# How does one go about proving the fundamental theorem of calculus?

## Question:

How does one go about proving the fundamental theorem of calculus?

## The Fundamental Theorem of Calculus:

The Fundamental Theorem of Calculus is used every time that an integral is evaluated, either as an indefinite integral without limits, or as a definite integral with specified limits. The theorem has two parts. The theorem is stated below:

{eq}\text{The Fundamental Theorem of Calculus:} {/eq}

1) Let {eq}f {/eq} be a continuous function defined on the real interval {eq}[a,b] {/eq}.

If the function {eq}F {/eq} is defined on {eq}[a,b] {/eq} as

{eq}\displaystyle F(x) = \int_a^x f(t) \; dt {/eq}

then {eq}F {/eq} is an antiderivative of {eq}f {/eq}. i.e. {eq}F'(x) = f(x) ,\; x \in[a,b] {/eq}.

2) If {eq}G {/eq} is an antiderivative of the continuous function {eq}f {/eq} defined on {eq}[a,b] {/eq} then,

{eq}\displaystyle \int_a^b f(x) \;dx = \bigg[ G(x) \bigg]_a^b = G(b) - G(a) {/eq}.

We use other theorems such as the Average Value theorem in our proof of the Fundamental Theorem of Calculus.

The proof ought to be in any Calculus textbook. However we repeat the proof as it is asked as a question.

Proof of the Fundamental Theorem of Calculus:

1) By the definition of the derivative of the function {eq}F(x) {/eq} we have,

{eq}\begin{align*} F'(x) &= \lim\limits_{h \to 0} \frac{ F(x+h) - F(x)}{x + h - x} \\ \end{align*} {/eq}

By definition of the function {eq}F(x) {/eq} in the statement of the theorem, we have,

{eq}\begin{align*} F(x) &= \int_a^x f(t) \; dt \\ \end{align*} {/eq}

Applying this to the definition of the derivative of {eq}F(x) {/eq} we get,

{eq}\begin{align*} F'(x) &= \lim\limits_{h \to 0} \frac{ 1}{h} \bigg[ \int_a^{x+h} f(t) \; dt - \int_a^{x} f(t) \; dt \bigg] \; &&\text{(1)}\\ \end{align*} {/eq}

The integrals over subintervals can be added to form the integral over the union of the intervals, therefore we have,

{eq}\begin{align*} \int_a^{x+h} f(t) \; dt &= \int_a^{x} f(t) \; dt +\int_x^{x+h} f(t) \; dt \end{align*} {/eq}

Hence we can write the equation (1) as,

{eq}\begin{align*} F'(x) &= \lim\limits_{h \to 0} \frac{ 1}{h} \bigg[ \int_a^{x} f(t) \; dt + \int_x^{x+h} f(t) \; dt - \int_a^{x} f(t) \; dt \bigg] \\ F'(x) &= \lim\limits_{h \to 0} \frac{ 1}{h} \bigg[ \int_x^{x+h} f(t) \; dt \bigg] \; &&\text{(2)}\\ \end{align*} {/eq}

By the Average Value Theorem, the integral

{eq}\displaystyle \frac{ 1}{h} \bigg[ \int_x^{x+h} f(t) \; dt \bigg] = f(\bar{t}) {/eq},

the average value of the function {eq}f(x) {/eq} for some {eq}\bar{t} \in [x, x+ h] {/eq}

Therefore we can write equation (2) as,

{eq}\begin{align*} F'(x) &= \lim\limits_{h \to 0} f(\bar{t}) \; &&\text{(3)}\\ \end{align*} {/eq}

Since, {eq}\bar{t} \in [x, x + h] {/eq}, we must have, {eq}h \to 0 \Longrightarrow \bar{t} \to x {/eq}.

Therefore equation (3) becomes,

{eq}\begin{align*} F'(x) &= \lim\limits_{\bar{t} \to x} f(\bar{t}) = f (x)\\ \end{align*} {/eq}

Hence we have proved part 1) of the theorem.

2) We are given that {eq}G(x) {/eq} is an antiderivative of the function {eq}f(x) , \; x \in [a,b] {/eq}. From part 1) we are also given that {eq}F(x) {/eq} is an antiderivative of the function {eq}f(x), \; x \in [a,b] {/eq}.

Hence we must have the relation,

{eq}G(x) = F(x) + C {/eq}

for some constant of integration, {eq}C {/eq}.

To find the value of {eq}C {/eq} we evaluate the function at the end points {eq}[a, b] {/eq}, using the definition of {eq}F(x) {/eq}.

We have,

{eq}\begin{align*} F(x) &= \int_a^x f(t) \; dt \\ F(a) &= \int_a^a f(t) \; dt = 0 \end{align*} {/eq}

Substituting the above result in the relation we get,

{eq}G(a) = F(a) + C \Longrightarrow G(a) = C {/eq}

From the definition of the function, we also have,

{eq}\begin{align*} F(x) &= \int_a^x f(t) \; dt \\ F(b) &= \int_a^b f(t) \; dt \end{align*} {/eq}

Substituting the above result in the relation we get,

{eq}\begin{align*} G(b) &= F(b) + C \Longrightarrow G(b) = \int_a^b f(t) \; dt + G(a) \\ G(b) - G(a) &= \int_a^b f(t) \; dt \end{align*} {/eq}

Thus we have proved the second part of the theorem.