The Fundamental Theorem of Calculus: Learn It 2

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The Fundamental Theorem of Calculus: Learn It 2

Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives

As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas.

The theorem is comprised of two parts. The first part, which is stated here, establishes the relationship between differentiation and integration.

Fundamental Theorem of Calculus, Part 1

If $f(x)$ is continuous over an interval $\left[a,b\right],$ and the function $F(x)$ is defined by

F (x) = \int_{a}^{x} f (t) d t,

$F(x)={\displaystyle\int }_{a}^{x}f(t)dt,$

then ${F}^{\prime }(x)=f(x)$ over $\left[a,b\right].$

Before we look at the proof, let’s clarify a few points:

Notation: We define $F(x)$ as the definite integral of $f(t)$ from $a$ to $x$ . This might seem confusing because we’ve said a definite integral is a number. However, $F(x)$ is a function that gives the value of the definite integral for each $x$ .
Implications: The Fundamental Theorem of Calculus is crucial because it shows that integration and differentiation are inverse processes. It guarantees that any continuous function $f(x)$ has an antiderivative $F(x)$ .

Proof

Applying the definition of the derivative, we have

\begin{matrix} F^{'} (x) & = lim_{h \to 0} \frac{F (x + h) - F (x)}{h} \\ = lim_{h \to 0} \frac{1}{h} [\int_{a}^{x + h} f (t) d t - \int_{a}^{x} f (t) d t] \\ = lim_{h \to 0} \frac{1}{h} [\int_{a}^{x + h} f (t) d t + \int_{x}^{a} f (t) d t] \\ = lim_{h \to 0} \frac{1}{h} \int_{x}^{x + h} f (t) d t . \end{matrix}

$\begin{array}{}{F}^{\prime }(x)\hfill & =\underset{h\to 0}{\text{lim}}\dfrac{F(x+h)-F(x)}{h}\hfill \\ & =\underset{h\to 0}{\text{lim}}\dfrac{1}{h}\left[{\displaystyle\int }_{a}^{x+h}f(t)dt-{\displaystyle\int }_{a}^{x}f(t)dt\right]\hfill \\ & =\underset{h\to 0}{\text{lim}}\dfrac{1}{h}\left[{\displaystyle\int }_{a}^{x+h}f(t)dt+{\displaystyle\int }_{x}^{a}f(t)dt\right]\hfill \\ & =\underset{h\to 0}{\text{lim}}\dfrac{1}{h}{\displaystyle\int }_{x}^{x+h}f(t)dt.\hfill \end{array}$

Looking carefully at this last expression, we see $\dfrac{1}{h}{\displaystyle\int }_{x}^{x+h}f(t)dt$ is just the average value of the function $f(x)$ over the interval $\left[x,x+h\right].$ Therefore, by the mean value theorem for integrals, there is some number $c$ in $\left[x,x+h\right]$ such that

\frac{1}{h} \int_{x}^{x + h} f (x) d x = f (c) .

$\dfrac{1}{h}{\displaystyle\int }_{x}^{x+h}f(x)dx=f(c).$

In addition, since $c$ is between $x$ and $x+h$ , $c$ approaches $x$ as $h$ approaches zero. Also, since $f(x)$ is continuous, we have $\underset{h\to 0}{\text{lim}}f(c)=\underset{c\to x}{\text{lim}}f(c)=f(x).$

Putting all these pieces together, we have

\begin{matrix} F^{'} (x) & = lim_{h \to 0} \frac{1}{h} \int_{x}^{x + h} f (x) d x \\ = lim_{h \to 0} f (c) \\ = f (x), \end{matrix}

$\begin{array}{}{F}^{\prime }(x)\hfill & =\underset{h\to 0}{\text{lim}}\frac{1}{h}{\displaystyle\int }_{x}^{x+h}f(x)dx\hfill \\ & =\underset{h\to 0}{\text{lim}}f(c)\hfill \\ & =f(x),\hfill \end{array}$

and the proof is complete.

$_\blacksquare$

Use the first part of the Fundamental Theorem of Calculus to find the derivative of

g (x) = \int_{1}^{x} \frac{1}{t^{3} + 1} d t .

$g(x)={\displaystyle\int }_{1}^{x}\dfrac{1}{{t}^{3}+1}dt.$

According to the Fundamental Theorem of Calculus, the derivative is given by

g^{'} (x) = \frac{1}{x^{3} + 1} .

