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The Fundamental Theorem of Calculus.

The two main concepts of calculus are integration and differentiation. The Fundamental Theorem of Calculus (FTC) says that these two concepts are es- sentially inverse to one another.

The fundamental theorem states that if $\mathit{F}$ has a continuous derivative on an interval $\mathrm{\left[}\mathit{a}\mathrm{,}\mathrm{}\mathit{b}\mathrm{\right]}$, then

${\int }_{\mathit{a}}^{\mathit{b}}\mathit{F}\mathrm{\prime }\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{=}\mathit{F}\mathrm{\left(}\mathit{b}\mathrm{\right)}\mathrm{-}\mathit{F}\mathrm{\left(}\mathit{a}\mathrm{\right)}$ .

This form allows one to compute integrals by finding anti-derivatives.

The FTC says that integration undoes differentiation (up to a constant which is irrevocably lost when taking derivatives), in the sense that

$\mathit{F}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}{\int }_{\mathit{a}}^{\mathit{x}}\frac{\mathit{d}}{\mathit{d}\mathit{t}}\mathrm{\left(}\mathit{F}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{+}\mathit{C}$

where $\mathit{C}\mathrm{=}\mathit{F}\mathrm{\left(}\mathit{a}\mathrm{\right)}$ .

The second part of the fundamental theorem says that differentiation undoes integration, in the sense that

$\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}\frac{\mathit{d}}{\mathit{d}\mathit{x}}{\int }_{\mathit{a}}^{\mathit{x}}\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{,}$

where $\mathit{f}$ is a continuous function on an open interval containing $\mathit{a}$ and $\mathit{x}\mathrm{.}$

Problems

1. Let $\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}\frac{\mathrm{1}}{\mathrm{1}\mathrm{+}{\mathit{x}}^{\mathrm{4}}}\mathrm{+}\mathit{a}$, and let $\mathit{F}$ be an antiderivative of $\mathit{f}$, so that $\mathit{F}\mathrm{\prime }\mathrm{=}\mathit{f}\mathrm{.}$ Find $\mathit{a}$ so that $\mathit{F}$ has exactly one critical point.

Solution. $\mathit{a}\mathrm{=}\mathrm{-}\mathrm{1}$. Clemson Calculus Competition. $\mathrm{\square }$

2. Let

$\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}{\int }_{\mathit{x}}^{\mathrm{2}}\frac{\mathrm{1}}{\sqrt{\mathrm{1}\mathrm{+}{\mathit{t}}^{\mathrm{3}}}}\mathit{d}\mathit{t}\mathrm{.}$

Find

${\int }_{\mathrm{0}}^{\mathrm{2}}\mathit{x}\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{d}\mathit{x}\mathrm{.}$

Solution. Integrate by parts to get

${\int }_{\mathrm{0}}^{\mathrm{2}}\mathit{x}\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{d}\mathit{x}\mathrm{=}\mathrm{-}{\int }_{\mathrm{0}}^{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{2}}{\mathit{x}}^{\mathrm{2}}\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{d}\mathit{x}\mathrm{=}{\int }_{\mathrm{0}}^{\mathrm{2}}\frac{{\mathit{x}}^{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{1}\mathrm{+}{\mathit{x}}^{\mathrm{3}}}}\mathit{d}\mathit{x}\mathrm{.}$

The rest is a straightforward integral after substituting for ${\mathit{x}}^{\mathrm{3}}$. The solu- tion is $\frac{\mathrm{2}}{\mathrm{3}}\mathrm{.}$ $\mathrm{\square }$

3. What function is defined by the equation

$\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}{\int }_{\mathrm{0}}^{\mathit{x}}\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{+}\mathrm{1}$?

Solution. Larson 6.9.7 $\mathrm{\square }$

4. Let $\mathit{f}$ be such that

$\mathit{x}\mathrm{sin}\mathrm{\left(}\mathit{\pi }\mathit{x}\mathrm{\right)}\mathrm{=}{\int }_{\mathrm{0}}^{{\mathit{x}}^{\mathrm{2}}}\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{.}$

Find $\mathit{f}\mathrm{\left(}\mathrm{4}\mathrm{\right)}$ .

