Fundamental Theorem of Calculus, Part 2

Tags

Calculus

Cegep/2

Word count

285 words

Reading time

2 minutes

Abbr. FTC2

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$,
then the definite integral of $f$ between $a$ and $b$ is the difference between the antiderivatives at $a$ and $b$, i.e.

$${\int }_a^{b}f(x)dx=F(b)-F(a)$$

$f(x)$ can be **any** antiderivative of $f$.

Subtraction notation: $F(b) - F(a) = \left. F(x) \right|_a^b$

Proof

We know that $f$ is continuous on $[a,b]$.
We want to show that ${\int }_a^{b}f(x)dx=F(b)-F(a)\mathrm{\forall }F$.
Let $A(x)={\int }_a^{x}f(t)dt$, then by FTC1 we know $A^{\prime }=f$.
Since $A$ and $f$ are both antiderivatives of $f^{\prime }$, $f(x)=A(x)+c$.
Then,

$$\begin{array}{rl}f(b)-f(a)& =A(b)+c-(A(a)+c)\\ & =A(b)-A(a)\\ & =A(b)-0\\ & ={\int }_a^{b}f(t)dt\end{array}$$

$◻$

Consequence

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}{\int }_{h(x)}^{g(x)}f(t)dt& =\frac{\mathrm{d}}{\mathrm{d}x}({\int }_a^{g(x)}f(t)dt-{\int }_a^{h(x)}f(t)dt)\\ & =\frac{\mathrm{d}}{\mathrm{d}x}(F(g(x))-F(a)-(F(h(x))-F(a)))\\ & =\frac{\mathrm{d}}{\mathrm{d}x}(F(g(x))-F(h(x)))\\ & =f(g(x))g^{\prime }(x)-f(h(x))h^{\prime }(x)\end{array}$$

Examples

$g(x)=\frac{1}{3}{x}^3$ is an antiderivative of $f(x)=x^2$.

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Evaluate the following integrals.

$${\int }_0^1{x}^{2}dx={\frac{1}{3}{x}^3|}_0^1=\frac{1}{3}{1}^3-0=\frac{1}{3}$$$${\int }_1^3{e}^{x}dx={e^x|}_1^3=e^3-e$$$${\int }_0^3(3x^2+x)dx={x^3+\frac{x^2}{2}|}_0^3=3^3+\frac{3^2}{2}-0=\frac{63}{2}$$

Contributors

Ajitani Hifumi

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