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The fundamental theorem relates accumulation and rate of change by stating that if $F'(x) = f(x)$, then the accumulated area $A(x)$ from a fixed point $a$ to $x$ is given by $A(x) = F(x) - F(a)$. It shows that differentiation and integration are inverse processes: the derivative of the accumulation function $A(x)$ recovers the integrand $f(x)$, and antiderivatives allow the evaluation of definite integrals.

Conditions: The function $f$ is continuous.; $F$ is an antiderivative of $f$ (i.e., $F' = f$).

The fundamental theorem relates accumulation and rate of change by stating that if $F'(x) = f(x)$, then the accumulated area $A(x)$ from a fixed point $a$ to $x$ is given by $A(x) = F(x) - F(a)$. It shows that differentiation and integration are inverse processes: the derivative of the accumulation function $A(x)$ recovers the integrand $f(x)$, and antiderivatives allow the evaluation of definite integrals.

Conditions: The function $f$ is continuous.; $F$ is an antiderivative of $f$ (i.e., $F' = f$).

The notation $df/dx$ motivates thinking of tiny vertical changes paired with tiny horizontal changes, and informally canceling $dx$ helps intuition. However, it is misleading because the derivative is defined by a limit, not by dividing two ordinary numbers.

Conditions: Interpreting the differential notation in the context of the theorem.; Understanding the formal definition of the derivative.

The notation $df/dx$ motivates thinking of tiny vertical changes paired with tiny horizontal changes, and informally canceling $dx$ helps intuition. However, it is misleading because the derivative is defined by a limit, not by dividing two ordinary numbers.

Conditions: Interpreting the differential notation in the context of the theorem.; Understanding the formal definition of the derivative.

The geometric meaning is that each small horizontal step uses its own local slope. The resulting vertical increments accumulate to the total rise from $f(a)$ to $f(b)$.

Conditions: Viewing the theorem through the lens of Riemann sums or local linear approximation.; The function is differentiable on the interval.

The geometric meaning is that each small horizontal step uses its own local slope. The resulting vertical increments accumulate to the total rise from $f(a)$ to $f(b)$.

Conditions: Viewing the theorem through the lens of Riemann sums or local linear approximation.; The function is differentiable on the interval.

Continuity ensures that as the interval width $h$ approaches zero, the average value of $f$ over $[x, x+h]$ converges to the instantaneous value $f(x)$. Without continuity, the local behavior might oscillate wildly or have jumps, preventing the limit of the difference quotient from settling on a single well-defined value $f(x)$.

Conditions: $A(x) = \int_a^x f(t) dt$; $f$ is continuous at $x$

Continuity ensures that as the interval width $h$ approaches zero, the average value of $f$ over $[x, x+h]$ converges to the instantaneous value $f(x)$. Without continuity, the local behavior might oscillate wildly or have jumps, preventing the limit of the difference quotient from settling on a single well-defined value $f(x)$.

Conditions: $A(x) = \int_a^x f(t) dt$; $f$ is continuous at $x$

The net-change formula states that for a continuously differentiable function on an interval, integrating its derivative gives the endpoint difference. Specifically, the integral of the derivative $f'$ from $a$ to $b$ equals $f(b) - f(a)$.

Conditions: The function is continuously differentiable on the interval.; The integration is over the interval from $a$ to $b$.

The net-change formula states that for a continuously differentiable function on an interval, integrating its derivative gives the endpoint difference. Specifically, the integral of the derivative $f'$ from $a$ to $b$ equals $f(b) - f(a)$.

Conditions: The function is continuously differentiable on the interval.; The integration is over the interval from $a$ to $b$.

The accumulation function $A(x)$ defines area by integrating from a fixed lower endpoint to a variable upper endpoint $x$. For $x \ge 0$, this represents ordinary geometric area.

Conditions: The lower endpoint is fixed.; The upper endpoint $x$ is variable.; Signed integration is used to handle direction and negative values.

The accumulation function $A(x)$ defines area by integrating from a fixed lower endpoint to a variable upper endpoint $x$. For $x \ge 0$, this represents ordinary geometric area.

Conditions: The lower endpoint is fixed.; The upper endpoint $x$ is variable.; Signed integration is used to handle direction and negative values.

The derivative $A'(x)$ equals $f(x)$ because moving the endpoint from $x$ to $x+h$ adds a thin strip whose area increment is approximately $f(x)h$. Dividing this increment by $h$ and taking the limit as $h$ approaches zero yields the local average height, which continuity ensures approaches the exact endpoint height $f(x)$.

