So far, differential calculus and integral calculus have seemed to be two completely separate branches of mathematics. Differential calculus arose from constructing the tangent line to a curve and studying rates of change. Integral calculus arose from calculating the area under a curve. The fundamental theorem of calculus connects these two branches of calculus. Also, we will see why we use $$\int$$ both for indefinite integrals (antiderivatives) and definite integrals.

When $$f(t)$$ is a known continuous function, and $$a$$ and $$b$$ are two constants, the value of the integral $$\int_{a}^{b}f(t)dt$$ is a definite number. Hence, if we replace the upper limit $$b$$ by a variable $$x$$, the definite integral is a function of $$x$$, say $$A(x)$$. Graphically, $A(x)=\int_{a}^{x}f(t)dt=\text{net area under graph of }f\text{ on }[a,x],$ (see Figure 1).

Suppose $$x$$ takes on an increment $$h$$ (Figure 2). Then

$A(x+h)=\int_{a}^{x+h}f(t)dt,$

and the value of $$A$$ changes by

\begin{aligned} A(x+h)-A(x) & =\int_{a}^{x+h}f(t)dt-\int_{a}^{x}f(t)dt\\ & =\int_{a}^{x+h}f(t)dt+\int_{x}^{a}f(t)dt\\ & \begin{equation*}=\int_{x}^{x+h}f(t)dt\small\tag{property 4 in Section 8.2}\end{equation*}\end{aligned}

For small $$h$$, we can see from Figure 3 that $\int_{x}^{x+h}f(t)dt\approx f(x)h.$ So $\frac{A(x+h)-A(x)}{h}\approx f(x)$ As $$h\to0$$, we can replace the approximation symbol by the equality. That is, $\underbrace{\lim_{h\to0}\frac{A(x+h)-A(x)}{h}}_{=A'(x)}=f(x).$ In other words, we have shown that $A'(x)=\frac{d}{dx}\int_{a}^{x}f(t)dt=f(x).$ This is the Fundamental Theorem of Calculus:

Theorem 1. (The Fundamental Theorem of Calculus, Part 1): If $$f$$ is a continuous function on the interval $$[a,b]$$, then the function $$A(x)$$ defined by $A(x)=\int_{a}^{x}f(t)dt\quad(a\leq x\leq b)$ is continuous on $$[a,b]$$ and differentiable on $$(a,b)$$, and $$A'(x)=f(x)$$.

• The Fundamental Theorem of Calculus is often written as $\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\int_{a}^{x}f(t)dt=f(x).}$

The Fundamental Theorem of Calculus provides a simple method for evaluating definite integral. The following theorem is sometimes called the second part of the Fundamental Theorem of Calculus.

Let $A(x)=\int_{a}^{x}f(t)dt.$ We know from the first part of the Fundamental Theorem of Calculus that $$A'(x)=f(x)$$; that is, $$A(x)$$ is an antiderivative of $$f(x)$$ or $$A(x)$$ is an integral function of $$f(x)$$. If $$F(x)$$ is any antiderivative of $$f(x)$$ on $$[a,b]$$, then it follows from this theorem that  $$F$$ and $$A$$ differ by a constant; that is, there is a constant $$C$$ such that $F(x)=A(x)+C.$ So \begin{aligned} F(b)-F(a) & =\left(A(b)+C\right)-\left(A(a)+C\right)\\ & =A(b)-A(a)\\ & =\int_{a}^{b}f(t)dt-\underbrace{\int_{a}^{a}f(t)dt}_{=0}\\ & =\int_{a}^{b}f(t)dt.\end{aligned} Therefore:

Theorem 2. (The Fundamental Theorem of Calculus, Part 2) Suppose $$f$$ is a continuous function on the interval $$[a,b]$$ and $$F$$ is any antiderivative of $$f$$; that is, $$F'(x)=f(x)$$. Then $\int_{a}^{b}f(x)dx=F(b)-F(a).$ .

