I need to solve this:


So, first I solve the integral for $(0,x)$, only need to evaluate in $x$:


Then I solve the derivative:



Is this the end?

  • 2
    $\begingroup$ You're making exactly the usual mistake that one makes if one is not sharply attentive. "First I solve the integral". That's not the best way, and sometimes that's exactly what prevents you from getting the problem done. (Also, you don't "solve" the integral; you evaluate it. You solve problems or solve equations; you evaluate expressions. This is an expression, not an equation. You seek a value not a solution. Except this time, you don't. $\endgroup$ Commented Aug 16, 2013 at 2:58

5 Answers 5


I think you're being asked to know the following fact. Let $f$ be any reasonably nice function — smooth is plenty good enough. Then

$$ \frac{\mathrm d}{\mathrm d x} \int_0^x f(t)\,\mathrm d t = f(x). $$

The proof is as follows. The left hand side asks:

Question: How much does $\int_0^x f$ change when you change $x$ to $x + \mathrm d x$?

Here $\mathrm d x$ is some very small amount, meaning the question is approximate, and you should ignore numbers that are much smaller than $\mathrm d x$.

Well, $\int_0^{x + \mathrm dx} = \int_0^x + \int_x^{x + \mathrm d x}$, so the answer is clearly:

Answer 1: $\int_0^x f$ changes by $\int_x^{x + \mathrm d x}f$.

But this isn't that useful. How much is $\int_x^{x + \mathrm d x}f$? Well, it's the area under a curve, from $x$ to $x+\mathrm{d}x$. So the width of the region whose area we're asking for is $\mathrm d x$. The height is roughly $f(x)$. The left-hand height is exactly $f(x)$. The right-hand height is $f(x+\mathrm d x) \approx f(x) + f'(x)\,\mathrm d x$. In any case, the point is that if $f$ is smooth, say, then the height of the curve doesn't vary that much over the interval $[x,x+\mathrm d x]$, and in particular varies by something much less than $1$. Which means we can approximate the area as $(\text{width}) \times (\text{height}) + $ an error which is $(\text{width}) \times (\text{something much less than 1})$. Evaluating gives an area which is $f(x)\,\mathrm d x + (\text{something much less than $\mathrm d x$})$.

Ok, so now we can update our answer:

Answer 2: $\int_0^x f$ changes by $f(x)\,\mathrm d x + (\text{something much smaller than $\mathrm d x$})$.

Dividing through by $\mathrm d x$ and ignoring the error in the estimate, which is very small (meaning it goes to $0$ as $\mathrm d x \to 0$) gives the original displayed formula.

Applying the displayed formula to your question gives:

$$ \frac{\mathrm d}{\mathrm d x} \int_0^x \sqrt{t^2 + 1}\,\mathrm d t = \sqrt{x^2 + 1} $$

This is justified because the function $f(x) = \sqrt{t^2+1}$ is smooth. There are functions that do not satisfy the displayed formula — you will find some if you think about discontinuous functions — but they are unlikely to come up in your homework.


Let $f(t)=\sqrt{t^2+1}$. Usually, to find $\int_0^x f(t)\,dt$ we would go through the following steps:

(i) Find a function $F(t)$ such that $F'(t)=f(t)$. So we would find an indefinite integral (antiderivative) of $f(t)$.

(ii) Then $\int_0^x f(t0\,dt=F(x)-F(0)$.

We are asked to find the derivative of $F(x)-F(0)$. This is $f(x)$. In our case $f(x)=\sqrt{x^2+1}$. We had to do almost no work to find the answer!

A more carefully stated version of the result is called the Fundamental Theorem of Calculus.

If $f$ is a continuous function, and $F(x)=\int_a^x f(t)\,dt$, then $F'(x)=f(x)$.

Remark: The proposed solution has problems. In principle, we could (in this case) first find $\int_0^x \sqrt{t^2+1}\,dt$, and then differentiate with respect to $x$. However, that would be much harder than the method that used the Fundamental Theorem.

It looks as if that's what you attempted. However, $\frac{x\sqrt{x^2}}{2}\,dt$ bears no relationship to $\int_0^x\sqrt{t^2+1}\,dt$. Even the shape is wrong. After we have finished integrating, there should not be a $dt$ in the answer.

  • $\begingroup$ In the first line $f(f)$ should be $f(t)$. I can't make one-character edits. $\endgroup$ Commented Aug 16, 2013 at 5:33
  • $\begingroup$ @in_wolfram_we_trust: Thank you for telling me about $f(f)$. Fixed! $\endgroup$ Commented Aug 16, 2013 at 5:36

You need to use the fact that $${d\over dx}\int_a^x f(t)\, dt = f(x),$$ for any constant $a$.


You have some issues with the calculation of your antiderivatives. The correct antiderivative is $$ \int_0^x\sqrt{t^2+1}\,dt=\frac12\,\left(t\,\sqrt{x^2+1}+\mbox{arcsinh}(x)\right). $$ You can go an differentiate that if you want. But the point of the question is that it is a straightforward application of the Fundamental Theorem of Calculus: $$ \frac{d}{dx}\,\int_0^x\sqrt{t^2+1}\,dt=\sqrt{x^2+1}. $$

  • $\begingroup$ Ooh, pain. A direct appeal to the fundamental theorem is WAY simpler. $\endgroup$ Commented Aug 16, 2013 at 0:33

This is an application of the Fundamental Theorem of Calculus, which states that if $f$ is a continuous function on an open interval $I$, and $a$ is a point in $I$, then if $F$ is defined to be $F(x) = \int\limits_{a}^{x} f(t)\,dt$, then $\frac{d}{dx} F(x) = f(x)$ for all $x \in I$.

As $f(t) = \sqrt{t^2 + 1}$ is continuous on the open interval $I = (-\infty, \infty)$, taking the point $0$ in $I$ to be $a$ gives that for $F(x) = \int\limits_{0}^{x} f(t)\,dt$, $\frac{d}{dx} F(x) = f(x)$, so

$$\frac{d}{dx} \int\limits_{0}^{x} \sqrt{t^2 + 1}\,dt = \sqrt{x^2 + 1}.$$


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