${g}^{\prime }(x)=\frac{1}{{x}^{3}+1}.$

Let $F(x)={\displaystyle\int }_{1}^{\sqrt{x}} \sin tdt.$ Find ${F}^{\prime }(x).$

Letting $u(x)=\sqrt{x},$ we have $F(x)={\displaystyle\int }_{1}^{u(x)} \sin tdt.$ Thus, by the Fundamental Theorem of Calculus and the chain rule,

\begin{matrix} F^{'} (x) & = \sin (u (x)) \frac{d u}{d x} \\ = \sin (u (x)) \cdot (\frac{1}{2} x^{- 1 / 2}) \\ = \frac{\sin \sqrt{x}}{2 \sqrt{x}} . \end{matrix}

$\begin{array}{}\\ {F}^{\prime }(x)\hfill & = \sin (u(x))\frac{du}{dx}\hfill \\ & = \sin (u(x))·(\frac{1}{2}{x}^{-1\text{/}2})\hfill \\ & =\frac{ \sin \sqrt{x}}{2\sqrt{x}}.\hfill \end{array}$

Let $F(x)={\displaystyle\int }_{x}^{2x}{t}^{3}dt.$ Find ${F}^{\prime }(x).$

We have $F(x)={\displaystyle\int }_{x}^{2x}{t}^{3}dt.$ Both limits of integration are variable, so we need to split this into two integrals. We get

\begin{matrix} F (x) & = \int_{x}^{2 x} t^{3} d t \\ = \int_{x}^{0} t^{3} d t + \int_{0}^{2 x} t^{3} d t \\ = − \int_{0}^{x} t^{3} d t + \int_{0}^{2 x} t^{3} d t . \end{matrix}

$\begin{array}{}\\ F(x)\hfill & ={\displaystyle\int }_{x}^{2x}{t}^{3}dt\hfill \\ & ={\displaystyle\int }_{x}^{0}{t}^{3}dt+{\displaystyle\int }_{0}^{2x}{t}^{3}dt\hfill \\ & =\text{−}{\displaystyle\int }_{0}^{x}{t}^{3}dt+{\displaystyle\int }_{0}^{2x}{t}^{3}dt.\hfill \end{array}$

Differentiating the first term, we obtain

\frac{d}{d x} [− \int_{0}^{x} t^{3} d t] = − x^{3} .

$\frac{d}{dx}\left[\text{−}{\displaystyle\int }_{0}^{x}{t}^{3}dt\right]=\text{−}{x}^{3}.$

Differentiating the second term, we first let $u(x)=2x.$ Then,

\begin{matrix} \frac{d}{d x} [\int_{0}^{2 x} t^{3} d t] & = \frac{d}{d x} [\int_{0}^{u (x)} t^{3} d t] \\ = {(u (x))}^{3} \frac{d u}{d x} \\ = {(2 x)}^{3} \cdot 2 \\ = 16 x^{3} . \end{matrix}

$\begin{array}{}\\ \frac{d}{dx}\left[{\displaystyle\int }_{0}^{2x}{t}^{3}dt\right]\hfill & =\frac{d}{dx}\left[{\displaystyle\int }_{0}^{u(x)}{t}^{3}dt\right]\hfill \\ & ={(u(x))}^{3}\frac{du}{dx}\hfill \\ & ={(2x)}^{3}·2\hfill \\ & =16{x}^{3}.\hfill \end{array}$

Thus,

\begin{matrix} F^{'} (x) & = \frac{d}{d x} [− \int_{0}^{x} t^{3} d t] + \frac{d}{d x} [\int_{0}^{2 x} t^{3} d t] \\ = − x^{3} + 16 x^{3} \\ = 15 x^{3} . \end{matrix}

$\begin{array}{}\\ \\ {F}^{\prime }(x)\hfill & =\frac{d}{dx}\left[\text{−}{\displaystyle\int }_{0}^{x}{t}^{3}dt\right]+\frac{d}{dx}\left[{\displaystyle\int }_{0}^{2x}{t}^{3}dt\right]\hfill \\ & =\text{−}{x}^{3}+16{x}^{3}\hfill \\ & =15{x}^{3}.\hfill \end{array}$

Watch the following video to see the worked solution to this example.

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