Solution. Put $\mathit{F}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}{\int }_{\mathrm{0}}^{\mathit{x}}\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}$. Then $\mathit{F}\mathrm{\left(}{\mathit{x}}^{\mathrm{2}}\mathrm{\right)}\mathrm{=}\mathit{x}\mathrm{sin}\mathrm{\left(}\mathit{\pi }\mathit{x}\mathrm{\right)}$ . Take the derivative of both sides to get $\mathrm{2}\mathit{x}\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}\mathrm{sin}\mathrm{\left(}\mathit{\pi }\mathit{x}\mathrm{\right)}\mathrm{+}\mathit{\pi }\mathit{x}\mathrm{cos}\mathrm{\left(}\mathit{\pi }\mathit{x}\mathrm{\right)}$ and plug in $\mathit{x}\mathrm{=}\mathrm{2}$ to find $\mathit{f}\mathrm{\left(}\mathrm{4}\mathrm{\right)}\mathrm{=}\mathit{\pi }\mathrm{/}\mathrm{2}$. Kenneth Roblee, Troy U. $\mathrm{\square }$

5. If $\mathit{a}\mathrm{,}$ $\mathit{b}\mathrm{,}$ $\mathit{c}\mathrm{,}$ $\mathit{d}$ are polynomials, show that

${\int }_{\mathrm{1}}^{\mathit{x}}\mathit{a}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{c}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{d}\mathit{x}{\int }_{\mathrm{1}}^{\mathit{x}}\mathit{b}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{d}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{d}\mathit{x}\mathrm{-}{\int }_{\mathrm{1}}^{\mathit{x}}\mathit{a}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{d}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{d}\mathit{x}{\int }_{\mathrm{1}}^{\mathit{x}}\mathit{b}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{c}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathit{d}\mathit{x}$

is divisible by $\mathrm{\left(}\mathit{x}\mathrm{-}\mathrm{1}{\mathrm{\right)}}^{\mathrm{4}}\mathrm{.}$

Solution. Denote the expression in question by $\mathit{F}\mathrm{\left(}\mathit{x}\mathrm{\right)}$ . $\mathit{F}$ is a polynomial. We know $\mathrm{\left(}\mathit{x}\mathrm{-}\mathrm{1}{\mathrm{\right)}}^{\mathrm{4}}$ divides $\mathit{F}$ if and only if $\mathit{F}\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }\mathrm{\left(}\mathrm{1}\mathrm{\right)}\mathrm{=}\mathrm{0}$. Check this by differentiation. (Larson 6.9.3). $\mathrm{\square }$

6. Suppose that $\mathit{f}$ is differentiable, and that $\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}$ is strictly increasing for $\mathit{x}\mathrm{\ge }\mathrm{0}$. If $\mathit{f}\mathrm{\left(}\mathrm{O}\mathrm{\right)}\mathrm{=}\mathrm{0}$, prove that $\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{/}\mathit{x}$ is strictly increasing for $\mathit{x}\mathrm{>}\mathrm{0}\mathrm{.}$

Solution. Larson 6.9.12 $\mathrm{\square }$

7. (MCMC 2005 II.5) Suppose that $\mathit{f}\mathrm{:}\mathrm{\left[}\mathrm{0}\mathrm{,}\mathrm{}\mathrm{\infty }$) $\mathrm{\to }\mathrm{\left[}\mathrm{0}\mathrm{,}\mathrm{}\mathrm{\infty }$) is a differentiable function with the property that the area under the curve $\mathit{y}\mathrm{=}\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}$ from $\mathit{x}\mathrm{=}\mathit{a}$ to $\mathit{x}\mathrm{=}\mathit{b}$ is equal to the arclength of the curve $\mathit{y}\mathrm{=}\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}$ from $\mathit{x}\mathrm{=}\mathit{a}$ to $\mathit{x}\mathrm{=}\mathit{b}$. Given that $\mathit{f}\mathrm{\left(}\mathrm{O}\mathrm{\right)}\mathrm{=}\mathrm{5}\mathrm{/}\mathrm{4}$, and that $\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}$ has a minimum value on the interval $\mathrm{\left(}\mathrm{0}\mathrm{,}\mathrm{}\mathrm{\infty }\mathrm{\right)}$ , find that minimum value.

