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Fundamental Theorem of Calculus, Part 1

Tags

Calculus

Cegep/2

Word count

309 words

Reading time

2 minutes

Abbr. FTC1

If $f(t)$ is continuous on $[a,b]$, let $A(x)={\int }_a^{x}f(t)dt$, then:

1. $A$ is continuous on $[a,b]$
2. $A$ is differentiable on $(a,b)$
3. ==$A^{\prime }=f$==

Proof

We know that $f$ is continuous on $[a,b]$.
We want to show that $A^{\prime }(x)=f(x)$.
Let $A(x)={\int }_a^{x}f(t)dt$, then

$$\begin{array}{rl}A(x+h)-A(x)& ={\int }_a^{x+h}f(t)dt-{\int }_a^{x}f(t)dt\\ & ={\int }_x^{x+h}f(t)dt\\ ⟹\frac{A(x+h)-A(x)}{h}& =\frac{1}{h}{\int }_x^{x+h}f(t)dt\end{array}$$

We want to show that $A^{\prime }(x)=\underset{h\to 0}{lim}\frac{A(x+h)-A(x)}{h}=\underset{h\to 0}{lim}\frac{1}{h}{\int }_x^{x+h}f(t)dt=f(x)$.
Since $f$ is continuous on $[x,x+h]$,
By E.V.T., there must exist an absolute minimum $m_h$ and an absolute maximum $M_h$ for $f$ on $[x,x+h]$.
Therefore,

$$\begin{array}{rlr}{m}_h& \le f(t)& \le M_h\\ (m_h)h& \le {\int }_x^{x+h}f(t)dt& \le (M_h)h\\ m_h& \le \frac{1}{h}{\int }_x^{x+h}f(t)dt& \le M_h\\ \underset{h\to 0}{lim}{m}_h& \le \underset{h\to 0}{lim}\frac{1}{h}{\int }_x^{x+h}f(t)dt& \le \underset{h\to 0}{lim}{M}_h\end{array}$$

Since $\underset{h\to 0}{lim}{m}_h=\underset{h\to 0}{lim}{M}_h=f(x)$,
By the Squeeze Theorem,

$$\underset{h\to 0}{lim}\frac{1}{h}{\int }_x^{x+h}f(t)dt=f(x)$$

$◻$

Examples

Solve $\frac{\mathrm{d}}{\mathrm{d}x}{\int }_5^x(1-\sqrt{\sin t})dt$.

$1-\sqrt{\sin t}$ is continuous at $t=5$.
$\therefore$ by FTC1,

$$\frac{\mathrm{d}}{\mathrm{d}x}{\int }_5^x(1-\sqrt{\sin t})dt=1-\sqrt{\sin x}$$

Contributors

Ajitani Hifumi

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