# The Fundamental Theorem of Calculus

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When we introduced definite integrals we computed them according to definition as a limit of Riemann sums and we saw that this procedure is not very easy. In fact there is a much simpler method for evaluating integrals.

We already discovered it when we talked about Area Problem first time.

There we introduced function P(x) whose value is area under function f on interval [a,x] (x can vary from a to b).

Now when we know about definite integrals we can write that P(x)=int_a^xf(t)dt (note that we changes x to t under integral in order not to mix it with upper limit).

Also we discovered Newton-Leibniz formula which states that P'(x)=f(x) and P(x)=F(x)-F(a) where F'=f.

Here we will formalize this result and give another proof because these fact are very important in calculus: they connect differential calculus with integral calculus.

The Fundamental Theorem of Calculus. Suppose f is continuous on [a,b].

1. If P(x)=int_a^x f(t)dt, then P'(x)=f(x).
2. int_a^b f(x)dx=F(b)-F(a) where F is any antiderivative of f, that is F'=f.

Part 1 can be rewritten as d/(dx)int_a^x f(t)dt=f(x), which says that if f is integrated and then the result is differentiated, we arrive back at the original function.

Part 2 can be rewritten as int_a^bF'(x)dx=F(b)-F(a) and it says that if we take a function F, first differentiate it, and then integrate the result, we arrive back at the original function F, but in the form F(b)-F(a).

Fundamental Theorem of Calculus says that differentiation and integration are inverse processes.

Proof of Part 1. Let P(x)=int_a^x f(t)dt. If x and x+h are in the open interval (a,b) then P(x+h)-P(x)=int_a^(x+h)f(t)dt-int_a^xf(t)dt.

Now use adjacency property of integral: int_a^(x+h)f(t)dt-int_a^x f(t)dt=(int_a^x f(t)dt+int_x^(x+h)f(t)dt)-int_a^x f(t)dt=int_x^(x+h)f(t)dt.

Now apply Mean Value Theorem for Integrals:

int_x^(x+h)f(t)dt=n(x+h-x)=nh, where m'<=n<=M' (M' is maximum value and m' is minimum values of f on [x,x+h]).

So, we obtained that P(x+h)-P(x)=nh. If we let h->0 then P(x+h)-P(x)->0 or P(x+h)->P(x).

This proves that P(x) is continuous function.

Without loss of generality assume that h>0.

Since f is continuous on [x,x+h], the Extreme Value Theorem says that there are numbers c and d in [x,x+h] such that f(c)=m and f(d)=M, where m and M are minimum and maximum values of f on [x,x+h].

By comparison property 5 we have m(x+h-x)<=int_x^(x+h)f(t)dt<=M(x+h-h) or mh<=int_x^(x+h)f(t)dt<=Mh.

This can be divided by h>0: m<=1/h int_x^(x+h)f(t)dt<=M or m<=(P(x+h)-P(x))/h<=M.

Finally, f(c)<=(P(x+h)-P(x))/h<=f(d).

This inequality can be proved for h<0 similarly.

Now we let h->0.

Then c->x and d->x since c and d lie between x and x+h.

So, lim_(h->0)f(c)=lim_(c->x)f(c)=f(x) and lim_(h->0)f(d)=lim_(d->x)f(d)=f(x) because f is continuous.

Therefore, from last inequality and Squeeze Theorem we conclude that lim_(h->0)(P(x+h)-P(x))/h=f(x).

But we recognize in left part derivative of P(x), therefore P'(x)=f(x).

Proof of Part 2. We divide interval [a,b] into n subintervals with endpoints x_0(=a),x_1,x_2,...,x_n(=b) and with width of subinterval Delta x=(b-a)/n. Let F be any antiderivative of f. By subtracting and adding like terms, we can express the total difference in the F values as the sum of the differences over the subintervals: F(b)-F(a)=F(x_n)-F(x_0)=

=F(x_n)-F(x_(n-1))+F(x_(n-2))+...+F(x_2)-F(x_1)+F(x_1)-F(x_0)=

=sum_(i=1)^n(F(x_i)-F(x_(i-1))).

Now F is continuous (because it’s differentiable) and so we can apply the Mean Value Theorem to F on each subinterval [x_(i-1),x_i].

Thus, there exists a number x_i^(**) between x_(i-1) and x_i such that F(x_i)-F(x_(i-1))=F'(x_i^(**))(x_i-x_(i-1))=f(x_i^(**)) Delta x.

Therefore, F(b)-F(a)=sum_(i=1)^n f(x_i^(**))Delta x .

Now we take the limit of each side of this equation as n->oo. The left side is a constant and the right side is a Riemann sum for the function f, so F(b)-F(a)=lim_(n->oo) sum_(i=1)^n f(x_i^(**)) Delta x=int_a^b f(x)dx .

This finishes proof of Fundamental Theorem of Calculus.

When using Evaluation Theorem following notation is used: F(b)-F(a)=F(x)|_a^b=[F(x)]_a^b .

We already talked about introduced function P(x)=int_a^x f(t)dt.

