Fundamental Theorem of Calculus (College Board AP® Calculus AB): Revision Note

You can use $f\left(x\right)=f\left(a\right)+{\int }_a^{x}f\text{'}\left(t\right)dt$ to write an expression for a function without needing to find the constant of integration. This is especially useful on a calculator question.

For example, suppose you are given $f\text{'}\left(x\right)={\sin }^{3}x$ and $f\left(0\right)=3$ and you are asked to find $f\left(\pi \right)$, then you can just type the following into your calculator:

$f\left(\pi \right)=3+{\int }_0^{\pi }{\sin }^{3}xdx=4.3333...$

Be sure you are familiar with the implications of the second fundamental theorem!

Exam questions will probably not refer explicitly to the theorem, but they will expect you to recognize that:

• ${\int }_a^{x}f\left(t\right)dt$ is an antiderivative of $f$

• and especially that $\frac{d}{dx}\left({\int }_a^{x}f\left(t\right)dt\right)=f\left(x\right)$

The chain rule with the second fundamental theorem of calculus is tested regularly as an MCQ. It can be extended to both limits being functions of $x$.

$\frac{d}{dx}\left({\int }_{h\left(x\right)}^{g\left(x\right)}f\left(t\right)dt\right)=f\left(g\left(x\right)\right)·g\text{'}\left(x\right)-f\left(h\left(x\right)\right)·h\text{'}\left(x\right)$

This can be shown by writing ${\int }_{h\left(x\right)}^{g\left(x\right)}f\left(t\right)dt={\int }_{h\left(x\right)}^{a}f\left(t\right)dt+{\int }_a^{g\left(x\right)}f\left(t\right)dt$.

This can then be rewritten as ${\int }_{h\left(x\right)}^{g\left(x\right)}f\left(t\right)dt={\int }_a^{g\left(x\right)}f\left(t\right)dt-{\int }_a^{h\left(x\right)}f\left(t\right)dt$.

However, this version is rarely tested.

$f$ is a differentiable function which satisfies $f\text{'}\left(x\right)={\left(\ln x\right)}^2$ and $f\left(1\right)=2$.

Use your calculator to find the value of $f\left(5\right)$.

Answer:

Use the fundamental theorem of calculus

${\int }_1^{5}f\text{'}\left(x\right)dx=f\left(5\right)-f\left(1\right)$

Rearrange and substitute the values and expressions into the formula

$\begin{array}{rcl}f\left(5\right)& =& f\left(1\right)+{\int }_1^5{\left(\ln x\right)}^{2}dx\\ & =& 2+4.8570...\\ & =& 6.8570...\end{array}$

$f\left(5\right)=6.857$ (to 3 decimal places)

The graph of a differentiable function $f$ is shown below.

If $h\left(x\right)={\int }_0^{x}f\left(t\right)dt$, which of the following is true?

(A) $h\left(4\right)<h^{\text{'}}\left(4\right)<h^{\text{'}\text{'}}\left(4\right)$

(B) $h\left(4\right)<h^{\text{'}\text{'}}\left(4\right)<h^{\text{'}}\left(4\right)$

(C) $h^{\text{'}}\left(4\right)<h\left(4\right)<h^{\text{'}\text{'}}\left(4\right)$

(D) $h^{\text{'}\text{'}}\left(4\right)<h\left(4\right)<h^{\text{'}}\left(4\right)$

(E) $h^{\text{'}\text{'}}\left(4\right)<h^{\text{'}}\left(4\right)<h\left(4\right)$

Answer:

We need to see what information we can deduce about $h\left(4\right)$, $h^{\text{'}}\left(4\right)$ and $h^{\text{'}\text{'}}\left(4\right)$

$h\left(4\right)={\int }_0^{4}f\left(t\right)dt$, which is calculating the accumulated change between 0 and 4

• $f$ is positive on $\left(0,4\right)$ as the graph is above the $x$-axis for this interval

• So that means that $h\left(4\right)={\int }_0^{4}f\left(t\right)dt>0$

$h^{\text{'}}\left(x\right)=\frac{d}{dx}\left(h\left(x\right)\right)=\frac{d}{dx}\left({\int }_0^{x}f\left(t\right)dt\right)$

• We are told that $f$ is differentiable

  • Therefore $f$ is continuous and the second fundamental theorem of calculus is valid

• By the second fundamental theorem of calculus, $\frac{d}{dx}\left({\int }_a^{x}f\left(t\right)dt\right)=f\left(x\right)$

• Therefore $h^{\text{'}}\left(x\right)=f\left(x\right)$

• So $h^{\text{'}}\left(4\right)=f\left(4\right)=0$, as we can see from the graph

$h^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left(h^{\text{'}}\left(x\right)\right)=\frac{d}{dx}\left(f\left(x\right)\right)=f^{\text{'}}\left(x\right)$

• From the graph we can see that $f$ is a decreasing function in the vicinity of 4

• Therefore $h^{\text{'}\text{'}}\left(4\right)=f^{\text{'}}\left(4\right)<0$

It follows that $h^{\text{'}\text{'}}\left(4\right)<h^{\text{'}}\left(4\right)<h\left(4\right)$

Option (E)