Solution. The area under the curve $\mathit{y}\mathrm{=}\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}$ from $\mathit{x}\mathrm{=}\mathit{a}$ to $\mathit{x}\mathrm{=}\mathit{b}$ is

${\int }_{\mathit{a}}^{\mathit{b}}\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{,}$

and the arclength of the curve $\mathit{y}\mathrm{=}\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}$ from $\mathit{x}\mathrm{=}\mathit{a}$ to $\mathit{x}\mathrm{=}\mathit{b}$ is

${\int }_{\mathit{a}}^{\mathit{b}}\sqrt{\mathrm{1}\mathrm{+}\mathrm{\left(}\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{t}\mathrm{\right)}{\mathrm{\right)}}^{\mathrm{2}}}\mathit{d}\mathit{t}\mathrm{.}$

Therefore,

${\int }_{\mathit{a}}^{\mathit{b}}\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{=}{\int }_{\mathit{a}}^{\mathit{b}}\sqrt{\mathrm{1}\mathrm{+}\mathrm{\left(}\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{t}\mathrm{\right)}{\mathrm{\right)}}^{\mathrm{2}}}\mathit{d}\mathit{t}$

for all nonnegative $\mathit{a}$ and $\mathit{b}$. In particular, we can write

${\int }_{\mathrm{0}}^{\mathit{x}}\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{=}{\int }_{\mathrm{0}}^{\mathit{x}}\sqrt{\mathrm{1}\mathrm{+}\mathrm{\left(}\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{t}\mathrm{\right)}{\mathrm{\right)}}^{\mathrm{2}}}\mathit{d}\mathit{t}$

for all nonnegative $\mathit{x}$. Both sides of the above equation define a function of $\mathit{x}$, and since they are equal, their derivatives are equal; their derivatives are given by the Second Fundamental Theorem of Calculus:

$\frac{\mathit{d}}{\mathit{d}\mathit{x}}\mathrm{\left(}{\int }_{\mathrm{0}}^{\mathit{x}}\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{\right)}\mathrm{=}\frac{\mathit{d}}{\mathit{d}\mathit{x}}\mathrm{\left(}{\int }_{\mathrm{0}}^{\mathit{x}}\sqrt{\mathrm{1}\mathrm{+}\mathrm{\left(}\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{t}\mathrm{\right)}{\mathrm{\right)}}^{\mathrm{2}}}\mathit{d}\mathit{t}\mathrm{\right)}$ ,

i. e.,

$\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}\sqrt{\mathrm{1}\mathrm{+}\mathrm{\left(}\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{x}\mathrm{\right)}{\mathrm{\right)}}^{\mathrm{2}}}\mathrm{.}$

So, we are looking for a function $\mathit{y}$ which satisifies the differential equation

$\mathit{y}\mathrm{=}\sqrt{\mathrm{1}\mathrm{+}\mathrm{\left(}\mathit{y}\mathrm{\prime }{\mathrm{\right)}}^{\mathrm{2}}}\mathrm{.}$

This equation is separable:

$\mathit{y}\mathrm{=}\sqrt{\mathrm{1}\mathrm{+}\mathrm{\left(}\mathit{y}\mathrm{\prime }{\mathrm{\right)}}^{\mathrm{2}}}\mathrm{\to }{\mathit{y}}^{\mathrm{2}}\mathrm{=}\mathrm{1}\mathrm{+}\mathrm{\left(}\mathit{y}\mathrm{\prime }{\mathrm{\right)}}^{\mathrm{2}}$ (1)

$\mathrm{\to }\mathrm{\left(}\mathit{y}\mathrm{\prime }{\mathrm{\right)}}^{\mathrm{2}}\mathrm{=}{\mathit{y}}^{\mathrm{2}}\mathrm{-}\mathrm{1}$ (2)

$\mathrm{\to }\mathit{y}\mathrm{\prime }\mathrm{=}\sqrt{{\mathit{y}}^{\mathrm{2}}\mathrm{-}\mathrm{1}}$ (3)

$\mathrm{\to }\frac{\mathit{d}\mathit{y}}{\sqrt{{\mathit{y}}^{\mathrm{2}}\mathrm{-}\mathrm{1}}}\mathrm{=}\mathit{d}\mathit{x}$. (4)

Integrating both sides yields

$\int \frac{\mathit{d}\mathit{y}}{\sqrt{{\mathit{y}}^{\mathrm{2}}\mathrm{-}\mathrm{1}}}\mathrm{=}\int \mathit{d}\mathit{x}\mathrm{\to }\mathrm{ln}\mathrm{|}\mathit{y}\mathrm{+}\sqrt{{\mathit{y}}^{\mathrm{2}}\mathrm{-}\mathrm{1}}\mathrm{|}\mathrm{=}\mathit{x}\mathrm{+}\mathit{C}$

