I need to find a function whose derivative is $\sin x^3$ and whose value at $0$ is $2$. I have the solution and understand its integral part formally. What I don't understand formally (but do understand intuitively) is why we add $2$ to the definite integral. See the correct solution below.

$$\int_{0}^x \sin(t^3)dt + 2 $$

The result I have obtained myself is the definite integral according to the Second Fundamental Theorem of Calculus.


Suppose you don't have the $+2$, and just the formula $$\int_0^x \sin(t^3) \, dt $$ Now, the requirement is that the "value at $0$ is $2$". In other words when you plug $x=0$ into this formula, you are supposed to get $2$. So, let's plug in $x=0$ and see what we get: $$\int_0^0 \sin(t^3) \, dt $$ Well, that's equal to $0$. In fact $\int_a^a f(t) \, dt = 0$ no matter what the function $f(t)$ is and no matter what $a$ is.

Oops! We were supposed to get $2$, not $0$.

How do we fix that? By adding $2$: $$\int_0^x \sin(t^3) \, dt + 2 $$ And now, when you plug in $x=0$ to this formula, you get $2$, as required.

  • $\begingroup$ this is super close to what my intuition here is. I was confused when integrating from 0 to 0 was giving me 2 in accordance with the conditions given, because it gives 0 at all times. Thanks for giving me a hand here. $\endgroup$ – dr_dronych Apr 14 '17 at 6:11

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