# 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.

## Answer and Explanation:

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.