# The Fundamental Theorem of Calculus

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 $g'(x)=f(x)$.
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, $int_1^3 (2t^4-8t^(-1/2)+7/(t^2+1))dt=(2/5 t^5-16sqrt(t)+7tan^(-1)(t))|_1^3=$

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

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