If F(x) is the antiderivative of f(x), is it true that \(\int_a^b\)f(x)dx=F(b)-F(a)?

This is part of a series on common misconceptions.

True or False?

If $F(x)$ is the antiderivative of $f(x),$ is it true that $\displaystyle \int _a^b f(x) \, dx = F(b) -F(a)?$

Why some people say it's true: It's what I was taught at school to calculate proper integrals. You just apply the second fundamental theorem of calculus. When I was asked to compute $\int_1^2 x\ dx,$ I used the formula and got the right answer, which was $\frac{3}{2}.$

Why some people say it's false: The functions might not be continuous over the entire interval.

The statement is $\color{#D61F06} {\textbf{false}}$. The claim is true if and only if the integrand is continuous over the entire domain of $(a,b)$. This is an important condition to satisfy before you apply the second fundamental theorem of calculus.

Rebuttal: It's true for the following example:

$$\int_1^2 x \,dx = \left. \frac{1}{2}x^2 \right|_1^2= \frac{3}{2},$$

so it must always work on all integrals. By the second fundamental theorem of calculus, our claim must be true.

Reply: The integration used in the rebuttal is true only because $f(x) = x$ is continuous on the interval $(1,2)$. That is why we can integrate like that. In other words, we have only shown that it's true for a specific case. If a function is not continuous over an interval, like $f(x)=\frac{1}{x}$ over the interval $(-1,1)$ $\big($note that $f(0)$ is undefined$\big),$ the fundamental theorem of calculus cannot be applied.

Rebuttal: If you can't use the second fundamental theorem of calculus, then how do you calculate integrals that have discontinuities?

Reply: If an integrand contains a discontinuity within the interval, then the integral is undefined. However, the integral can be interpreted as a Cauchy principal value by finding the sum of improper integrals. For example,

$$\int_{-1}^{1}\frac{1}{x}\ dx.$$

The integrand has a point of discontinuity at $x=0,$ and thus the integral is undefined. However, the Cauchy principal value will be computed as a sum of improper integrals:

$$\begin{aligned} PV\ \int_{-1}^{1}\frac{1}{x}\ dx &= \int_{-1}^{0}\frac{1}{x}\ dx + \int_{0}^{1}\frac{1}{x}\ dx \\ \\ &= \left. \lim_{a \rightarrow 0^-}\ln|x| \right|_{-1}^{a} + \left. \lim_{a \rightarrow 0^+} \ln|x| \right|_{a}^{1} \\ \\ &= \left(\lim_{x \rightarrow 0^+}\ln(x) \right) - 0 + 0 - \left(\lim_{x \rightarrow 0^+}\ln(x) \right) \\ \\ &= \lim_{x \rightarrow 0^+}\big(\ln(x)-\ln(x) \big) \\ \\ &= 0. \end{aligned}$$

In this case, the Cauchy principal value matches what we would expect if it was a proper integral. However, this is not always the case. See the improper integrals page for more examples of the Cauchy principal value.

Want to make sure you've got this concept down? Try these problems:

See Also

• List of Common Misconceptions
• Fundamental Theorem of Calculus