How can one explain that $$\frac{d}{dx}\left(\int_0^x{\cos(t^2+t)dt}\right) = \cos(x^2+x)$$ Without solving the integral?

I know it's related to the fundamental theorem of calculus, but here we have a derivative with respect to $x$, while the antiderivative is with respect to $t$.

Thank you.


Say $f(t)=\cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus, $$F(x)-F(0)$$ $F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $\cos(x^2+x)$.

  • 2
    $\begingroup$ An antiderivative $F$ exists because $f$ is continuous on the closed interval $[0, x]$. This proposition is sometimes referred to as the first part of the fundamental theorem of calculus, namely $x \mapsto \int_0^x f(t) \mathrm{d}t$ is such an antiderivative. $\endgroup$ – ComFreek Jan 13 '19 at 17:12

Perhaps you are mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC). First part of the theorem deals exactly with finding derivative of things of the form $\int_{a} ^{x} f(t) \, dt$.

More formally, let the function $f:[a, b] \to\mathbb {R} $ be Riemann integrable on $[a, b] $. The intent of the first part of FTC is to study the properties of a related function $F:[a, b] \to\mathbb {R} $ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ The function $F$ is defined by means of an integral and is not necessarily an anti-derivative of $f$.

FTC says that this function $F$ is continuous on $[a, b] $. And more importantly $F$ is differentiable at those points where $f$ is continuous and at such a point $c\in[a, b] $ we have $F'(c) =f(c) $.

FTC does not say anything about $F'(c) $ when $f$ is discontinuous at $c$ and it may be possible that in such cases

  • $F'(c) $ does not exist
  • or it exists but does not equal $f(c) $
  • or it may exist and be equal to $f(c) $

In any case one should observe that first part of FTC does not deal with anti-derivatives.

Now your function under the integral namely $\cos(t^2+t)$ is continuous everywhere and hence the integral $\int_{0}^{x}\cos(t^2+t)\,dt$ is differentiable everywhere with derivative $\cos(x^2+x)$. Thus the result in your question is an immediate consequence of the first part of FTC.

There is a second part of FTC which deals with anti-derivatives. Like the first part it begins with a function $f:[a, b] \to \mathbb {R} $ which is Riemann integrable on $[a, b] $ but its intent is to evaluate the integral $\int_{a} ^{b} f(x) \, dx$ in an easy manner. But to achieve this goal it makes an additional and strong assumption: it assumes that there is an anti-derivative $F$ of $f$ on $[a, b] $. In other words we assume the existence of a function $F:[a, b] \to\mathbb {R} $ such that $F'(x) =f(x) \, \forall x\in[a, b] $ and then FTC says that the integral $\int_{a} ^{b} f(x) \, dx$ equals $F(b) - F(a) $.


By the fundamental theorem of calculus, $\int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case: $$\int_0^x{\cos(t^2+t)dt}=F(x)-F(0)$$ $$\frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=\cos(x^2+x)$$ $F(0)$ disappears after differentiation because it is a constant


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.