(where the first integral is evaluated using the trig substitution $\mathit{y}\mathrm{=}\mathrm{sec}\mathit{\theta }$ and the two arbitrary constants of integration are combined into one con- stant on the right hand side). Next, since $\mathit{f}\mathrm{\left(}\mathrm{O}\mathrm{\right)}\mathrm{=}\mathrm{5}\mathrm{/}\mathrm{4}$ is positive, we can drop the absolute value, and solve for $\mathit{y}$:

$\mathrm{ln}\mathrm{\left(}\mathit{y}\mathrm{+}\sqrt{{\mathit{y}}^{\mathrm{2}}\mathrm{-}\mathrm{1}}\mathrm{\right)}\mathrm{=}\mathit{x}\mathrm{+}\mathit{C}\mathrm{\to }\mathit{y}\mathrm{+}\sqrt{{\mathit{y}}^{\mathrm{2}}\mathrm{-}\mathrm{1}}\mathrm{=}{\mathit{e}}^{\mathit{x}\mathrm{+}\mathit{C}}\mathrm{=}\mathit{A}{\mathit{e}}^{\mathit{x}}$ (where $\mathit{A}\mathrm{=}{\mathit{e}}^{\mathit{C}}$)

$\mathrm{\to }\sqrt{{\mathit{y}}^{\mathrm{2}}\mathrm{-}\mathrm{1}}\mathrm{=}\mathit{A}{\mathit{e}}^{\mathit{x}}\mathrm{-}\mathit{y}$

(5)

$\mathrm{\to }{\mathit{y}}^{\mathrm{2}}\mathrm{-}\mathrm{1}\mathrm{=}\mathrm{\left(}\mathit{A}{\mathit{e}}^{\mathit{x}}\mathrm{-}\mathit{y}{\mathrm{\right)}}^{\mathrm{2}}\mathrm{=}{\mathit{A}}^{\mathrm{2}}{\mathit{e}}^{\mathrm{2}\mathit{x}}\mathrm{-}\mathrm{2}\mathit{A}\mathit{y}{\mathit{e}}^{\mathit{x}}\mathrm{+}\mathrm{y}\mathrm{6}\mathrm{\right)}$

$\mathrm{\to }\mathrm{-}\mathrm{1}\mathrm{=}{\mathit{A}}^{\mathrm{2}}{\mathit{e}}^{\mathrm{2}\mathit{x}}\mathrm{-}\mathrm{2}\mathit{A}\mathit{y}{\mathit{e}}^{\mathit{x}}$

(7)

$\mathrm{\to }\mathrm{2}\mathit{A}\mathit{y}{\mathit{e}}^{\mathit{x}}\mathrm{=}{\mathit{A}}^{\mathrm{2}}{\mathit{e}}^{\mathrm{2}\mathit{x}}\mathrm{+}\mathrm{1}$ (8)

$\mathrm{\to }\mathit{y}\mathrm{=}\frac{{\mathit{A}}^{\mathrm{2}}{\mathit{e}}^{\mathrm{2}\mathit{x}}\mathrm{+}\mathrm{1}}{\mathrm{2}\mathit{A}{\mathit{e}}^{\mathit{x}}}\mathrm{=}\frac{\mathit{A}}{\mathrm{2}}{\mathit{e}}^{\mathit{x}}\mathrm{+}\frac{\mathrm{1}}{\mathrm{2}\mathit{A}}{\mathit{e}}^{\mathrm{-}\mathit{x}}$. (9)

Using $\mathit{f}\mathrm{\left(}\mathrm{O}\mathrm{\right)}\mathrm{=}\mathrm{5}\mathrm{/}\mathrm{4}$, we find

$\frac{\mathrm{5}}{\mathrm{4}}\mathrm{=}\frac{\mathit{A}}{\mathrm{2}}\mathrm{+}\frac{\mathrm{1}}{\mathrm{2}\mathit{A}}\mathrm{\to }\mathit{A}\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}}$ or 2.

This gives two possible functions:

$\mathit{y}\mathrm{=}\frac{\mathrm{1}}{\mathrm{4}}{\mathit{e}}^{\mathit{x}}\mathrm{+}{\mathit{e}}^{\mathrm{-}\mathit{x}}$ or $\mathit{y}\mathrm{=}{\mathit{e}}^{\mathit{x}}\mathrm{+}\frac{\mathrm{1}}{\mathrm{4}}{\mathit{e}}^{\mathrm{-}\mathit{x}}\mathrm{.}$

