If $F(x) = \int_0^x e^{\sin(t)} dt$ , so $F'(0)$ is equal to:

Question

If $F(x) = \int_0^x e^{\sin(t)} dt$ , so $F'(0)$ is equal to:

(A) $-2$ $\quad$ (B) $-1$ $\quad$ (C) $0$ $\quad$ (D) $\mathbf1$ $\quad$ (E) $2$

My attempt:

$$\frac{d}{dt}F(x) = \frac{d}{dt}\int_0^x e^{\sin(t)} dt\Rightarrow F'(x) = \int_0^x \frac{d}{dt}e^{\sin(t)} dt$$

$$F'(x) = e^{\sin(t)}|_{0}^{x} = e^{x} - 1 \implies F'(0) = e^0-1 = 0$$

(wrong answer)

Where am I missing?

Answer 1

The variable in the function

$F(x) = \displaystyle \int_0^x e^{\sin t} \; dt \tag 1$

is $x$, not $t$; the variable "$t$" is "integrated out"; we could just as well have written

$F(x) = \displaystyle \int_0^x e^{\sin s} \; ds, \tag 2$

or used any other variable (besides $x$) under the integral sign.

Therefore,

$F'(x) = \dfrac{d}{dx} \displaystyle \int_0^x e^{\sin t} \; dt = e^{\sin x} \tag 3$

by the Fundamental Theorem of Calculus; evaluating at $x = 0$ yields

$F'(0) = e^{\sin 0} = e^0 = 1; \tag 4$

the correct answer is thus (D).

Answer 2

When you have $g(x) = \int_0^xf(t)dt$ then $g'(x) = \frac{\delta g}{\delta x} = (\int_0^xf(t)dt)^{'}= f(x)$

So in your function: $$F'(x) = e^{sin(x)}$$ and for x = 0: $$F'(0) = e^{sin(0)} = e^0 = 1$$

That's completely different from $g(x)=\int f(x)dx$ where $g'(x) = f(x)$. That's your mistake in your approach of the problem.

Answer 3

We have $(\int_0^xf)'=f(x)$ by the Fundamental Theorem of Calculus, but you applied the rule for integration $$\int_a^bf=F(b) -F(a) $$ where $F'=f$.