Foreshadowing the FTC

This is an example to help prepare students to tackle the Fundamental Theorem of Calculus (FTC). Use it after the lesson on Riemann sums and the definition of the definite integral, but before the FTC derivation.

Consider the area, A, between the graph of f\left( x \right)=\cos \left( x \right) and the x-axis on the interval \left[ 0,\tfrac{\pi }{2} \right]. Set up a Riemann sum using the general partition:

 0={{x}_{0}}<{{x}_{1}}<{{x}_{2}}<{{x}_{3}}<\cdots <{{x}_{n-2}}<{{x}_{n-1}}<{{x}_{n}}=\tfrac{\pi }{2}

\displaystyle A=\int_{0}^{{\scriptstyle{}^{\pi }\!\!\diagup\!\!{}_{2}\;}}{\cos \left( x \right)dx}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\cos \left( {{c}_{i}} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right)}

Since we can choose the value of {{c}_{i}} any way we want, let’s take the same intervals and use {{c}_{i}} the number guaranteed by the Mean Value Theorem for the function F\left( x \right)=\sin \left( x \right) on the each sub-interval interval.  That is, on each sub-interval at x={{c}_{i}}

\left. \frac{d}{dx}\sin \left( x \right) \right|_{x={{c}_{i}}}^{{}}=\cos \left( {{c}_{i}} \right)=\frac{\sin \left( {{x}_{i}} \right)-\sin \left( {{x}_{i-1}} \right)}{\left( {{x}_{i}}-{{x}_{i-1}} \right)}

Then, substituting into the Riemann sum above

\displaystyle A=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\cos \left( {{c}_{i}} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right)}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\frac{\sin \left( {{x}_{i}} \right)-\sin \left( {{x}_{i-1}} \right)}{\left( {{x}_{i}}-{{x}_{i-1}} \right)}\left( {{x}_{i}}-{{x}_{i-1}} \right)}

\displaystyle A=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\left( \sin \left( {{x}_{i}} \right)-\sin \left( {{x}_{i-1}} \right) \right)}

Now writing out the terms we have a telescoping series:

\displaystyle A=\left( \sin \left( {{x}_{1}} \right)-\sin \left( {{x}_{0}} \right) \right)+\left( \sin \left( {{x}_{2}} \right)-\sin \left( {{x}_{1}} \right) \right)+\left( \sin \left( {{x}_{3}} \right)-\sin \left( {{x}_{2}} \right) \right)+

\displaystyle \cdots +\left( \sin \left( {{x}_{n-1}} \right)-\sin \left( {{x}_{n-2}} \right) \right)+\left( \sin \left( {{x}_{n}} \right)-\sin \left( {{x}_{n-1}} \right) \right)

\displaystyle A=\sin \left( {{x}_{n}} \right)-\sin \left( {{x}_{0}} \right)

\displaystyle A=\sin \left( \tfrac{\pi }{2} \right)-\sin \left( 0 \right)=1

As you can see, this is really just the derivation of the FTC applied to a particular function. Now the students should have a better idea of what’s going on when you solve the problem in general, i.e. when you prove the FTC.








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