TBIL-Calc FTC for derivatives of integrals (IN6)

Section 4.6 FTC for derivatives of integrals (IN6)

Learning Outcomes

Find the derivative of an integral using the Fundamental Theorem of Calculus.

Note 4.6.1.

In this section we extend the Fundamental Theorem of Calculus discussed in Section 4.5 to include taking the derivatives of integrals. We will call this addition to the Fundamental Theorem of Calculus (FTC) part II. First we will introduce part II and then discuss the implications of this addition.

Theorem 4.6.2. The Fundamental Theorem of Calculus (Part II).

If a function \(f\) is continuous on the closed interval \([a,b]\text{,}\) then the area function

\begin{equation*} A(x) = \int_a^x f(t)\,dt \,\,\,\,\, \mathrm{for}\,\,\, a\leq x\leq b, \end{equation*}

is continuous on \([a,b]\) and differentiable on \((a,b)\text{.}\) The area function satisfies \(A'(x) = f(x)\text{.}\) Equivalently,

\begin{equation*} A'(x) = \frac{d}{dx}\int_0^x f(t)\,dt = f(x), \end{equation*}

which means that the area function of \(f\) is an antiderivative of f on \([a,b]\text{.}\)

Activity 4.6.3.

For the following activity we will explore the Fundamental Theorem of Calculus Part II.

(a)

Given that \(A(x) = \int_a^xt^3\,dt\text{,}\) then by the Fundamental Theorem of Calculus Part I,

\(\displaystyle A(x) = x^3-a^3\)
\(\displaystyle A(x) = a^4 - x^4\)
\(\displaystyle A(x) = \frac{1}{4}(x^4 - a^4)\)
\(\displaystyle A(x) = 3x^2\)

(b)

Using what you found for \(A(x)\text{,}\) what is \(A'(x)\)

\(\displaystyle A'(x) = 3x^2\)
\(\displaystyle A'(x) = 4a^3 - 4x^3\)
\(\displaystyle A'(x) = x^3\)
\(\displaystyle A'(x) = 6x\)

(c)

Use the Fundamental Theorem of Calculus Part II to find \(A'(x)\text{.}\) What do you notice between what you got above and using FTC Part II? Which method do you prefer?

\(\displaystyle A'(x) = 3x^2\)
\(\displaystyle A(x) = 4a^3 - 4x^3\)
\(\displaystyle A(x) = x^3\)
\(\displaystyle A(x) = 6x\)

Activity 4.6.4.

Given \(A(x) = \int_x^be^t\,dt\text{,}\) what is \(A'(x)\text{?}\)

\(\displaystyle A'(x) = -e^x\)
\(\displaystyle A'(x) = e^x\)
\(\displaystyle A'(x) = e^b-e^x\)
\(\displaystyle A'(x) = e^x-e^b\)

Observation 4.6.5.

For the first two activities we have only explored when the function of the limits of the integrand are \(x\text{.}\) Now we want to see what happens when the limits are more complicated. To do this we will follow a similar procedure as that done in activity 1.

Activity 4.6.6.

Recall that by the Fundamental Theorem of Calculus Part I, \(\int_a^bf(t)\,dt = F(b)-F(a)\text{.}\)

(a)

Let \(A(x) = \int_x^{x^2}f(t)\,dt\) and re-write using FTC Part I.

(b)

Using what you got find \(A'(x)\text{.}\) Explain what derivative rule(s) you used.

(c)

Using what you found what is the derivative of \(A(x) = \int_x^{x^2}(t+2)\,dt\text{?}\)

\(\displaystyle A'(x) = 2x(x+2)-(x+2)\)
\(\displaystyle A'(x) = (x+2)-2x(x^2+2)\)
\(\displaystyle A'(x) = (x^2+2)-(x+2)\)
\(\displaystyle A'(x) = 2x(x^2+2)-(x+2)\)

Remark 4.6.7.

Now we have some thoughts of how to generalize the FTC Part II when the limts are more complicated.

Theorem 4.6.8. The Fundamental Theorem of Calculus (Part II): Modified.

\begin{equation*} A(x) = \int_{g(x)}^{h(x)} f(t)\,dt \end{equation*}

then

\begin{equation*} A'(x) = \frac{d}{dx}\int_{g(x)}^{h(x)} f(t)\,dt = f(h(x))h'(x) - f(g(x))g'(x), \end{equation*}

where \(g(x)\) and \(h(x)\) are continous differentiable functions.

Activity 4.6.9.

Given \(A(x) = \int_{x^3}^{x^5}(\sin(t) - 2)\,dt\text{,}\) what is \(A'(x)\text{?}\)

\(\displaystyle A'(x) = -e^x\)
\(\displaystyle A'(x) = e^x\)
\(\displaystyle A'(x) = e^b-e^x\)
\(\displaystyle A'(x) = e^x-e^b\)