6.4 The Fundamental Theorem of Calculus and Accumulation Functions

Learning objectives

—FUN-5.A—Represent accumulation functions using definite integrals• FUN-5.A Represent accumulation functions using definite integrals.• FUN-5.A.1 The definite integral can be used to define new functions.• FUN-5.A.2 If f is a continuous function on an interval containing a, then (d/dx)(∫ₐˣ f(t)dt) = f(x), where x is in the interval.• Enduring understanding FUN-5: The Fundamental Theorem of Calculus connects differentiation and integration.

An accumulation function uses a fixed starting point and a moving endpoint: $A(x)=\int_a^x f(t)\,dt$ . For each input $x$ , $A(x)$ is the signed change accumulated from $a$ to $x$ . The letter $t$ is a dummy integration variable, keeping it distinct from the endpoint $x$ .

A(x)=\int_a^x f(t),dt \quad\Longrightarrow\quad A'(x)=f(x)

If $f$ is continuous on an interval containing $a$ and $x$ , increasing the endpoint by a small amount $\Delta x$ adds approximately $f(x)\Delta x$ to the accumulation. Dividing by $\Delta x$ and taking the limit leaves $f(x)$ . This is the Fundamental Theorem connection between integration and differentiation.

Let $A(x)=\int_0^x(2t+1)\,dt$ . Then $A(0)=0$ , because no interval has yet been accumulated. The Fundamental Theorem gives $A'(x)=2x+1$ . Evaluating directly, $A(x)=x^2+x$ , so $A(2)=6$ and differentiating this expression confirms the same derivative.

$A(x)$ is signed accumulation, not necessarily total geometric area: negative values of $f$ subtract. Also, the stated derivative rule assumes continuity on the relevant interval and a variable upper limit exactly equal to $x$ ; do not silently drop those conditions.

Next Topic

—6.5 Interpreting the Behavior of Accumulation Functions Involving AreaRead