• A Useful Notation: The difference $$F(b)-F(a)$$ is often denoted by $\left.F(x)\right|_{a}^{b}\qquad\text{or}\qquad\left.F(x)\right]_{a}^{b}.$ If $$F$$ has more than one term and there might be a confusion about the involved terms, we use $\left[F(x)\right]_{a}^{b}.$ To emphasize that $$a$$ and $$b$$ are values for the variable $$x$$, we may write $\left[F(x)\right]_{x=a}^{x=b}.$

The second part of the Fundamental Theorem of Calculus says to compute the definite integral of a function:

First: Find the indefinite integral of the given function $$f$$.

Second: Substitute in the indefinite integral first the upper limit and then the lower limit for the variable of integration, and subtract the last result from the first.

So the second part of the Fundamental Theorem of Calculus can be written as $\bbox[#F2F2F2,5px,border:2px solid black]{\int_{a}^{b}f(x)dx=\left[\int f(x)dx\right]_{a}^{b}.}$ Now it makes sense why we use the same symbol $$\int$$ for both the indefinite integral and the definite integral.

Notice that

In the indefinite integral, $$\int$$ has no upper and lower limits. The indefinite integral is a function.

In the definite integral, $$\int$$ has upper and lower limits. The definite integral is a number.

Example 1

Find $${\displaystyle \int_{1}^{3}x^{2}dx}$$.

Solution 1

Because the indefinite integral of $$x^{2}$$ is $\int x^{2}dx=\frac{1}{3}x^{3}+C,$ it follows from the second part of the Fundamental Theorem of Calculus that $\int_{1}^{3}x^{2}dx=\left[\frac{1}{3}x^{3}\right]_{1}^{3}=\frac{1}{3}(3^{3}-1^{3})=\frac{26}{3};$ this is exactly the same result that we obtained using the Riemann sum in the first section of this chapter.

Notice that $$x^{3}/3$$ is a particular antiderivative (also called an integral function) of $$x^{2}$$. Instead of $$x^{3}/3$$ we could use another antiderivative such as $$x^{3}/3+2$$, $$x^{3}/3-\sqrt{2}$$, and so on, but the constant does not affect the result. Let’s try this out and see what will happen if instead of $$x^{3}/3$$ we use $$x^{3}/3-\sqrt{2}$$: \begin{aligned} \left[\frac{1}{3}x^{3}-\sqrt{2}\right]_{1}^{3} & =\left(\frac{1}{3}\times3^{3}-\sqrt{2}\right)-\left(\frac{1}{3}\times1^{3}-\sqrt{2}\right)\\ & =\frac{1}{3}\times3^{3}-\frac{1}{3}\times1=\frac{26}{3}.\end{aligned}

To apply the second part of the Fundamental Theorem of Calculus, we just need to use a particular antiderivative (also called a particular integral function). Using the most general antiderivative is not necessary.

Example 2

Find $${\displaystyle {\displaystyle \int_{0}^{\pi}\sin x\,dx}}$$.

Solution 2

\begin{aligned} \int_{0}^{\pi}\sin x\:dx & =-\cos x\big|_{0}^{\pi}\\ & =-\cos\pi-(-\cos0)\\ & =-(-1)-(-1)\\ & =2.\end{aligned}

Example 3

Find $${\displaystyle \int_{1}^{\sqrt{3}}\frac{1}{1+x^{2}}dx}.$$

Solution 3

Because $\int\frac{1}{1+x^{2}}dx=\arctan x+C,$ we have \begin{aligned} \int_{1}^{\sqrt{3}}\frac{1}{1+x^{2}}dx & =\left.\arctan x\right|_{1}^{\sqrt{3}}\\ & =\arctan\sqrt{3}-\arctan1\\ & =\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}.\end{aligned}

[Some books use $$\tan^{-1}$$ instead of $$\arctan$$ for the inverse of the tangent function.]
Example 4

Find $${\displaystyle \int_{e}^{8}\frac{1}{x}\,dx}.$$

Solution 4

Because $\int\frac{1}{x}dx=\ln|x|+C,$ we have \begin{aligned} \int_{e}^{8}\frac{1}{x}dx & =\ln|x|{\Large |}_{e}^{8}\\ & =\ln8-\ln e\\ & =\ln2^{3}-\underbrace{\ln e}_{1}\\ & =3\ln2-1\qquad\qquad\qquad\small{(\ln a^{b}=b\ln a)}\end{aligned}

