# Can you apply the fundamental theorem of calculus with the variable inside the integrand? For examp

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