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.


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

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



It is


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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.