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It sounds like such a simple thing to do and yet i'm stuck. I appreciate any help or guidance.

$$\frac{\mathrm{d}}{\mathrm{d}x} \int_{0}^{x} e^{(kt)} f(t) \,\mathrm{d}t $$

I tried this method: $$\int_{0}^{x} e^{(kt)} f(t) \,\mathrm{d}t =F(x)-F(0)$$ $$\frac{\mathrm{d}}{\mathrm{d}x} \left(F(x)-F(0)\right) = \frac{\mathrm{d}}{\mathrm{d}x} (F(x)) \overset{?}{=} \frac{\mathrm{d}}{\mathrm{d}x} \left(\int e^{(kx)} f(x) \,\mathrm{d}x\right)= e^{(kx)} f(x)$$ But I'm not sure if the one but last step is "ok" to do. I'm looking for a better way to justify this.

Blue
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David
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    It looks like a straight forward application of the Fundamental Theorem of Calculus. (Your final answer is correct.) – paw88789 Jan 25 '22 at 16:39
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    You might point out that $F(0)$ will be a constant, which is why it disappears under differentiation. – johnnyb Jan 25 '22 at 17:10

2 Answers2

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According to the fundamental theorem of calculus, if $g:[0,x]\rightarrow\mathbb{R}$ is continuous, then the function $F:[0,x]\rightarrow\mathbb{R}$ defined by $$F(x):=\int_0^xg(t)\,\mathrm{d}t$$ is differentiable on $(0,x),$ and $F'=f$ on $(0,x).$ In this case, you have $g(t)=e^{kt}f(t),$ hence $F'(x)=e^{kx}f(x).$

Angel
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A general version of differentiating an integral is described by Leibniz's integral rule. There are multiple proofs of this rule that can be found in the article itself.

Here's an MSE post that also talks about the proof.

DatBoi
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