State and apply the Fundamental Theorem of Calculus to evaluate definite integrals and to differentiate integral functions

What this dot point is asking

SCSA Unit 4 emphasises the Fundamental Theorem of Calculus as the bridge between the definite integral (a limit of sums) and antidifferentiation. This dot point asks you to state and apply both forms, examined in both sections of the WACE written examination.

The two forms

The evaluation form lets you compute an area without summing rectangles: find any antiderivative and subtract its values at the limits. The constant of integration cancels in $F(b)-F(a)$, so it is omitted in definite integrals. The derivative form says that integrating then differentiating returns the original function.

It does not matter which antiderivative you choose for the evaluation form. If $F(x)$ and $F(x)+5$ are both antiderivatives of $f$, then $\bigl(F(b)+5\bigr)-\bigl(F(a)+5\bigr)=F(b)-F(a)$: the added constant cancels in the subtraction. This is the formal reason the constant of integration is dropped in a definite integral, and it means you can always pick the simplest antiderivative without worrying about $c$.

Evaluating a definite integral

Differentiating an integral function

The derivative form is examined directly when a function is defined by an integral with a variable upper limit. Differentiating simply substitutes the upper limit into the integrand.

Reverse problems with the evaluation form

The evaluation form also supports problems that give the value of a definite integral and ask for an unknown, such as a limit of integration or a constant in the integrand. The method is to carry out the integration in terms of the unknown, set the result equal to the given value, and solve the resulting equation. For instance, finding $k$ in $\int_0^k 3x^2\,dx=8$ gives $k^3=8$, so $k=2$. These reverse questions test whether you can apply the theorem flexibly rather than only computing a numerical area.

A note on notation

The vertical-bar notation $\bigl[F(x)\bigr]_a^b$ is shorthand for $F(b)-F(a)$ and keeps the working tidy. Write the antiderivative inside the brackets, mark the limits, then substitute the upper limit first and subtract the lower. Keeping this order avoids the sign error of computing $F(a)-F(b)$, which negates the answer. When the integrand has several terms, antidifferentiate each term and apply the limits to the whole expression at once rather than term by term, which reduces arithmetic slips.

Why the two processes are inverse

The theorem shows that the area-accumulation function $g(x)=\int_a^x f(t)\,dt$ has the original $f$ as its derivative, so accumulating area then measuring its rate of change recovers $f$. Conversely, the evaluation form expresses the net accumulated area as the change in an antiderivative. This duality is the central result of single-variable calculus.