Leland Morrow

2022-07-01

Can you apply the fundamental theorem of calculus with the variable inside the integrand?
For example: $\frac{d}{dx}\left({\int }_{a}^{x}xf\left(t\right)dt\right)$

hopeloothab9m

By definition
$\frac{d}{dx}{\int }_{a}^{x}f\left(x,t\right)dt=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{1}{\mathrm{\Delta }x}\left[{\int }_{a}^{x+\mathrm{\Delta }x}f\left(x+\mathrm{\Delta }x,t\right)dt-{\int }_{a}^{x}f\left(x,t\right)dt\right]$
$=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{1}{\mathrm{\Delta }x}\left[{\int }_{a}^{x+\mathrm{\Delta }x}\left(f\left(x,t\right)+\frac{\mathrm{\partial }f}{\mathrm{\partial }x}\mathrm{\Delta }x\right)dt-{\int }_{a}^{x}f\left(x,t\right)dt\right]$
$=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{1}{\mathrm{\Delta }x}\left[{\int }_{a}^{x+\mathrm{\Delta }x}f\left(x,t\right)dt-{\int }_{a}^{x}f\left(x,t\right)dt+{\int }_{a}^{x+\mathrm{\Delta }x}\frac{\mathrm{\partial }f}{\mathrm{\partial }x}\mathrm{\Delta }xdt\right]$
$=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{1}{\mathrm{\Delta }x}\left[{\int }_{x}^{x+\mathrm{\Delta }x}f\left(x,t\right)dt+{\int }_{a}^{x+\mathrm{\Delta }x}\frac{\mathrm{\partial }f}{\mathrm{\partial }x}\mathrm{\Delta }xdt\right]$
$=f\left(x,x\right)+\underset{\mathrm{\Delta }x\to 0}{lim}{\int }_{a}^{x+\mathrm{\Delta }x}\frac{\mathrm{\partial }f}{\mathrm{\partial }x}dt$
$=f\left(x,x\right)+{\int }_{a}^{x}\frac{\mathrm{\partial }}{\mathrm{\partial }x}f\left(x,t\right)dt$
In conclusion, you will only get the first term if you apply the FTOC blindly. Because the integrand contains $x$, you will have the second extra term.

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