Example 5

Find ${\displaystyle \int_{0}^{1}\left(x^{3}-2\sqrt{x}+\frac{4}{\sqrt{1-x^{2}}}-3e^{2x}\right)dx}.$

Solution 5
First, we need to find the indefinite integral of $x^{3}-2\sqrt{x}+\frac{4}{\sqrt{1-x^{2}}}-3e^{2x}$:
\begin{align}
\int\left(x^{3}-2\sqrt{x}+\frac{4}{\sqrt{1-x^{2}}}-3e^{2x}\right)&=\int x^{3}dx-2\int x^{1/2}dx+4\int\frac{1}{\sqrt{1-x^{2}}}dx-3\int e^{2x}dx\\
&=\frac{1}{4}x^{4}-2\times\frac{2}{3}x^{3/2}+4\arcsin x-\frac{3}{2}e^{2x}.
\end{align}
Therefore,
\begin{align}
\int_{0}^{1}&\left(x^{3}-2\sqrt{x}+\frac{4}{\sqrt{1-x^{2}}}-3e^{2x}\right)dx=\left[\frac{1}{4}x^{4}-2\times\frac{2}{3}x^{3/2}+4\arcsin x-\frac{3}{2}e^{2x}\right]_{0}^{1}\\&=\left(\frac{1}{4}-\frac{4}{3}+4\underbrace{\arcsin1}_{\pi/2}-\frac{3}{2}e^{2}\right)-\left(0-0+4\underbrace{\arcsin0}_{0}-\frac{3}{2}\underbrace{e^{0}}_{1}\right)\\&=\frac{5}{12}+2\pi-\frac{3}{2}e^{2}
\end{align}

To use the second part of the Fundamental Theorem of Calculus, the integrand $$f$$ must be continuous between $$a$$ and $$b$$: No holes, jumps, or vertical asymptotes (where the distance between the graph and a vertical line constantly gets smaller) are allowed.

Therefore, because $$f(x)=1/x^{2}$$ has a vertical asymptote at $$x=0$$, we cannot write $\xcancel{\int_{-2}^{1}\frac{1}{x^{2}}dx=\left[\frac{x^{-1}}{-1}\right]_{-2}^{1}=-\frac{1}{1}-\left(-\frac{1}{-2}\right)=-\frac{3}{2}.}$

If there is a jump in the graph of $$f$$ at $$x=c$$ ($$a<c<b$$) as shown in Figure 4, we can simply write

$\int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx,$ and apply the second part of the Fundamental Theorem of Calculus to each of the integrals on the right hand side.

This technique does not work if $$f$$ has a vertical asymptote at $$x=c$$ (Figure 5).

Example 6

Given $f(x)=\begin{cases} 2x-1 & -1\leq x\leq1\\ -x+1 & 1<x\leq2 \end{cases},$ find $${\displaystyle \int_{-1}^{2}f(x)dx}$$.

Solution 6

We write \begin{aligned} \int_{-1}^{2}f(x)dx & =\int_{-1}^{1}f(x)dx+\int_{1}^{2}f(x)dx\\ & =\int_{-1}^{1}(2x-1)dx+\int_{1}^{2}(-x+1)dx\\ & =\left[x^{2}-x\right]_{-1}^{1}+\left[-\frac{1}{2}x^{2}+x\right]_{1}^{2}\\ & =(1-1)-(1-(-1))+(-2+2)-(-1/2+1)\\ & =-\frac{5}{2}\end{aligned} Also, from the graph of $$f$$, we get the same result (Figure 6): \begin{aligned} \int_{-1}^{2}f(x)dx & =-A_{1}+A_{2}-A_{3}\\ & =-\frac{1.5\times3}{2}+\frac{0.5\times1}{2}-\frac{1\times1}{2}\\ & =-\frac{5}{2}.\end{aligned}