First time poster here, thanks in advance!

To get right to my question, it concerns solving for x in the following equation: $$\int_{0}^{x}(t^2+1) dt=x^2$$ Where the "standard" approach would be to apply the fundamental theorem of calculus like this: $$\frac{\mathrm{d} }{\mathrm{d} x} \int_{0}^{x}(t^2+1) dt= \frac{\mathrm{d} }{\mathrm{d} x} x^2$$ $$x^2+1=2x$$ Then solving for x would return x=1

However, why do you get a different result by first computing the definite integral with respect to t, then putting the bounds (x and 0) and getting this: $$\frac{x^3}{3}+x=x^2$$ And clearly, here x must equal 0. Furthermore evaluating x=1 on the first equation does not hold true while evaluating x=0 does. Perhaps it is wrong to evaluate integrals with variable bounds like that? Then how come double and triple integrals regularly use variable bounds that way?. Is there something fundamentally wrong with taking the derivative of both sides?

Again, any help and explanation is very much appreciated!

EDIT: Thanks for the very clear answers everyone, makes complete sense now.

  • 3
    $\begingroup$ Suppose you're solving the equation $x^2=x+1$. Do you take the derivatives of both sides and set them equal to each other? It's basically the same mistake. $\endgroup$
    – bjorn93
    Mar 28, 2020 at 6:55

2 Answers 2


There's no reason to think that solutions to $f'(x)=g'(x)$ will be solutions to $f(x)=g(x)$ (or vice versa). For instance, the equation $x^2+1=x$ has no solutions, but, after differentiating both sides, we get $2x=1$, which clearly has a solution.

  • 5
    $\begingroup$ Even simpler example: $x=0$ clearly has a solution, but taking derivatives we get $1=0$, which clearly does not. $\endgroup$
    – user169852
    Mar 28, 2020 at 7:37

You need to first realize what the integral $\int_0^x(t^2+1)dt$ is. First thing to note is that $t$ is a dummy variable and that the integral is in essential a function $f(x)=\int_0^x(t^2+1)dt=x^3/3+x$. Let us write $g(x)=x^2$ as the RHS of your original equation.

Now let's rewrite you equation as $f(x)=g(x)$, and check what you are doing with you different methods.

Method 1: You are actually solving $f'(x)=x^2+1=2x=g'(x)$.

Method 2: You are actually solving $f(x)=g(x)$.

See the difference?

Note that $f(x)=g(x)$ and $f'(x)=g'(x)$ mean totally different things and neither implies the other.


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.