This latter has a minimum at $\mathit{x}\mathrm{=}\mathrm{-}\mathrm{ln}\mathrm{2}$, which is not positive, so we reject that function. The former has a minimum at $\mathit{x}\mathrm{=}\mathrm{ln}\mathrm{2}$, and the $\mathit{y}$ value is 1. Note: One could also deduce from the differential equation $\mathit{y}\mathrm{\prime }\mathrm{=}\sqrt{{\mathit{y}}^{\mathrm{2}}\mathrm{-}\mathrm{1}}$ that at the minimum value, since $\mathit{y}\mathrm{\prime }\mathrm{=}\mathrm{0}$, the $\mathit{y}$-value must be 1. $\mathrm{\square }$

8. (MCMC 2006 I.5) Let $\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}$ and $\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}$ be differentiable on $\mathrm{\left[}\mathit{a}\mathrm{,}\mathrm{}\mathit{x}\mathrm{\right]}$ and for each $\mathit{x}$ suppose there is a number ${\mathit{c}}_{\mathit{x}}$ such that $\mathit{a}\mathrm{<}{\mathit{c}}_{\mathit{x}}\mathrm{<}\mathit{x}$ and

${\int }_{\mathit{a}}^{\mathit{x}}\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{=}\mathit{f}\mathrm{\left(}{\mathit{c}}_{\mathit{x}}\mathrm{\right)}\mathrm{\left(}\mathit{x}\mathrm{-}\mathit{a}\mathrm{\right)}$ .

Assume that $\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{a}\mathrm{\right)}\mathrm{\ne }\mathrm{0}$. Then prove that

$\underset{\mathit{x}\mathrm{\to }\mathit{a}}{\mathrm{lim}}\frac{{\mathit{c}}_{\mathit{x}}\mathrm{-}\mathit{a}}{\mathit{x}\mathrm{-}\mathit{a}}\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{.}$

Solution. Let

$\mathit{F}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}{\int }_{\mathit{a}}^{\mathit{x}}\mathit{f}\mathrm{\left(}\mathit{t}\mathrm{\right)}\mathit{d}\mathit{t}\mathrm{.}$

Using Taylor's expansion of $\mathit{F}\mathrm{\left(}\mathit{x}\mathrm{\right)}$ , we have

$\mathit{F}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}\mathit{F}\mathrm{\left(}\mathit{a}\mathrm{\right)}\mathrm{+}\mathrm{\left(}\mathit{x}\mathrm{-}\mathit{a}\mathrm{\right)}\mathit{F}\mathrm{\prime }\mathrm{\left(}\mathit{a}\mathrm{\right)}\mathrm{+}\frac{\mathrm{\left(}\mathit{x}\mathrm{-}\mathit{a}{\mathrm{\right)}}^{\mathrm{2}}}{\mathrm{2}}\mathit{F}\mathrm{\prime }\mathrm{\prime }\mathrm{\left(}{\mathit{\theta }}_{\mathit{x}}\mathrm{\right)}$ ,

where ${\mathit{\theta }}_{\mathit{x}}$ lies strictly between $\mathit{a}$ and $\mathit{x}$, and as $\mathit{x}$ goes to $\mathit{a}\mathrm{,}$ ${\mathit{\theta }}_{\mathit{x}}$ also goes to $\mathit{a}$. We also have $\mathit{F}\mathrm{\left(}\mathit{a}\mathrm{\right)}\mathrm{=}\mathrm{0}\mathrm{,}$ $\mathit{F}\mathrm{\prime }\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}\mathit{f}\mathrm{\left(}\mathit{x}\mathrm{\right)}$ , and $\mathit{F}\mathrm{\prime }\mathrm{\prime }\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{x}\mathrm{\right)}$ . Thus,

$\mathit{F}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}\mathrm{0}\mathrm{+}\mathrm{\left(}\mathit{x}\mathrm{-}\mathit{a}\mathrm{\right)}\mathit{f}\mathrm{\left(}\mathit{a}\mathrm{\right)}\mathrm{+}\frac{\mathrm{\left(}\mathit{x}\mathrm{-}\mathit{a}{\mathrm{\right)}}^{\mathrm{2}}}{\mathrm{2}}\mathit{f}\mathrm{\prime }\mathrm{\left(}{\mathit{\theta }}_{\mathit{x}}\mathrm{\right)}$ .

