The Fundamental Theorem of Calculus

This theorem bridges the antiderivative concept with the area problem. Indeed, let f (x) be a function defined and continuous on [a, b]. Consider the function

F(x) = $$\int_{a}^{x}$$f (t)dt

defined on [a, b]. Then we have

$${\frac{F(x+h) - F(x)}{h}}$$ = $${\frac{1}{h}}$$$$\int_{x}^{x+h}$$f (t)dt  .

Since $$\int_{x}^{x+h}$$dt = h, we get

$${\frac{F(x+h) - F(x)}{h}}$$ - f (x) = $${\frac{1}{h}}$$$$\int_{x}^{x+h}$$$$\Big($$f (t) - f (x)$$\Big)$$dt  .

Since F(t) is continuous, we can easily prove that

$$\lim_{h \rightarrow 0}^{}$$$${\frac{F(x+h) - F(x)}{h}}$$ - f (x) = 0  .

This means that the function F(x) is differentiable and F '(x) = f (x). In other words, the function F(x) is an antiderivative of f (x). From this and what we learned about antiderivatives, we obtain the following fundamental result:

The Fundamental Theorem of Calculus Let f (x) be continuous on [a, b]. If F(x) is any antiderivative of f (x), then

$$\int_{a}^{b}$$f (x)dx = F(b) - F(a)  .

Remark. The number F(b) - F(a) is also denoted by [F(x)]^b_a (or $$\left.\vphantom{ F(x)\,}\right.$$F(x) $$\left.\vphantom{ F(x)\,}\right\vert _{a}^{b}$$). So

$$\int_{a}^{b}$$f (x)dx = [F(x)]^b_a

Example. We have

$$\int_{0}^{x}$$cos(t)dt = [sin(t)]₀^x = sin(x) - sin(0) = sin(x)  .

So

$$\int_{0}^{\pi/2}$$cos(t)dt = sin$$\left(\vphantom{\frac{\pi}{2}}\right.$$$${\frac{\pi}{2}}$$$$\left.\vphantom{\frac{\pi}{2}}\right)$$ = 1  .

Example. We have

$$\int_{0}^{x}$$$${\frac{1}{1 + t^2}}$$dt = [arctan(t)]₀^x = arctan(x)  .

In particular, we have

$$\int_{0}^{1}$$$${\frac{1}{1 + x^2}}$$dt = arctan(1) = $${\frac{\pi}{4}}$$   ^.

Combining the Chain Rule with the Fundamental Theorem of Calculus, we can generate some nice results. Indeed, let f (x) be continuous on [a, b] and u(x) be differentiable on [a, b]. Define the function

F(x) = $$\int_{a}^{u(x)}$$f (t)dt  .

Then the Chain Rule implies that F(x) is differentiable and

F '(x) = f$$\Big($$u(x)$$\Big)$$u '(x)  .

We can generalize this a little bit more to find the derivative of a function of the form

H(x) = $$\int_{u(x)}^{v(x)}$$f (t)dt

where u(x) and v(x) are both differentiable on [a, b]. We have

H '(x) = f$$\Big($$v(x)$$\Big)$$v '(x) - f$$\Big($$u(x)$$\Big)$$u '(x)  .

Example. Find the derivative of

$$\int_{0}^{x^2}$$cos(t²)dt  .

Set

F(x) = $$\int_{0}^{x}$$cos(t²)dt  .

From the Fundamental Theorem of Calculus, we know that F(x) is an antiderivative of cos(x²). We have

$$\int_{0}^{x^2}$$cos(t²)dt = F(x²)  .

The Chain Rule then implies that

$${\frac{d}{dx}}$$$$\left(\vphantom{\int_0^{x^2} \cos(t^2) dt}\right.$$$$\int_{0}^{x^2}$$cos(t²)dt$$\left.\vphantom{\int_0^{x^2} \cos(t^2) dt}\right)$$ = F '(x²)2x = 2x cos$$\Big($$(x²)²$$\Big)$$ = 2x cos(x⁴)  .

Note that F(x) does not have an explicit form. So it is quite amazing that even if F(x) is defined via some theoretical result, we are still able to find the derivative of the given function.

Example. Find the derivative of

$$\int_{x}^{x^2}$$e^-t2dt  .

Set

F(x) = $$\int_{0}^{x}$$e^-t2dt  .

From the Fundamental Theorem of Calculus, we know that F(x) is an antiderivative of e^-x2. We have

$$\int_{x}^{x^2}$$e^-t2dt = F(x²) - F(x)  .

The Chain Rule implies

$${\frac{d}{dx}}$$$$\left(\vphantom{\int_x^{x^2} e^{-t^2} dt}\right.$$$$\int_{x}^{x^2}$$e^-t2dt$$\left.\vphantom{\int_x^{x^2} e^{-t^2} dt}\right)$$ = F '(x²)2x - F '(x) = 2xe^-x4 - e^x2  .

Exercise 1. Let f (x) be a function defined and continuous on [a, b]. Compare

$${\frac{d}{dx}}$$$$\left(\vphantom{\int_a^x f(t) dt}\right.$$$$\int_{a}^{x}$$f (t)dt$$\left.\vphantom{\int_a^x f(t) dt}\right)$$    and    $$\int_{a}^{x}$$$${\frac{d}{dx}}$$$$\Big($$f (t)$$\Big)$$dt  .

Answer.

Exercise 2. Find a function F(x) such that F '(x) = sin(x²) and F(1) = 2.

Answer.

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