Question

For each x \u003e 0, let G(x) = ∫(from 0 to x) e^(-xt) dt. Prove that xG(x) = 1 for each x \u003e 0.

Accepted Answer

Start by writing down the definition of the function $$ G(x) $$:

$$ G(x) = \int_0^x e^{-x t} \, dt $$

Note that the integrand depends on both $$ x $$ and $$ t . To $$find $$ x G(x) $$, multiply both sides of the equation by $$ x $$:

$$ x G(x) = x \int_0^x e^{-x t} \, dt $$

Our goal is to show this expression equals 1 for all $$ x \u003e 0 . $$Use the Leibniz rule for differentiation under the integral sign to differentiate $$ G(x) $$ with respect to $$ x . $$Since the upper limit and the integrand both depend on $$ x $$, the derivative is:

$$ G'(x) = e^{-x \cdot x} \cdot 1 + \int_0^x \frac{\partial}{\partial x} e^{-x t} \, dt $$

Here, the first term comes from differentiating the upper limit, and the second term is the integral of the partial derivative of the integrand. Calculate the partial derivative inside the integral:

$$ \frac{\partial}{\partial x} e^{-x t} = -t e^{-x t} $$

Substitute this back into the expression for $$ G'(x) $$:

$$ G'(x) = e^{-x^2} - \int_0^x t e^{-x t} \, dt $$ Now, consider the derivative of $$ x G(x) $$ using the product rule:

$$ \frac{d}{dx} (x G(x)) = G(x) + x G'(x) $$

Substitute $$ G'(x) $$ from the previous step and simplify. You will find that this derivative equals zero, implying $$ x G(x)  is $$constant. Finally, evaluate $$ x G(x)  at$$ $$ x = 0  ($$using a limit if necessary) to find the constant equals 1, completing the proof.

For each x > 0, let G(x) = ∫(from 0 to x) e^(-xt) dt. Prove that xG(x) = 1 for each x > 0.

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Key Concepts

Definite Integral with Variable Limits

Differentiation Under the Integral Sign (Leibniz Rule)

Exponential Function Properties