Conditions: The function $f$ is continuous at $x$.; $A(x)$ is defined as the accumulation of $f$ from a fixed lower limit to $x$.

The derivative $A'(x)$ equals $f(x)$ because moving the endpoint from $x$ to $x+h$ adds a thin strip whose area increment is approximately $f(x)h$. Dividing this increment by $h$ and taking the limit as $h$ approaches zero yields the local average height, which continuity ensures approaches the exact endpoint height $f(x)$.

Conditions: The function $f$ is continuous at $x$.; $A(x)$ is defined as the accumulation of $f$ from a fixed lower limit to $x$.

Continuity ensures that the local average height of the function over a small interval approaches the exact height at the endpoint as the interval width shrinks to zero. This property is crucial for proving that the derivative of the accumulation function $A(x)$ is exactly the integrand $f(x)$, as it allows the replacement of the average value with the point value in the limit.

Conditions: The function $f$ is continuous at the point $x$.; The interval width $h$ approaches zero.

Continuity ensures that the local average height of the function over a small interval approaches the exact height at the endpoint as the interval width shrinks to zero. This property is crucial for proving that the derivative of the accumulation function $A(x)$ is exactly the integrand $f(x)$, as it allows the replacement of the average value with the point value in the limit.

Conditions: The function $f$ is continuous at the point $x$.; The interval width $h$ approaches zero.

The accumulation idea applies to motion by integrating velocity over time. Integrating signed velocity gives net displacement, while integrating the absolute value of velocity gives total distance traveled.

Conditions: The variable of integration is time.; The integrand is velocity (signed or absolute).

The accumulation idea applies to motion by integrating velocity over time. Integrating signed velocity gives net displacement, while integrating the absolute value of velocity gives total distance traveled.

Conditions: The variable of integration is time.; The integrand is velocity (signed or absolute).

The factor is essential because the Fundamental Theorem of Calculus combined with the Chain Rule requires multiplying the integrand evaluated at the upper limit by the derivative of that limit. Without this factor, the rate of change of the accumulated area with respect to the outer variable $t$ would be incorrect.

Conditions: The upper limit is a function of $t$, not just $t$ itself.; The integrand is continuous.

The factor is essential because the Fundamental Theorem of Calculus combined with the Chain Rule requires multiplying the integrand evaluated at the upper limit by the derivative of that limit. Without this factor, the rate of change of the accumulated area with respect to the outer variable $t$ would be incorrect.

Conditions: The upper limit is a function of $t$, not just $t$ itself.; The integrand is continuous.

The final exact value obtained for the displayed arc-length integral is $\frac{208}{27}$. This is calculated by multiplying the outside constant $\frac{8}{27}$ by the evaluated bracket difference 26.

Conditions: The integral has been fully substituted and evaluated up to the step $\frac{8}{27}\cdot 26$.; The arithmetic simplification $8 \cdot 26 = 208$ is performed exactly.

The final exact value obtained for the displayed arc-length integral is $\frac{208}{27}$. This is calculated by multiplying the outside constant $\frac{8}{27}$ by the evaluated bracket difference 26.

Conditions: The integral has been fully substituted and evaluated up to the step $\frac{8}{27}\cdot 26$.; The arithmetic simplification $8 \cdot 26 = 208$ is performed exactly.

To evaluate $\int_2^3 \sqrt{x} dx$, first find the antiderivative $\frac{2}{3}x^{3/2}$. Then, apply the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower bound from its value at the upper bound: $\left[\frac{2}{3}x^{3/2}\right]_2^3 = \frac{2}{3}(3)^{3/2} - \frac{2}{3}(2)^{3/2}$.

Conditions: The integral is $\int_2^3 \sqrt{x} dx$.; The antiderivative is $\frac{2}{3}x^{3/2}$.; The bounds are 2 and 3.

To evaluate $\int_2^3 \sqrt{x} dx$, first find the antiderivative $\frac{2}{3}x^{3/2}$. Then, apply the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower bound from its value at the upper bound: $\left[\frac{2}{3}x^{3/2}\right]_2^3 = \frac{2}{3}(3)^{3/2} - \frac{2}{3}(2)^{3/2}$.

Conditions: The integral is $\int_2^3 \sqrt{x} dx$.; The antiderivative is $\frac{2}{3}x^{3/2}$.; The bounds are 2 and 3.