4.3 · Accumulation functions and the Fundamental Theorem

An accumulation function has the integrand as its rate

If $\displaystyle f$ is continuous, $\displaystyle A(x)=\int_a^x f(t)\,dt$ has derivative $\displaystyle A'(x)=f(x)$. The integration variable $\displaystyle t$ is a placeholder; the upper bound $\displaystyle x$ controls how much accumulation is included.

If the upper bound is $\displaystyle g(x)$, the chain rule adds a factor $\displaystyle g'(x)$. A variable lower bound contributes a negative sign. With two variable bounds, differentiate the upper contribution minus the lower contribution.

$$\frac{d}{dx}\int_a^{g(x)}f(t)\,dt=f(g(x))g'(x)$$

Evaluate with an antiderivative

If $\displaystyle F'=f$ on the interval, $\displaystyle \int_a^b f(x)\,dx=F(b)-F(a)$. Evaluate the full antiderivative at each bound before subtracting. A definite integral is a number, so no arbitrary constant remains.

Reversing the bounds changes the sign. Splitting an interval adds the integrals. The bounds must lie within a region where the integral and any antiderivative argument are valid.

$$\int_a^b f(x)\,dx=F(b)-F(a)$$

Combine an initial amount with net change

If $\displaystyle r(t)$ is an inflow rate minus an outflow rate, the amount at $\displaystyle b$ is its initial amount plus $\displaystyle \int_a^b r(t)\,dt$. A negative net change decreases the initial amount.

The signs of $\displaystyle f$ determine whether its accumulation function increases or decreases; the signs of $\displaystyle f'$ determine accumulation concavity. Do not confuse the accumulation function’s height with the integrand’s height.