How to calculate $f'(0)$ and $f'$ $(\sqrt{2})$ while $f(x)$= $\int_{x}^{x^3} e^{t^{2}}dt\ $?

I thought about using the fundamental theorem of calculus, but im not sure im fully aware of how to use it in this case.

any kind of help/directing, would be appreciated.


2 Answers 2


Yes, you use the fundamental theorem. Let $F(x) = \int_a^x e^{t^2}\ dt$. Then

$$f(x) = F(x^3)-F(x) = \int_x^{x^3} e^{t^2} \ dt$$


$$f'(x) = F'(x^3)\cdot 3x^2 - F'(x)$$

where by the fundamental theorem,

$$F'(x) = e^{x^2}$$

  • $\begingroup$ Lets suggest that $f(x)$ = $\int_{0}^{x^2-3x} (2 + sint^2) dt$ , what function would u suggest at the beginning for $F(x)$? $\endgroup$ Apr 13, 2015 at 22:51
  • $\begingroup$ It's the same idea, $F(x)=\int_a^x (2+\sin t^2)\ dt$. Then $f(x) = F(x^2-3x)-F(0)$. Or, you could just pick $a=0$ and use $f(x)=F(x^2-3x)$. $\endgroup$
    – David P
    Apr 14, 2015 at 1:19

Some hints. The Fundamental Theorem says that $$ \frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t=e^{u^2}\tag{1} $$ Let $u=x^3$, then $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\int_0^ue^{t^2}\,\mathrm{d}t &=\frac{\mathrm{d}u}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t\\ &=3x^2\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t\tag{2} \end{align} $$ Then note that $$ \int_x^ue^{t^2}\,\mathrm{d}t=\int_0^ue^{t^2}\,\mathrm{d}t-\int_0^xe^{t^2}\,\mathrm{d}t\tag{3} $$


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