval (0, ) , find that minimum value.

Solution. The area under the curve y=f(x) from x=a to x=b is

abf(t)dt,

and the arclength of the curve y=f(x) from x=a to x=b is

ab1+(f(t))2dt.

Therefore,

abf(t)dt=ab1+(f(t))2dt

for all nonnegative a and b. In particular, we can write

0xf(t)dt=0x1+(f(t))2dt

for all nonnegative x. Both sides of the above equation define a function of x, and since they are equal, their derivatives are equal; their derivatives are given by the Second Fundamental Theorem of Calculus:

ddx(0xf(t)dt)=ddx(0x1+(f(t))2dt) ,

i. e.,

f(x)=1+(f(x))2.

So, we are looking for a function y which satisifies the differential equation

y=1+(y)2.

This equation is separable:

y=1+(y)2y2=1+(y)2 (1)

(y)2=y2-1 (2)

y=y2-1 (3)

dyy2-1=dx. (4)

Integrating both sides yields

dyy2-1=dxln|y+y2-1|=x+C

(where the first integral is evaluated using the trig substitution y=secθ and the two arbitrary constants of integration are combined into one con- stant on the right hand side). Next, since f(O)=5/4 is positive, we can drop the absolute value, and solve for y:

ln(y+y2-1)=x+Cy+y2-1=ex+C=Aex (where A=eC)

y2-1=Aex-y

(5)

y2-1=(Aex-y)2=A2e2x-2Ayex+y6)

-1=A2e2x-2Ayex

(7)

2Ayex=A2e2x+1 (8)

y=A2e2x+12Aex=A2ex+12Ae-x. (9)

Using f(O)=5/4, we find

54=A2+12AA=12 or 2.

This gives two possible functions:

y=14ex+e-x or y=ex+14e-x.

This latter has a minimum at x=-ln2, which is not positive, so we reject that function. The former has a minimum at x=ln2, and the y value is 1. Note: One could also deduce from the differential equation y=y2-1 that at the minimum value, since y=0, the y-value must be 1.

8. (MCMC 2006 I.5) Let f(t) and f(t) be differentiable on [a, x] and for each x suppose there is a number cx such that a<cx<x and

axf(t)dt=f(cx)(x-a) .

Assume that f(a)0. Then prove that

limxacx-ax-a=12.

Solution. Let

F(x)=axf(t)dt.

Using Taylor's expansion of F(x) , we have

F(x)=F(a)+(x-a)F(a)+(x-a)22F(θx) ,

where θx lies strictly between a and x, and as x goes to a, θx also goes to a. We also have F(a)=0, F(x)=f(x) , and F(x)=f(x) . Thus,

F(x)=0+(x-a)f(a)+(x-a)22f(θx) .

By definition,

f(cx)=1x-aF(x)=f(a)+x-a2f'(θx) .

Therefore,

f(cx)-f(a)x-a=12f(θx) .

On the other hand we can write

f(cx)-f(a)x-a

as a product

f(cx)-f(a). cx-a.

cx-a x-a

On taking the limits of these as x goes to a, we get

limxa12f(θx)=limxaf(cx)-f(a)cx-acx-ax-a.

This gives

12f(a)=limxaf(cx)-f(a)cx-alimxacx-ax-a.

In other words,

12f(a)=f(a)limxacx-ax-a.

This shows

limxacx-ax-a=12.