# For which $a,b$ is $F(x)=\int_a^b\sqrt{t^4+1}\!\,\,dt$ given some initial conditions?

The antiderivative $F(x)$ of $f(x)=\sqrt{x^4+1}$ satisfying the initial condition $F(1)=0$ is given by $F(x)=\int_a^b\sqrt{t^4+1}\!\,\,dt$.
Find $a,b$.

I don't understand how to find a and b. Ive never done a question like this before.

-

You'll want to make use of the power of the Fundamental Theorem of Calculus:

Applied to your problem, we have that for any constant $\,a,\,$ the derivative of $$\int_a^x \sqrt{t^4+1}\,dt\;\;\text{ with respect to x is given by:}\quad\sqrt{x^4+1}$$

So for $$\int_a^b \sqrt{t^4+1}\,dt$$

we let $b = x$, and choose $a$ such that $\,F(a) = F(1) = 0$: i.e. choose $a = 1$.

-

Hint: By the Fundamental Theorem of Calculus, for any constant $k$ the derivative of $$\int_k^x \sqrt{t^4+1}\,dt$$ with respect to $x$ is $\sqrt{x^4+1}$.

So take $b=x$. What $k$ will take care of the condition $F(1)=0$?

-