section 4.3
Table of Contents
Section 4.3 The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is conveyed in two parts. Just like subtraction is the inverse of addition, and division is the inverse of multiplication, the Fundamental Theorem tells use that integration is the inverse of differentiation.
Integral as a function #
In the previous section, we say the definite integral, which evaluates to a number. We can interpret that number as the area under the function between endpoints, for example.
Consider the following integral with its upper bound replaced by a variable:
$$ F(x) = \int_a^x f(t) \, dt $$
The value of $F(x)$ will tell you the area under the curve $f(t)$ on the interval $[a, x]$ with $x \ge a$.
The Fundamental Theorem of Calculus, part 1 #
(The core idea: the derivative is the inverse of the integral)
If $f$ is a continuous function on $[a, b]$, then the function $F$ defined by $$ F(x) = \int_a^x f(t) \, dt \hspace{2em} a \le x \le b $$ is continuous on $[a, b]$, and differentiable on $(a, b)$ and $F’(x) = f(x)$
Repeating for emphasis: $$ \dfrac{d}{dx} \int_a^x f(t) \, dt = f(x) $$ The derivative of an integral, with respect to its variable upper limit is the integrand evaluated at the upper limit.
Example #
Find the derivative of the function $\displaystyle g(x) = \int_a^x \sqrt{1-t^2}\, dt$
Click to reveal the answer.
$g'(x) = \sqrt{1-x^2}Example #
Find $\displaystyle \frac{d}{dx} \int_1^{x^4} \sec t\, dt$ Hint: use the chain rule
Click to reveal the answer.
$\begin{align*} \frac{d}{dx} \int_1^{x^4} \sec t\\, dt &= \sec(x^4) \cdot \dfrac{d}{dx}(x^4) \\ &= 4x^3 \sec(x^4) \end{align*}$Example #
Find the derivative of $\displaystyle h(x) = \int_x^a t \, dt$ Hint: Look at the order of the bounds.
Click to reveal the answer.
The FTC, part 1 requires that the *upper* bound be the variable. Using properties of the integral, we write: $\begin{align*} \dfrac{d}{dx} \int_x^a t \\, dt &= \dfrac{d}{dx}\left( - \int_a^x t \\, dt \right) \\ &= - \dfrac{d}{dx} \int_a^x t \\, dt \\ &= -x \end{align*}$The Fundamental Theorem of Calculus, part 2 #
(The core idea: the integral is the inverse of the derivative)
If $f$ is continuous on $[a, b]$, then $$ \int_a^b f(x) \, dx = F(b) - F(a) $$ where $F$ is any antiderivative of $f$, that is, a function such that $F’ = f$
Example #
Evaluate the integral $\displaystyle \int_{-1}^2 x^3\, dx$
Solution Since $\dfrac{x^4}{4}$ is an antiderivative of $x^3$, we apply the fundamental theorem of calculus to find: $\begin{align*} \int_{-1}^2 x^3\, dx &= \left. \dfrac{x^4}{4} \right|_{x=-1}^2 \ &= \left(\frac{2^4}{4}\right) - \left(\frac{(-1)^4}{4}\right) \ &= \frac{16}{4} - \frac{1}4 \ &= \frac{15}{4} \end{align*}$
Comment on notation #
If $F$ is an antiderivative of $f$, then we write:
$$ \int_a^b f(x) \, dx = F(x) \Big|_{x = a}^b $$
where the
$$ F(x)\Big|_{x = a}^{b} $$
means $F(b) - F(a)$. Plug in upper value, plug in lower value, and subtract them.
Our textbook omits the $x=\dots$ part on the lower bound, but I do it and encourage you as well. Later it will be very beneficial when we’re trying to keep track of what the actual variable of integration is.
Example #
Evaluate the integral $\displaystyle \int_{0}^3 x^3\, dx$
Example #
Find the area under the sine curve between 0 and $\pi$.
Example #
What is wrong with this work? $$ \int_{-1}^3 \frac{1}{x^2} \, dx = \left. \dfrac{x^{-1}}{-1} \right|_{x=-1}^3 = -\dfrac13 - 1 = -\dfrac43 $$
Hint: Is the integrand continuous?