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Using the fundamental theorem of calculus, find the second derivative of $\int_{{\sqrt(x)}}^{x}x-e^t\,dt$

I've looked up the theorem on wikipedia but I can't really see what I'm meant to do.

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$$I:=\int_{\sqrt x}^x x-e^t \mathrm{d}t = \int_{\sqrt x}^x \mathrm{d}t -\int_{\sqrt x}^x e^t \mathrm{d}t=x\int_{\sqrt x}^x \mathrm{d}t-e^t|_{\sqrt x}^x=x(x-\sqrt{x})-e^t|_{\sqrt x}^x= x(x-\sqrt x)-e^x+e^{\sqrt x}$$

Can you calculate $\frac{\mathrm{d}^2I}{\mathrm{d}x^2}$ by yourself?

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Hint

It is

$$F(x)=\int_{\sqrt{x}}^x(x-e^t)dt=x(x-\sqrt{x})-\int_{\sqrt{x}}^xe^tdt.$$

To get the derivative of the first part must be easy. So, consider $G(x)=\int_{\sqrt{x}}^xe^tdt.$ The theorem says that

$$G'(x)=e^x (x)'-e^{\sqrt{x}} (\sqrt{x})',$$ where $'$ means derivative.

Can you continue from here?

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