Question

Prove the Fundamental Theorem of Calculus in your own words. Use the proof in this section as a guide.

Short answer

Ans: If $f$is continuous on $[a,b]$and F is any antiderivative of $f$ , then${\int }_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right).$

Step-by-step solution

Step 1. Given Information: 

The Fundamental Theorem of Calculus .

Step 2. Proving Inverse Fundamental Theorem of Calculus :

Before we get to the proofs, let’s first state the Fundamental Theorem of Calculus and the Inverse Fundamental Theorem of Calculus. When we do prove them, we’ll prove ${FTC}^{-1}$before we prove FTC.

role="math" localid="1648633044440" $$\begin{gathered} \mathrm{If}\mathrm{F}0\mathrm{is}\mathrm{continuous}\mathrm{on}[\mathrm{a},\mathrm{b}],\mathrm{then} \\ {\int }_{\mathrm{a}}^{\mathrm{b}}{\mathrm{F}}^{\text{'}}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{F}\left(\mathrm{b}\right)-\mathrm{F}\left(\mathrm{a}\right) \\ \mathrm{In}\mathrm{other}\mathrm{words},\mathrm{if}\mathrm{F}\mathrm{is}\mathrm{an}\mathrm{antiderivative}\mathrm{of}\mathrm{f},\mathrm{then} \\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{F}\left(\mathrm{b}\right)-\mathrm{F}\left(\mathrm{a}\right) \\ \mathrm{A}\mathrm{common}\mathrm{notation}\mathrm{for}\mathrm{F}\left(\mathrm{b}\right)-\mathrm{f}\left(\mathrm{a}\right)\mathrm{is}\overline{)\mathrm{F}\left(\mathrm{x}\right)}\begin{array}{c}\mathrm{b}\\ \mathrm{a}\end{array} \\ \mathrm{There}\mathrm{are}\mathrm{stronger}\mathrm{statements}\mathrm{of}\mathrm{these}\mathrm{theorems}\mathrm{that}\mathrm{don}’\mathrm{t}\mathrm{have} \\ \mathrm{the}\mathrm{continuity}\mathrm{assumptions}\mathrm{stated}\mathrm{here},\mathrm{but}\mathrm{these}\mathrm{are}\mathrm{the}\mathrm{ones}\mathrm{we}’\mathrm{ll} \\ \mathrm{prove}. \end{gathered}$$

Step 3. Proving the Fundamental Theorem of Calculus :

$$\begin{gathered} Iffisacontinuousfunctionontheclosedinterval[a,b], \\ andFisitsaccumulationfunctiondefinedby \\ F\left(x\right)={\int }_a^{x}f\left(t\right)dt \\ forxin[a,b],thenFisdifferentiableon[a,b]anditsderivativeisf, \\ thatis,F\text{'}\left(x\right)=f\left(x\right)forx\in [a,b]. \\ \frac{d}{dx}{\int }_a^{x}f\left(t\right)dt=f\left(x\right) \\ ProofoftheFT{C}^{-1}.Firstofall,\sin cefiscontinuous, \\ it’sintegrable,thatistosay, \\ F\left(x\right)={\int }_a^{x}f\left(t\right)dt \\ WeneedtoshowthatF\text{'}\left(x\right)=f\left(x\right).Bythedefinitionofderivatives, \\ F\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{F(x+h)-F\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{1}{h}\left({\int }_a^{x+h}f\left(t\right)dt-{\int }_a^{x}f\left(t\right)dt\right) \\ \underset{h\to 0}{=\lim }\frac{1}{h}{\int }_x^{x+h}f\left(t\right)dt \end{gathered}$$

Step 4. Proving F '(x) = f(x)

$$\begin{gathered} \mathrm{We}’\mathrm{ll}\mathrm{now}\mathrm{go}\mathrm{on}\mathrm{to}\mathrm{prove}\mathrm{the}\mathrm{FTC}\mathrm{from}\mathrm{the}{\mathrm{FTC}}^{-1} \\ G\left(x\right)={\int }_a^{x}F\text{'}\left(t\right)dt \\ \mathrm{Then}\mathrm{by}\mathrm{ftc}-1,\mathrm{G}0\left(\mathrm{x}\right)=\mathrm{F}0\left(\mathrm{x}\right).\mathrm{Therefore},\mathrm{G}\mathrm{and}\mathrm{F}\mathrm{differ}\mathrm{by}\mathrm{a}\mathrm{constant}\mathrm{C}, \\ \mathrm{that}\mathrm{is},\mathrm{G}\left(\mathrm{x}\right)-\mathrm{F}\left(\mathrm{x}\right)=\mathrm{C}\mathrm{for}\mathrm{all}\mathrm{x}\in [\mathrm{a},\mathrm{b}] \\ \mathrm{but}, \\ \mathrm{G}\left(\mathrm{a}\right)={\int }_{\mathrm{a}}^{\mathrm{a}}\mathrm{F}\text{'}\left(\mathrm{t}\right)\mathrm{dt}=0 \\ \mathrm{and}\mathrm{G}\left(\mathrm{a}\right)-\mathrm{F}\left(\mathrm{a}\right)=\mathrm{C},\mathrm{so}\mathrm{C}=-\mathrm{F}\left(\mathrm{a}\right).\mathrm{Hence},\mathrm{G}\left(\mathrm{x}\right)-\mathrm{F}\left(\mathrm{x}\right)=-\mathrm{F}\left(\mathrm{a}\right)\mathrm{for}\mathrm{all}\mathrm{x}\in [\mathrm{a},\mathrm{b}]. \\ \mathrm{In}\mathrm{particular},\mathrm{G}\left(\mathrm{b}\right)-\mathrm{F}\left(\mathrm{b}\right)=-\mathrm{F}\left(\mathrm{a}\right),\mathrm{so}\mathrm{G}\left(\mathrm{b}\right)=\mathrm{F}\left(\mathrm{b}\right)-\mathrm{F}\left(\mathrm{a}\right),\mathrm{that}\mathrm{is}, \\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{F}\text{'}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{F}\left(\mathrm{b}\right)-\mathrm{F}\left(\mathrm{a}\right). \end{gathered}$$