21. Antiderivatives, Areas and the FTC

f. Fundamental Theorem of Calculus for Area

On the previous page we learned that the area function, \(A_a(x)\), is an antiderivative of the original function, \(f(x)\). This is actually the first version of the Fundamental Theorem of Calculus:

Given a positive function \(f(x)\) defined on an interval \([a,b]\), then its area function, \(A_a(x)\), is an antiderivative of the original function, \(f(x)\). In a formula: \[ \dfrac{d}{dx}A_a(x)=f(x) \]

The area function, \(A_a(x)\) is a specific antiderivative of \(f(x)\). So it differs from an arbitrary antiderivative \(F(x)\) by a constant. This constant is found by using the initial condition \(A_a(a)=0\). On the previous page, we did this for several specific functions \(f(x)\). Let's do this now for a general function \(f(x)\). In the process we will derive a general formula for \(A(a,b)\).

Find the area below a positive function \(y=f(x)\) above the \(x\)-axis between \(x=a\) and \(x=b.\)

Given a positive function \(f(x)\) defined on an interval \([a,b]\), then the area under \(f(x)\) above the interval \([a,b]\) is: \[ A(a,b)=F(b)-F(a) \] where \(F(x)\) is any antiderivative of \(f(x)\).

We frequently write the right hand side of the last equation as \(\left.F(x)\rule{0pt}{10pt}\right|_a^b\) to mean that the function \(F(x)\) is evaluated at \(x=b\) and at \(x=a\) and then these values are subtracted. So we have: \[ A(a,b)=\left.F(x)\rule{0pt}{10pt}\right|_a^b=F(b)-F(a) \]

The following are the same example and exercise as on the previous page except solved by the Fundamental Theorem of Calculus.

Let \(f(x)=x^2\). An antiderivative is \(F(x)=\dfrac{x^3}{3}\). So the area is \[ A(1,4)=\left.\dfrac{x^3}{3}\right|_1^4 =\dfrac{4^3}{3}-\dfrac{1^3}{3}=21 \]