We will talk about it again because it is new type of function. It is just like any other functions (power or exponential): for any x int_a^xf(t)dt gives definite number. Sometimes we can represent P(x) in terms of functions we know, sometimes not.

For example, we know that (1/3x^3)'=x^2, so according to Fundamental Theorem of calculus P(x)=int_0^x t^2dt=1/3x^3-1/3*0^3=1/3x^3. Here we expressed P(x) in terms of power function.

But we can't represent in terms of elementary functions, for example, function P(x)=int_0^x e^(x^2)dx, because we don't know what is antiderivative of e^(x^2). What we can do is just to value of P(x) for any given x.

Geometrically P(x) can be interpreted as the net area under the graph of f from a to x, where x can vary from a to b. (Think of g as the "area so far" function).

Example 1. Graph of f is given below. If P(x)=int_0^xf(t)dt, find P(0), P(1), P(2), P(3), P(4), P(6), P(7). Sketch the rough graph of P. We immediately have that P(0)=int_0^0f(t)dt=0. We can see that P(1)=int_0^1 f(t)dt is area of triangle with sides 1 and 2. Therefore, P(1)=1/2 *1*2=1.

We see that P(2)=int_0^2f(t)dt is area of triangle with sides 2 and 4 so P(2)=1/2*2*4=4.

Area from 0 to 3 consists of area from 0 to 2 and area from 2 to 3 (triangle with sides 1 and 4):

P(3)=int_0^3f(t)dt=int_0^2f(t)dt+int_2^3f(t)dt=4+1/2*1*4=6.

Similarly P(4)=P(3)+int_3^4f(t)dt. But area of triangle on interval [3,4] lies below x-axis so we subtract it: P(4)=6-1/2*1*4=4. Now P(5)=P(4)+int_4^5 f(t)dt=4-1/2*1*4=2.

P(6)=P(5)+int_5^6f(t)dt=2+1/2*1*4=4.

Finally, P(7)=P(6)+int_6^7 f(t)dt where int_7^6 f(t)dt is area of rectangle with sides 1 and 4. So, P(7)=4+1*4=8.

Sketch of P(x) is shown below. Example 2. If P(x)=int_1^x t^3 dt , find a formula for P(x) and calculate P'(x).

Using part 2 of fundamental theorem of calculus and table of indefinite integrals we have that P(x)=int_1^x t^3 dt=(t^4/4)|_1^x=x^4/4-1/4.

Now, P'(x)=(x^4/4-1/4)'=x^3. We see that P'(x)=f(x) as expected due to first part of Fundamental Theorem.

Example 3. Find derivative of P(x)=int_0^x sqrt(t^3+1)dt.

Using first part of fundamental theorem of calculus we have that g'(x)=sqrt(x^3+1).

Example 4. Find d/(dx) int_2^(x^3) ln(t^2+1)dt.

Here we have composite function P(x^3). To find its derivative we need to use Chain Rule in addition to Fundamental Theorem.

Let u=x^3 then (du)/(dx)=(x^3)'=3x^2.

d/(dx) int_2^(x^3) ln(t^2+1)dt=d/(du) int_2^u ln(t^2+1) *(du)/(dx)=d/(du) int_2^u ln(t^2+1) *3x^2=

=ln(u^2+1) *3x^2=ln((x^3)^2+1) *3x^2=3x^2ln(x^6+1).

Now, a couple examples concerning part 2 of Fundamental Theorem.

Example 5. Calculate int_0^5e^xdx.

Using part 2 of fundamental theorem of calculus and table of indefinite integrals we have that int_0^5e^x dx=e^x|_0^5=e^5-e^0=e^5-1.

Example 6. Calculate int_0^(pi/2)cos(x)dx.

Using part 2 of fundamental theorem of calculus and table of indefinite integrals (antiderivative of cos(x) is sin(x)) we have that int_0^(pi/2)cos(x) dx=sin(x)|_0^(pi/2)=sin(pi/2)-sin(0)=1.

Example 7. Find int_0^2 (3x^2-7)dx.

Using properties of definite integral we can write that int_0^2(3x^2-7)dx=int_0^2 3x^2dx-int_0^2 7dx=3 int_0^2 x^2dx-7 int_0^2 7dx=

=3 (x^3/3)|_0^2-7*(2-0)=3 (8/3 -0/3)-14=-6.

Example 8. Find int_1^3 ((2t^5-8sqrt(t))/t+7/(t^2+1))dt .

First rewrite integral a bit: int_1^3 ((2t^5-8sqrt(t))/t+7/(t^2+1))dt=int_1^3 (2t^4-8t^(-1/2)+7/(t^2+1))dt

So, int_1^3 (2t^4-8t^(-1/2)+7/(t^2+1))dt=(2/5 t^5-16sqrt(t)+7tan^(-1)(t))|_1^3=

=(2/5 (3)^5-16sqrt(3)+7tan^(-1)(3))-(2/5 (1)^5-16sqrt(1)+7tan^(-1)(1))=

=564/5-16sqrt(3)-(7pi)/4+7tan^(-1)(3)~~88.3327.