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What's the derivative of the integral $$\int_1^x\sin(t) dt$$

Any ideas? I'm getting a little confused.

5 Answers 5

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You can use the fundamental theorem of calculus, but if you have not yet covered that theorem, in short, you'll be taking the derivative - with respect to x - of the integral of $\sin(t)dt$ when the integral is evaluated from $1$ to $x$:

$$\frac{d}{dx}\left(\int_1^x \sin(t) \text{d}t\right) = \frac{d}{dx} [-\cos t]_1^x = \frac{d}{dx}\left(-\cos(x) - (-\cos(1))\right) = \sin(x).$$

and you'll no doubt be encountering the Fundamental Theorem of Calculus very, very soon:

For any integrable function $f$, and constant $a$: $$\frac{d}{dx} \left(\int_a^x f(t)dt \right)= f(x),$$

(provided $f$ is continuous at $x$).

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    I think I've got it. d/dt(-cos(x)-(-cos(1)) = d/dt(-cos(x)+0) = sin(x)2012-11-21
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    Yes, you got it.2012-11-21
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    Your very welcome.2012-11-21
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    Just be careful, $\cos(1) \neq 0$.2012-11-21
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    Why do all the answers that explicitly write the derivative operator use d/dt? In this problem, t appears to be a bound variable in the integration, with the actual free independent variable being x. Just as d/dt (cos(1)) is 0, so is d/dt (cos(x))... I'm not asking if this is wrong (I know it is), I'm just commenting that it's weird that two of the answers have the same issue and, since there's already been the confusion with cos(1), I am sure this only adds confusion.2012-11-21
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    You should be differentiating with respect to $x$, not $t$.2012-11-21
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    @exodu5 thanks for pointing that out. I corrected my careless typos.2012-11-21
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    Well, we only know $$\frac{d}{dx} \left(\int_1^x f(t)dt \right)= f(x)$$ at points $x$ where $f$ is continuous or so.2012-11-21
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    It may be important to note that, with $a$ being a constant, $$\frac{d}{dx}\int_{a}^{x}f(t)dt=f(x)$$ since, with $\int f(x) dx= F(x)$, we have: $$\frac{d}{dx}\int_{a}^{x}f(t)dt=\frac{d}{dx}\left[F(x)-F(a)\right]=\frac{d}{dx}F(x)-0=f(x).$$ This is just my opinion, as I was confused as to why the aforementioned was true and why there are so many different versions of it.2012-11-22
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    @Limitless I think that given the OP's interval (1 to $x$), I simply used $1$. But you are right, $a$ representing a constant is standard.2012-11-22
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    @amWhy, yes, I wasn't criticizing your work at all. I was just adding this little tidbit in case it helps someone. I was confused at first because I had seen other values of $a$ being used and it took me a moment to realize the idea works for any $a$, provided it is constant.2012-11-22
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    @Limitless No criticism taken. It is really *good* to point such things out, and I will likely revise one more time to be sure I don't confuse. thanx!2012-11-22
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    You're welcome, @amWhy.2012-11-22
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Using the fundamental theorem of calculus we know that the answer is $\sin(x)$

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$ \frac{d}{dt}\int_1^x\sin(t)dt = \frac{d}{dt} [-\cos t]_1^x = \frac{d}{dt}[-\cos x+\cos(1)] = \sin x $

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If $f$ is any function at all that can be integrated, then the derivative of the integral of $f(t)dt$ from $1$ to $x$ is $f(x)$. This wonderful fact is the Fundamental Theorem of Calculus.

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    I'm not so sure. For a trivial counter-example, if $\frac{d}{dx} \int_a^x f(t) dt = f(x)$ for every $x \in [a,b]$, then we can modify f on a set of measure 0 (say some countable set) and obtain a function for which this does not hold. The Fundamental theorem of Calculus holds in general only if f is continuous.2012-11-23
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You can use a nice theorem called the Fundamental Theorem of Calculus . Here, we're mainly worried about FTC part 1. Below is a summary of what FTC part 1 says.

Let $f$ be a continuous, function defined on $[a,b]$ and $$F(x) := \int_a^x{f(t)} \ dt \quad \quad \forall x \in [a,b] $$

Then, $F$ is continuous on the closed interval and differentiable on the open interval $(a,b)$ and $F'(x) = f(x) \ \forall x \in (a,b)$.

So, for your problem:

$$\frac{d}{dx}\int_1^x\sin(t) \ dt = \sin(x) \cdot \frac{d}{dx} (x) = \sin(x)$$

Note that you have to replace the $t's$ in the integrand with an $x$ and multiply by the derivative of the upper bound, assuming your lower bound is constant (which it is here.)

As a demonstration of this, suppose we want to calculate

$$\frac{d}{dx}\int_1^{x^2}\sin(t) \ dt $$

This is the same as $$\sin(x) \cdot \frac{d}{dx}(x^2) = 2x \sin(x)$$