By definition,

$\mathit{f}\mathrm{\left(}{\mathit{c}}_{\mathit{x}}\mathrm{\right)}\mathrm{=}\frac{\mathrm{1}}{\mathit{x}\mathrm{-}\mathit{a}}\mathit{F}\mathrm{\left(}\mathit{x}\mathrm{\right)}\mathrm{=}\mathit{f}\mathrm{\left(}\mathit{a}\mathrm{\right)}\mathrm{+}\frac{\mathit{x}\mathrm{-}\mathit{a}}{\mathrm{2}}\mathit{f}\mathrm{\text{'}}\mathrm{\left(}{\mathit{\theta }}_{\mathit{x}}\mathrm{\right)}$ .

Therefore,

$\frac{\mathit{f}\mathrm{\left(}{\mathit{c}}_{\mathit{x}}\mathrm{\right)}\mathrm{-}\mathit{f}\mathrm{\left(}\mathit{a}\mathrm{\right)}}{\mathit{x}\mathrm{-}\mathit{a}}\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}}\mathit{f}\mathrm{\prime }\mathrm{\left(}{\mathit{\theta }}_{\mathit{x}}\mathrm{\right)}$ .

On the other hand we can write

$\frac{\mathit{f}\mathrm{\left(}{\mathit{c}}_{\mathit{x}}\mathrm{\right)}\mathrm{-}\mathit{f}\mathrm{\left(}\mathit{a}\mathrm{\right)}}{\mathit{x}\mathrm{-}\mathit{a}}$

as a product

$\underset{\mathrm{‾}}{\mathit{f}\mathrm{\left(}{\mathit{c}}_{\mathit{x}}\mathrm{\right)}\mathrm{-}\mathit{f}\mathrm{\left(}\mathit{a}\mathrm{\right)}}\mathrm{.}\mathrm{}\underset{\mathrm{‾}}{{\mathit{c}}_{\mathit{x}}\mathrm{-}\mathit{a}}\mathrm{.}$

${\mathit{c}}_{\mathit{x}}\mathrm{-}\mathit{a}\mathrm{}\mathit{x}\mathrm{-}\mathit{a}$

On taking the limits of these as $\mathit{x}$ goes to $\mathit{a}$, we get

$\underset{\mathit{x}\mathrm{\to }\mathit{a}}{\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}}\mathit{f}\mathrm{\prime }\mathrm{\left(}{\mathit{\theta }}_{\mathit{x}}\mathrm{\right)}\mathrm{=}\underset{\mathit{x}\mathrm{\to }\mathit{a}}{\mathrm{lim}}\frac{\mathit{f}\mathrm{\left(}{\mathit{c}}_{\mathit{x}}\mathrm{\right)}\mathrm{-}\mathit{f}\mathrm{\left(}\mathit{a}\mathrm{\right)}}{{\mathit{c}}_{\mathit{x}}\mathrm{-}\mathit{a}}\mathrm{\cdot }\frac{{\mathit{c}}_{\mathit{x}}\mathrm{-}\mathit{a}}{\mathit{x}\mathrm{-}\mathit{a}}\mathrm{.}$

This gives

$\frac{\mathrm{1}}{\mathrm{2}}\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{a}\mathrm{\right)}\mathrm{=}\underset{\mathit{x}\mathrm{\to }\mathit{a}}{\mathrm{lim}}\frac{\mathit{f}\mathrm{\left(}{\mathit{c}}_{\mathit{x}}\mathrm{\right)}\mathrm{-}\mathit{f}\mathrm{\left(}\mathit{a}\mathrm{\right)}}{{\mathit{c}}_{\mathit{x}}\mathrm{-}\mathit{a}}\underset{\mathit{x}\mathrm{\to }\mathit{a}}{\mathrm{lim}}\frac{{\mathit{c}}_{\mathit{x}}\mathrm{-}\mathit{a}}{\mathit{x}\mathrm{-}\mathit{a}}\mathrm{.}$

In other words,

$\frac{\mathrm{1}}{\mathrm{2}}\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{a}\mathrm{\right)}\mathrm{=}\mathit{f}\mathrm{\prime }\mathrm{\left(}\mathit{a}\mathrm{\right)}\mathrm{\cdot }\underset{\mathit{x}\mathrm{\to }\mathit{a}}{\mathrm{lim}}\frac{{\mathit{c}}_{\mathit{x}}\mathrm{-}\mathit{a}}{\mathit{x}\mathrm{-}\mathit{a}}\mathrm{.}$

This shows

$\underset{\mathit{x}\mathrm{\to }\mathit{a}}{\mathrm{lim}}\frac{{\mathit{c}}_{\mathit{x}}\mathrm{-}\mathit{a}}{\mathit{x}\mathrm{-}\mathit{a}}\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{.}$

$\mathrm{\square }$