Introduction

Welcome to this guide on the Fundamental Theorem of Calculus, this article will guide you through the Fundamental Theorem of Calculus and its applications:

Fundamental Theorem of Calculus

Consider a continuous function $f(x)$ on the interval of $[a,b]$, then

${\int }_a^{b}f(x)dx=F(b)-F(a)$

Where $\frac{d}{dx}[F(x)]=f(x)$, function $F(x)$ is known as the antiderivative of $f(x)$. This word will appear multiple times in this lesson and lessons after it.

The theorem above is called the Fundamental Theorem of Calculus (FTC for short). This is a very powerful theorem, as it links differentiation and integration, the two fundamental operations in calculus.

Note this theorem can also be written as 

${\int }_a^{b}f(x)dx=F(x){|}_a^b$

Here $F(x){|}_a^b$ is equivalent to $F(b)-F(a)$

Example 1:

Evaluate ${\int }_0^{\pi /2}\cos xdx$

To evaluate this definite integral, we first need to find a function that has a derivative of $f(x)=\cos x$, recall from Unit 2 that $\sin x$ has a derivative of $\cos x$, thus by the Fundamental Theorem of Calculus:

${\int }_0^{\pi /2}\cos xdx=\sin (\frac{\pi }{2})-\sin (0)=1$

Applications of FTC

Application of Fundamental Theorem comes in various different questions, first we will take a look at finding a function defined with an variable integration bound.

Example 2:

Find the polynomial function defined by $f(x)={\int }_2^x(3t^2-2)dt$

At first glance it seems weird that the variable $x$ is on the upper bound, which sounds absurd. But let's pretend that $x$ is a number and use FTC to solve integral

${\int }_2^x(3t^2-2)dt=t^3-2t{|}_2^x=x^3-2x-(2^3-2\cdot 2)=x^3-2x-4$

Therefore $f(x)=x^3-2x-4$.

This question inspires us to recognize that when a variable is on the integration bound, it means the integral will be a function of the variable, or:

${\int }_a^{x}f(t)dt=F(x)-F(a)$

Where again $\frac{d}{dx}F(x)=f(x)$ and $f(a)$ is a number.

On the exam, you might also encounter questions that ask you to find the value of a function at a point.

Example 3:

$F(x)$ is an antiderivative of $f(x)$, if $F(5)=10$, find an expression that equals to $F(11)$

In order to solve this problem, let's first use FTC:

${\int }_5^{11}f(x)dx=F(11)-F(5)$

Notice here $F(11)$ appears, so we can easily find an expression for $F(11)$:

$F(11)=F(5)+{\int }_5^{11}f(x)dx=10+{\int }_5^{11}f(x)dx$

This type of question can also come in tabular form

Example 4:

Consider a function $f(x)$ and its derivative $f^{\prime }(x)$:

$\begin{array}{ccccc}x& 0& 2& 3& 5\\ f(x)& -15& -8& 4& 7\\ f^{\prime }(x)& 12& 8& 3& -2\end{array}$ 

Find the value of ${\int }_0^5{f}^{\prime }(x)dx$

We know that by FTC:

${\int }_0^5{f}^{\prime }(x)dx=f(5)-f(0)$

From the table, $f(5)=7$ and $f(0)=-15$, thus

${\int }_0^5{f}^{\prime }(x)dx=f(5)-f(0)=7+15=22$

This type of question can also come in the form of graphs.

Example 5:

Consider a function $f(x)$ with a derivative of $f^{\prime }(x)$. If this is the graph of $f^{\prime }(x)$ and $f(-2)=3$, find $f(0)$.

First, by FTC we have

${\int }_{-2}^0{f}^{\prime }(x)dx=f(0)-f(-2)$

Thus

$f(0)=f(-2)+{\int }_{-2}^0{f}^{\prime }(x)dx$

By the geometric meaning of integrals, ${\int }_{-2}^0{f}^{\prime }(x)dx$ is the area under the curve of $f^{\prime }(x)$, which is the triangle formed by the graph of $f^{\prime }(x)$ and the coordinate axis. 

This offers a way to calculate the integral, the area of the triangle is simply $S=\frac{1}{2}\cdot 2\cdot 5=5$, thus the integral also equals to $5$, meaning

$f(0)=f(-2)+{\int }_{-2}^0{f}^{\prime }(x)dx=3+5=8$

Practice