Fundamental Theorem of Calculus

(1) $\begin{equation*}\frac{d}{dx} \left (\int_2^{f(x)} g(t) dt \right )=g(f(x))f'(x)\end{equation*}$

My Year 12 Mathematics Methods students are getting ready for their exam, and questions using the above idea have created a bit of consternation. I am going to work through an example, and show why the ‘formula’ works.

Example

Find $\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )$ .

$\begin{equation*}=\frac{d}{dx} \left (\frac{4t^3}{3}+\frac{3t^2}{2} \right ]_2^{x^2}\end{equation}$

$\begin{equation*}=\frac{d}{dx} \left ( \frac{4(x^2)^3}{3}+\frac{3(x^2)^2}{2} -(\frac{4(2^3)}{3}+\frac{3(2^2)}{2} \right )\end{equation}$

$\begin{equation*}=\frac{d}{dx} \left (\frac{4x^6}{3}+\frac{3x^4}{2}-\frac{50}{3} \right) \end{equation}$

$\begin{equation*}=\frac{24x^5}{3}+\frac{12x^3}{2}\end{equation}$

(2) $\begin{equation*}=8x^5+6x^3\end{equation*}$

If we used ‘formula’ $1$

$\begin{equation*}\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )=(4(x^2)^2+3(x^2))(2x)\end{equation}$

(3) $\begin{equation*}\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )=8x^5+6x^3 \end{equation*}$

We can see equation $2$ and $3$ are the same.

More formally

$\begin{equation*}\frac{d}{dx} \left ( \int_c^{f(x)} g(t) dt \right )=\frac{d}{dx} \left (G(f(x))-G(c) \right ) \end{equation}$

Remember $\frac{d}{dx} \left (f(g(x)) \right )=f'(g(x))\times g'(x)$

$\begin{equation*}\frac{d}{dx} \left (G(f(x))-G(c) \right ) =G'(f(x))f'(x)=g(f(x))\times f'(x)\end{equation}$

Fundamental Theorem of Calculus

Example

More formally

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