# Fundamental Theorem of Calculus

### Later, on this page

Here we present two related fundamental theorems involving differentiation and integration, followed by an applet where you can explore what it means.

You can see some background on the Fundamental Theorem of Calculus in the Area Under a Curve and Definite Integral sections.

For this section, we assume that:

Some function `f` is continuous on a closed interval `[a,b]`

This means the curve has no gaps within the interval `x=a` and `x=b`, and those endpoints are included in the interval.

## First Fundamental Theorem

Given the condition mentioned above, consider the function `F` (upper-case "F") defined as:

`F(x) = int_a^xf(t)dt`

(Note in the integral we have an upper limit of `x`, and we are integrating with respect to variable `t`.)

The first Fundamental Theorem states that:

(1) Function `F` is also continuous on the closed interval `[a,b]`;

(2) Function `F` can be differentiated on the open interval `(a,b)`; and

(3) `F'(x)=f(x)` That is, the derivative of `F(x)` is `f(x)`.

Proof

Suppose `x` and `x+h` are values in the open interval `(a,b)`.

Since we defined `F(x)` as `int_a^xf(t)dt`, we can write:

`F(x+h)-F(x) ` `= int_a^(x+h)f(t)dt - int_a^xf(t)dt`

We can re-express the first integral on the right as the sum of 2 integrals (note the upper and lower limits), and simplify the whole thing as follows:

`F(x+h)-F(x) = (int_a^x f(t)dt + int_x^(x+h)f(t)dt) ` `- int_a^xf(t)dt`

`= int_x^(x+h)f(t)dt`

Now divide both sides by `h>0`:

`(F(x+h)-F(x))/h = 1/h int_x^(x+h)f(t)dt`

Now, for any curve in the interval `(x,x+h)` there will be some value `c` such that `f(c)` is the absolute minimum value of the function in that interval, and some value `d` such that `f(d)` is the absolute maximum value of the function in that interval. (This is a consequence of what is called the Extreme Value Theorem.)

So we can write:

`f(c) <= (F(x+h)-F(x))/h <= f(d)`

Now if `h` becomes very small, both `c` and `d` approach the value `x`. (They get "squeezed" closer to `x` as `h` gets smaller).

So we can write the following limits:

`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)`

Since our expressions are being squeezed on both sides to the value `f(x)`, we can conclude:

`lim_(h->0)(F(x+h)-F(x))/h = f(x)`

But we recognize the limit on the left is the definition of the derivative of `F(x)`, so we have proved that `F(x)` is differentiable, and that `F'(x) = f(x)`. Also, since `F(x)` is differentiable at all points in the interval `(a,b)`, it is also continuous in that interval.

**Note:** When integrating, it doesn't really make any difference what variable we use, so it's OK to use `t` or `x` interchangeably, as long as we are consistent.

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## Second Fundamental Theorem

We continue to assume `f` is a continuous function on `[a,b]` and `F` is an antiderivative of `f` such that `F'(x)=f(x)`.

The Second Fundamental Theorem of Calculus states that:

`int_a^bf(x)dx = F(b) - F(a)`

This part of the Fundamental Theorem connects the powerful algebraic result we get from integrating a function with the graphical concept of areas under curves.

To find the area we need between some lower limit `x=a` and an upper limit `x=b`, we find the total area under the curve from `x=0` to `x=b` and subtract the part we don't need, the area under the curve from `x=0` to `x=a`.

`G(b) + G(a) = (F(b)+K)` ` - (F(a) + K)`

`=F(b) - F(a)`

`=int_a^bf(t)dt - int_a^af(t)dt`

`=int_a^bf(t)dt - 0`

`=int_a^bf(t)dt`

Proof

From the First Fundamental Theorem, we had that `F(x) = int_a^xf(t)dt` and `F'(x) = f(x)`.

Suppose `G(x)` is any antiderivative of `f(x)`. (Remember, a function can have an infinite number of antiderivatives which just differ by some constant, so we could write `G(x) = F(x) + K`.)

So we'll have:

`G'(x) = F'(x)`

Now, since `G(x) = F(x) + K`, we can write:

`G(b) - G(a) = (F(b)+K)` ` - (F(a) + K)`

`=F(b) - F(a)`

`=int_a^bf(t)dt - int_a^af(t)dt`

`=int_a^bf(t)dt - 0`

`=int_a^bf(t)dt`

So we've proved that `int_a^bf(x)dx = F(b) - F(a)`.

**Note:** Once again, when integrating, it doesn't really make any difference what variable we use, so it's OK to use `t` or `x` interchangeably, as long as we are consistent.

## Fundamental Theorem of Calculus Applet

You can use the following applet to explore the Second Fundamental Theorem of Calculus.

### Things to Do

This applet has two functions you can choose from, one linear and one that is a curve. You can:

- Choose either of the functions.
- Drag the sliders left to right to change the lower and upper limits for our integral.
- Observe the resulting integration calculations underneath the graph slider. to the right of the graph.

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### Fundamental Theorem of Calculus

`int_a^b f(x)dx = [F(x)]_a^b`

`=F(b)-F(a)`

In this case, the function:

## Exercises

Find the derivatives:

1. `d/dx int_5^x (t^2 + 3t - 4)dt`

Answer

Recall from the First Fundamental Theorem, that if `F(x) = int_a^xf(t)dt`, then `F'(x)=f(x)`.

In this example, `f(t) = t^2+3t-4`.

So:

`d/dx int_5^x (t^2 + 3t - 4)dt = x^2 + 3x - 4`.

That's all there is too it. We don't need to integrate the expression after the integral sign (the integrand) first, then differentiate the result. We can write down the derivative immediately.

### Doing it from scratch

However, let's do it the long way round to see how it works.

`int_5^x (t^2 + 3t - 4)dt = [t^3/3 + (3t^2)/2 - 4t]_5^x`

`=[x^3/3 + (3x^2)/2 - 4x ] -` ` [5^3/3 + (3(5)^2)/2 - 4(5)]`

`=[x^3/3 + (3x^2)/2 - 4x] - [59.167]`

`=x^3/3 + (3x^2)/2 - 4x - 59.167`

Next, we take the derivative of this result, with respect to `x`:

`d/dx(x^3/3 + (3x^2)/2 - 4x - 59.167) ` `= x^2 +3x - 4`

This is the same result we obtained before. Notice it doesn't matter what the lower limit of the integral is (in this case, `5`), since the constant value it produces (in this case, `59.167`) will disappear during the differentiation step.

2. `d/dx int_m^x t sin(t^t)dt`

Answer

In this question, `f(t) = t sin(t^t)`

So the derivative is just:

`d/dx int_m^x t sin(t^t)dt = x sin(x^x)`

We haven't learned to integrate cases like `int_m^x t sin(t^t)dt`, but we don't need to know how to do it. (Actually, this integral is impossible using ordinary functions, but we can find its derivative easily.)

Note the constant `m` doesn't make any difference to the final derivative.

3. `d/dx int_0^x t sqrt(1+t^3)dt`

Answer

This time, `f(t) = t sqrt(1+t^3)`

So `d/dx int_0^x t sqrt(1+t^3)dt = x sqrt(1+x^3)`

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