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

**Why some people say it's true:**

This is true for \(F(x) =\dfrac 12 x^2 \) and \(f(x) = x\). The working is as follows:

\[\displaystyle \int_1^2 x \,dx = \left . \dfrac{x^{1+1}}{1+1} \right |_1^2 = \dfrac12 \left(2^2 - 1^2\right) = \dfrac32. \]

By definition, this checks out that the area enclosed is a trapezium. Calculating the area without using calculus also confirms this value.

**Why some people say it's false:**

This is true for \(F(x) =\ln|x| \) and \(f(x) = \dfrac1x\). The working is as follows:

\[\displaystyle \int_{-1}^1 \dfrac{dx}{x} = \bigg. \ln|x| \bigg|_{-1}^1 = 0 \] is not true, because \( \dfrac1x\) is not defined when \(x=0 \).

The statement is \( \color{red} {\textbf{false}}\). The claim is true if and only if the integrand is defined over the entire domain of \((a,b) \). This is due to the fact that we are applying the Fundamental Theorem of Calculus.

Rebuttal:But it's true for \( \displaystyle \int_1^2 x \,dx = \dfrac32 \) as shown above. So it must always work on all integrals. Thus by the definition of the fundamental theorem of calculus, our claim must be true.

Reply:The integration used in the rebuttal is true only because \(f(x) = x\) is well defined on the interval \((1,2) \). That's why we can integrate like that. In other words, we have only shown that it's true for a specific case. Furthermore, the fundamental theorem of calculus cannot be applied here because \(f = \frac1x\) is not continuous in that given domain of \((-1,1) \).

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

**See Also**

**Cite as:**If the derivative of \(F(x)\) is \(f(x)\), is it true that \(\int _a^b f(x) \, dx = F(b) -F(a) \)?.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/if-the-derivative-of-fx-is-fx-is-it-true-that-int/