The time has finally come to see how antiderivatives and the area between a function and the \(x\)-axis relate. The first step is to examine what are called area functions. These will help solidify a major concept of definite integrals.
Consider an example for this entire section instead of just explaining theory. Let's find the area between the \(t\)-axis and \(f(t)=4t+1\) from \(a=2\) out to an arbitray value \(x\). The area function here is defined by the following integral: $$A(x)=\int_{2}^{x}(4t+1)dt$$ This function depends on the variable \(x\), that is why the integrand function is using the "dummy" variable \(t\).
The shape is a trapezoid, and the area formula from Geometry is $$A=\frac{1}{2}(h)\left(b_1+b_2\right)$$ A couple of example calculations:
| $$\begin{array}{l} A(2)=\int_{2}^{2}(4t+1)dt \\ \implies A(2)=0\;\color{green}{\checkmark} \end{array}$$limits of integration are equal | $$\begin{array}{l} A(5)=\int_{2}^{5}(4t+1)dt \\ \implies A(5)=\frac{1}{2}(5-2)(f(2)+f(5)) \\ \implies A(5)=\frac{1}{2}(3)(9+21) \\ \implies A(5)=45\;\color{green}{\checkmark} \end{array}$$ |
These examples show a pattern to find \(A(x)\) as a formula in terms of \(x\). The base of the trapezoid is \(x-2,\;b_1=f(2)\implies b_1=9\) and \(b_2=f(x)\implies b_2=4x+1\). Plugging into the area formula:
$$\begin{array}{l} A(x)=\frac{1}{2}(x-2)(9+4x+1) \\ \implies A(x)=\frac{1}{2}(x-2)(4x+10) \\ \implies A(x)=\frac{1}{2}(4x^2+2x-20) \\ \implies A(x)=2x^2+x-10 \end{array}$$
Putting all this together: \(A(x)=\int_{2}^{x}(4t+1)dt=2x^2+x-10\). Taking the derivative of both sides of the above equation yields: \(A^\prime(x)=4x+1\). This is exactly the integrand function \(4t+1\), just using the dummy variable of \(t\). This shows that \(A(x)\) is an antiderivative of \(f(x)=4x+1\).
The following graphical "proof" will demonstrate the first part of the Fundamental Theorem of Calculus:
Using area functions, the approximate total area (green and blue): $$A(x+h)\approx\color{green}{A(x)}+\color{blue}{(x+h-x)f(x)}$$ Performing some algebra, and using limits gives the exact total area: $$\lim_{h\rightarrow0}\frac{A(x+h)-A(x)}{h}=\lim_{h\rightarrow0}f(x)$$ The left side of the equation is the definition of derivative from Chapter 3, here it is exactly \(A^\prime(x)\). The right side is just \(f(x)\) since the limit variable is \(h\) and \(f\) has the variable \(x\).
The above discussion leads to the result: Fundamental Theorem of Calculus Part#1: $$A^\prime(x)=\frac{d}{dx}\left(\int_a^xf(t)dt\right)=f(x)$$ In words, when a function \(f\) is "nice" (continuous, differentiable, etc.), the area function \(A(x)\) is also an antiderivative of \(f\).
Apply the FTC Part#1 to each:
| $$\frac{d}{dx}\left(\int_1^x\sin^2tdt\right)$$ | $$\frac{d}{dx}\left(\int_x^5\sqrt{t^2+1}dt\right)$$ | $$\frac{d}{dx}\left(\int_0^{x^2}\cos t^2dt\right)$$ |
| Straight application of Part#1: $$\frac{d}{dx}\left(\int_1^x\sin^2tdt\right)=\sin^2x\;\color{green}{\checkmark}$$ | Switch limits rule: $$\begin{array}{l} \frac{d}{dx}\left(\int_x^5\sqrt{t^2+1}dt\right) \\ \implies -\frac{d}{dx}\left(\int_5^x\sqrt{t^2+1}dt\right) \\ \implies -\sqrt{x^2+1}\;\color{green}{\checkmark} \end{array}$$ | Use of Chain Rule needed since upper limit is \(x^2\): $$\begin{array}{l} \frac{d}{dx}\left(\int_0^{x^2}\cos t^2dt\right) \\ \implies \cos(x^2)^2\cdot\frac{d}{dx}(x^2) \\ \implies 2x\cos x^4\;\color{green}{\checkmark} \end{array}$$ |
We are now ready for the final piece of the puzzle. From Section 4.9, if two functions \(g,\;h\) are antiderivatives of function \(f\), then they differ by a constant: \(g(x)=h(x)+C\), for some constant \(C\). Let \(F\) be any antiderivative of \(f\) on \([a,b]\) , then \(F(x)=A(x)+C\) , where \(A\) is the area function. Recall, \(A(a)=0\) , it follows that:
$$\begin{array}{l} F(b)-F(a)=(A(b)+C)-(A(a)+C) \\ \implies A(b)+C-A(a)-C \\ \implies A(b) \end{array}$$
Since \(A\) is an area function of \(f\): \(A(b)=\int_{a}^{b}f(t)dt\).
This result is the Fundamental Theorem of Calculus Part#2: for \(f\) continuous on \([a,b]\) and \(F\) any antiderivative of \(f\) on \([a,b]\)
$$\int_{a}^{b}f(x)dx=F(b)-F(a)$$